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Series 
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Technical  and  Bibliographic  Notes  /  Notes  techniques  et  bibliographiques 


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d'images  nAcassaire.  Les  diagrammes  sulvants 
illustrant  la  mAthode. 


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MICROCOPY    RESOIUTION    TEST   CHART 

ANSI  and  ISO  TEST  CHART  No    2 


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THEORY  OF  STRUCTURES 


AITD 


STRENGTH  OF  MATERIALS 


BT 


HENRY  T.  BOVEY,  M.A.,  LL.D.,  D.C.L.,  F.R.S. 

Pro/eMor  o/  CTrtJ  B»urin«<rino  and  Av^litA  ir«eftanie«,  UcQVi,  DMvenOv.  J 
JTember  o/  UW  JruMtutton  o/  Oictt  Bufllnwm ;  Form«rJw  fleUoic  o/  Qmcm*  Coitan, 

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FOUBTH  EDITION,  RBWRITTBN  AND  ENLARGED 


FIB8T  THOUSAJKft.^^     ^^.irC,  t^^P''' 


Ciii' 


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UnWev 

NEW  York' 

JOHN    WTLEY    &    SONS 

London  :    CHAPMAN    &    HALL,    Limits* 

1906 


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IIOBSBT  DRCmtOHT),  PRniTIR,  HtW  TORE 


DEDICATED 

TO 

TlOiUtam  C.  Acl>onaI^ 

Wbobi    Benefactionb    to    McGill   Uhivemitt 

Hats  Dohb  so  Much  to  Advance  the  Cause  of 

Scientific  Education 


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PREFACE  TO  THE  FOURTH  EDITION. 


The  present  edition  of  this  work  has  been  so  completely  revised 
and  enlarged   that   it   is   practically   a   new  treatise.    Numerous 
examples,  drawn  for  the  most  part  from  actual  practice,  have  been 
added  to  the  several  chapters,  and  a  large  number  have  been  worked 
out  in  detail  throughout  the  text  so  as  to  illustrate  the  principles 
of  strength,  construction,  and  design.    In  determining  the  stresses 
in  framed  structures  Bow's  method  has  been  used,  and  the  manner 
of  its  application  has  been  shown  in  various  examples  of  roofs  and 
bridges.    Chapter  X  on  bridges  presents  the  subject  in  a  somewhat 
new  manner,  giving  the  principles  now  generally  adopted  in  engineer- 
ing design.    The  tables  of  strengths,  elasticities,  weights,  etc.,  have 
been  carefully  revised,  increased  in  number,  and  brought  up  to  date. 
Many  examples  have  been  carefully   selected  from  the  works  of 
well-known  authors,  but  my  special   thanks  are  due  to  Professor 
Greenhill  for  permission  to  make  use  of  his  train  problems  in  Chap- 
ter III,  and  also  to  Messrs.  Waddell  and  Hedrick  for  important 
information  on  the  preliminaries  of  bridge  design,  and  for  notes  on 
problems  which  have  arisen  in  the  course  of  actual  construction. 
To  my  colleague,  Professor  H.  M.  McKay,  I  desire  to  express  my 
sincere  thanks  for  assistance  in  revising  proof-sheets  and  for  many 

valuable  suggestions. 

Henry  T.  Bovky. 

McGiLL  UNivBBaiTT,  October,  1«08. 


DeRni 
Franw 
Funici 
Polygi 
Line  c 
Mansa 
Non-c 
Result 
Funici 
Centrt 
Mome 
Bow's 
Roofn 
Loads 
Wind 
=stril 
Crane 


Shear 

Bridgi 

King-] 

Incom 

Queen 

Bent( 

Lock-i 

Dowst 

Warre 

Pratt 

Petit  ' 

Piers, 

Fink  ' 

lioUm 


CONTENTS. 


CHAPTER  L 

Framxo  Stbuctttrbs. 

Dennitions , j 

Frames  of  Two  or  More  Members 2 

Funicular  Polygon 3  y 

Polygon  of  Forces 4  g 

Line  of  Loads 5 

Mansard  Roof A. 

Non-closing  Polygons g 

Resultant  of  System  of  Forces j 

Funicular  2!urve ^ 

Centre  of  Gravity jq 

Moment  of  Inertia j  j 

Bow's  Method 22 

Roof-trusses j  j 

Loads,  Live  and  Dead 15 

Wind  Pressure jg 

■stribution  of  Loads jy- 

Crane,  Jib Ig, 

' '      Derrick 21 

Shear  Legs 22 

Bridge-  and  Roof-trusses 23  36  46 

King-post  Truss 27  34 

Incomplete  Frames 28 

Queen-post  Truss 3q 

Bent  Crane 35 

Lock-joint  Trusses 3g 

Bowstring  Truss 44  49, 

Warren  and  Howe  Trusses 47 

Pratt  Truss !!.!!!!!!!!!!  48 

Petit  Truss 4» 

'"'«'" ' .  52 

Fink  Truss 54 

BoUman  Truss 5^ 

vii 


vm 


CONTENTS. 


Method  of  Sections •  c^ 

Three-hinged  Arch eg 

Tables  of  Roof  Weights .64  65 

Table  of  Wind  Pressures '  gc 

Examples !!.!!!!!         66 

CHAPTER  II. 
Srearino  Forces  and  Uendino  Moments. 

Equilibrium  of  ReaniH jq. 

Shearing  Force '   .q™ 

Ik'iuiiiig  Moment jq^ 

Examples  of  Shearing  Force  and  liending  Moment 105 

Relation  between  Shearing  Force  and  Iknding  Moment 114 

Effect  of  Live  (or  Rolling)  Load 116  j25 

Moment  of  Forces  with  Resjiert  to  a  Point '  122 

Relation  l)et ween  Rending  Moment.i  and  Funicular  Polygon 124 

Maximum  Shear  and  .Maximum  Rending  Moment 127 

Graphical  Determination  of  Maximum  Rending  Moment 130 

Hinged  (lirdere 13^ 

Examples 141 


CHAPTER  III. 
Momentum,  1!.nerot,  Balanciro. 

Velocity  Acceleration jm 

^'°'^ '.'.''.''''.'.['.'.'.'.'.'.'.'.  155 

*;|'*''"«>- 155, 168 

«  ork  against  Oblique  Resistance I57 

Grapiiical  Method 15jj 

Impulse  and  Momentum l_r^ 

Triangle  and  Parallelogram  of  Velocities 163 

Relative  Velocity I64 

Ek)uations  of  Motion  and  Energy I64 

Angiilar  Velocity 1 66 

Centrifugal  Force 166 

E^'iinp'o" 169-179 


Inertia. 


180 

Ral.'inring Ig^ 

Curves  of  Pi.Mton  Velocity ^f^J^ 

Curve  of  Crank-elTort jfjy 

Examples 189-214 


CHAPTER  IV. 
Stress,  Strain,  Elabticitt,  Vihhation,  TmN  Ctlindem. 

Definitions 21 5 

217 


ilOokr's  IjiW. 


CONTENTS. 


Poisson's  Ratio 217 

Rosilience -lo 

Theory  of  Vibration " 221 

Forced  ( Mcillntion 22"; 

Oicillation  of  a  Vertical  Elastic  Rod 225  229 

Specific  Weight 231 

Coefhcicnt  of  Klasticity 232' 

Limit  of  Elasticity 236 

Breaking  Stress 239 

Factor  of  Safety 240 

Wdhler'a  Experiments 24'i 

V&iime ■■■...............[[[[  245 

L.iunhardt's  Formula 247 

^^■py^.•luch's  Formula 248 

I'nwin's  Formula 251 

Hctnarks  on  Values  of  t,  u,  s,  and  / 252 

Flow  of  Solids 254 

Thin  Hollow  Cylinders 258 

Spherical  Shells 26o 

Tractical  Remarks 261 

Table  of  Strength,  Elasticity,  etc, ,  of  Alloys 2(Q 

"  "     of  Iron  and  Steel 264 

"      "         "  "  "     of  Timbers '.'.'.'.'.""  266 

"      "         "         of  Wire  Ropes 264 

of  Copper  Wire ^  ^.. ...... ......  2M 

"      "        "         of  Manila  Rope '       26f 

"      "  Transverse  Strength  of  Timbers | '    _  265  267 

Coefficients  of  Linear  E.xpan8ion 268 

".     "  Crushing  Strength  of  Rocks 26a-270 

ot  Cement  and  Concrete 270 

"      "  Values  of  J?,  C,  /r,  and  « 27, 

^"^P'«* '.'.'.'.'.'.'.'.'.'.'  271-289 


CHAPTER  V. 

Stkehubs,  Stkains,  Eakthwokk,  and  RBTAiNiNn-WALLa. 

Tnlpmal  Stresswi _-- 

('■iiiipoimd  Strain _«. 

, .      .      .  ,,  .  . .  /W) 

<  oiist.-jnt  f  nnipnnrnt.s  of  p, ___ 

1'no.|iirtl  Principal  Stresses jo.. 

riaii(>x  t)f  Principal  Stress ~_. 

Flli|«e  of  Stress •■■  ^  ^  ^ '!"!'!  ^ !!!!!!!!! !"""  2W 

M.xhilus  of  Rigidity ] " „„„ 

<  onjiigate  Stnifww 299 

S'i'iftiiiff 

«  ^;j 

Rttiikine's  Earthwork  Theory '"'^ -q- 


CONTENTS. 

IVsesure  of  Earthwork. . , , »*<»• 

Foundation  in  Earth .'.'.'.'.... ^' 

Line  of  Rupture ** 

Retaining-walls 311 

Oondition.s  of  Equilibrium'  of  Retaining-waik ^*^ 

Practical  Rules  Rapecting  Retaining- walls    ^^^ 

Reservoir  Walls 324 

Relations  between  Elastic  Constants *^ 

General  Equations  of  Stress 332 

Relation  between  Stress  and  Strain.  ..  ^® 

Applications  to  Traction  and  Torsion ^^ 

Clapeyron's  Tlieorem ['] 348 

Transmission  of  Energy  353 

Thick  Cylinders '.". 354 

Exiuuplee " 355 

362-375 


CHAPTER  VI. 

Friction. 

Friction 

Laws  of  Friction "       376 

Inclined  Plane "  | 375 

^Vedge 377 

Screws 378 

Endless  Screw 381 

Rolling  Friction 384 

Journal  Friction 385 

Pivots 388 

Cylindrical  Pivot 390 

Wear  of  Pivots " " '    390 

Conical  Pivot 392 

Scliifle'a  Pivot  (anti-fritium). 393 

Holts  and  Ropea "  "  " 394 

Brakes 39S 

Effective  Tension  of  a  Belt    .      896 

Effect  of  High  Speed. . . 398 

Slip  of  Belts 398 

Prony'g  Dynamometer \[\[ 399 

Stiffness  of  BelU  and  Ropes. «» 

Wheel  and  Axle *^ 

Toothed  (Sparing [       402 

Bevel  whwln 404 

Efficiency  of  Nfechanisms *>8 

Table  of  Coeffjcienta  of  Journal  Friction *^ 

Examplw 408 

409 


CONTENTS. 


CHAPTER  VII. 
TiUNaTXRSE  Strknoth  of  Bkams. 

Moment  oT  Resistance '**■ 

General  Equations ^1® 

Strength  Modulus ■*'* 

T.il.le  of  Moments  of  Inertia  and  Strength  Moduli ^i!i 

\Vork  of  Flexure *22 

Equalization  of  Stress ^^^ 

Examples ^^ 

,   Reinforced  Concrete  Beams *^* 

Formula  Wl~Chd\ ^^ 

Beams  of  Uniform  Strength ^* 

Stiffness  of  Beams *^ 

s  Distribution  of  Shearing  Stress.  .  ...!!!!!!.... *"^' 

Deflection  Due  to  Shear ^^^ 

Curves  of  Maximum  Normal  and  Tangential  Stress *''^ 

Vanable  Moment  of  Inertia *^^ 

Springs " 462 

Allowance  for  Weight  of  Beam. .  *^ 

Beam  Acted  Upon  by  Oblique  Forces '*^ 

Bending  Moment  in  a  Plane  which  is  not  a'Principai  Wane VJt 

r  langed  Girdera 478 

.Moments  of  Resistance  of  Flanged  Girdera '*^ 

Built  Beams 482 

Design  of  a  Girder  of  an  I  Section. '**"'' 

Design  of  a  50-ft.  Plat*  Girder *^ 

Beams  Supported  at  More  Than  Two  Points  '. ■*** 

("onf  inuous  Ginien *12 

Comparative  Merits  of  Continuous  Girders ''® 

eZS.  ^"^''.  !^™''  **  ''"'"^  "'  ^^  °'  «^t'«--  Oirdera'.  [ ."  [  [  [  St 
M6-50a 


CHAPTER  VIII. 

Pillars. 

n.Tislfieation 

I  nilnre  of  RHar* "•* 

lliHlirkinson's  Formulip _ ' SM 

'"iriliin's  Formula SOU 

\iilu(s  of  o,  o,,  and  / WB 

^'Hlll.■s  of  o  and  o,  for  shapes *>1 

Striiglit-line  Formula *** 

CiiI.tV  Formulie  for  StruU WW 

l'(T,.et  on  a  Strut  of  a  Lateral  Load "0" 

f.'>ng  StruU "  * «10-«I4 

I  niiormiy  Var>1ng  Stra«<. .   . .              •"« 

618 


^^  CONTENTS. 

620 

Weyrauch's  Theor>'  of  BuckUng ','.'.'...'. 625 

Flexure  of  Columns 631 

Elliptic  Integral  Tables !!!!,!!!...  632-640 

Examples 

CHAPTER  IX. 

Torsion. 

641 

Laws  of  Torsion g42 

Torsional  Strength  of  Shafta M7 

Non-cireular  Sections 64» 

Variable  Resistance 851 

Distance  between  Bearings 652 

Efficiency  of  Shafting 655 

Combined  Bending  and  Torsion GST 

Centrifugal  Whirling  of  Shafts WD 

Helical  Springs ggj 

Open  Coiled  Helical  Spring 664-675 

Examples 


CHAPTER  X. 

BiUoaKa. 

67« 

dassification " ' " 676 

Position  of  Hatform g^y 

Types  of  Bridge  Truss •    •  •  •  ■  •.' '  •  " gi^<v 

Co^arat.ve  Advantages  of  Two,  Three,  or  More  Crdef.  •  ;    ; ; ;    ;    ;    ; ; ; ; ; ;  ]  ^J 

Dead  Uad 682 

Live  l-oad 08% 

Ecjuivalent  Uniform  Load ggg 

Stringers 686 

Camiyer ■ ,„; ' ' 687 

Rivet  Connection  Ijetween  Flanges  and  Web ^^ 

Eve-bars  and  Pins ggg 

Determination  of  Bridge  Stresses ^j^ 

Single-  and  Double-intersection  Truss 717 

Bowstring  Tniss 726 

Bowstring  Truss  with  Isosceles  Bracmg '".".'.'..'.'.'.'.".....  731 

Throe-liinged  Bridge.  .  .  747 

Wind  Pn-ssure 75I 

<'an»ilever  Tnisses ,,..,...'..'. 7fil 

DeflectidU  of  Tnisses 7^4 

Rivets 767 

Rivotf'd  Joints 773 

Ertiriencv  of  Riveted  Joints y^g 

Design  of  a  127-rt.  Riveted  Span '  ^^ 

Dwign  of  a  520-ft.  Swing  Span 


CONTENTS. 


XUl 


814 


Tables  of  Bridge  Weights 820-841 

Examples ' 


CHAPTER  XI. 

Suspension  Bridobs. 

842 

Cfthles 843 

Anchorage,  etc g^^ 

Suspenders q<  = 

Curve  of  Cable gg^ 

Pressure  upon  Piers ^^ 

Weight  of  Cable ogl 

Curve  of  Cable  with  Sloping  Suspenders ^^ 

StifTeiiing  Trusses g_g 

Suspension-bridge  Loads • 

Mwlificationa  of  Simple  Suspension  Bridge 876-882 

Examples 


CHAPTER  XII. 

Abcrbs  AMD  Archcd  Ribs. 

883 

Arches ogn 

Conditions  of  Equilibrium ^^2 

Joint  of  Rupture ggg 

Empirical  Formulse ^^ 

Moseley 's  Principle g^^ 

Fuller's  Method g^ 

Linear  Arches ojq 

Arched  Ribs -.q 

Arched  Rib  with  Pin  Ends 

Arched  Rib  with  Fixed  Ends ■•. 

V  ,      „f  If  885,922 

\  iilue  of  « 927  B36 

Effect  of  Temperature  Changes 93o'  945 

IVflection  of  axi  Arched  Rib ^^ 

Deformation  of  a  Loaded  Arched  Rib ^^ 

I Kiicral  Equations ^g 

St  resses  in  posts  and  diagonals 

Clerk  Maxwell's  method Ml-eW 

iuiamples 


THEORY  OF  STRUCTURES. 


CHAITER  I. 
FRAMES  LOADED  AT  THE  JOINTS. 

I.  Definitions.— Frfimes  arc  rigid  structures  composed  of 
straight  struts  and  ties,  jointed  together  by  means  of  bolts,  straps, 
mortises,  and  tenons,  etc.  Stmts  are  members  in  compression,  ties 
incinhers  in  tension,  and  the  ternx  brace  is  applied  to  either. 

riie  external  forces  upon  a  frame  are  the  loads  and  the  reac- 
tions at  the  points  of  support,  from  which  may  be  found  the  result- 
ant forces  at  the  joints.  The  greatest  care  should  be  exercised 
in  llio  design  of  tlie  joints.  The  resultant  forces  should  severally 
coincide  in  direction  with  the  axes  of  the  members  upon  which 
tliiy  act,  and  shouhl  intersect  the  joints  in  their  centres  of  gravity. 
Owing  to  a  want  of  homogeneity  in  the  material,  errors  of  work- 
manship, etc.,  this  coincidence  is  not  always  practicable,  but  it 
should  be  remembered  that  the  smallest  deviation  introduces  a 
landing  action.  Such  an  action  will  also  be  cau.sed  by  joint  fric- 
tion when  the  frame  is  insufficiently  braced.  The  points  in  which 
tli(  lines  of  action  of  the  resultants  intersect  the  joints  are  also 
ill(  il  the  centres  of  resistance,  anil  the  figure  formed  by  joining  the 
r.  Hires  of  resistance  in  order  is  usually  a  fKilygon,  which  is  desig- 
rialiil  the  line  of  resistance  of  tho  frame. 

The  position  of  the  centres  should  on  no  account  be  allowed 
to  vary.  It  is  assumed,  and  is  practically  true,  that  the  joints 
of  a  frame  are  flexible,  and  that  the  frame  under  a  given  load  does 
not  sensibly  change  in  form.  Thus  an  individual  member  is  merely 
stretched  or  compressed  in  the  direction  of  its  length,  i.e.,  along 


2  THEORY  OF  STRUCTURES. 

its  line  of  rosistance,  while  the  frame  as  a  whole  may  be  subjected 

to  a  bending  action. 

The  term  trus,  is  often  applied  to  a  frame  supporting  a  weight 
2.  Frame  of  Two  Members.-0.4 ,  OB  are  two  bars  jointed  at  O 

and  supported  at  the  ends  A,  B.     The  frame  in  Fig.  1  consists  of 

two  ties,  m  Fig.  3  of  two  struts,  and  in  Fig.  2  of  a  strut  and  a  tie 


Fio.  1. 


Fia.  2. 


Fio.  3. 


Let  P  be  the  resultant  force  at  the  joint,  and  let  it  act  in  the  direc- 
tion OC.  Take  OC  equal  to  P  in  magnitude,  and  draw  CD  parallel 
to  OB.    OD  IS  the  stress  along  OA,  and  CD  is  that  along  OB. 

Let  the  angle  AOB  =a,  and  the  angle  COD  =^. 

Let  Sh  S-jhe  the  stresses  along  OA,  OB,  respectively.    Then 


§]_^0D    .sin(a-^) 
P     OC^     sina 


and     ^^=^=$^^. 
r      OC    sm  a 


3.  Frame  of  Three  or  More  Members.— Let  A1A2A3  .  .     be  a 
polygonal  frame  jointed  at  .4,,  .I2,  .43,  .  .  .     Let  P,,  P^   p'3 

be  the  resultant  forces  at  the  joints  .4,,  .42,  .4, ,  respectively! 

Let  Si,  S2,  S3,  ...  bo  the  forces  along  .4i.42,  ,42^3,  .  .  .  ,  respect- 
ively. ' 

Consider  the  joint  .4,. 

The  lines  of  action  of  throe  forces,  P,,  Si,  and  5«,  intersect  in 
this  joint,  and  tlio  forces,  l,oi„g  i„  o.,uilibrium,  mav  he  represented 
m  direction  ari.l  inagnitu.lo  |,y  the  .sides  of  the  triangle  0.s,..e  in 
which  ,«,..,,  is  parallel  to  /^,  O.s-,  to  .S',,  and  Os^  to  S^. 

Similarly,  P2,  .S,.  S^  may  be  repro,s(>nted  bv  the  .sides  6f  the 
triangle  Osis.,  which  has  one  side,  C/v,,  common  to  the  triangle  O.s-,.?.. 
and  so  on. 

Thus  every  joint  furnisln-s  a  triangle  having  a  side  common 
to  each  of  the  two  a.ijacent  triangles,  and  all  the  triangles  together 
from  a  cloml  polygon  s^s,,,  .  .  .  The  .sides  of  this  polvgon  represent 


FRAMES  WADED  AT  THE  JOINTS.  3 

in  magnitutij  and  direction  the  resultant  forces  at  the  joints,  and 
the  radii  from  the  pole  0  to  the  angles  S1S2S3,  .  .  .  represent  in 
magnitude,  direction,  and  character  the  forces  along  the  several 
sides  of  the  frame  A1A2A3  .  .  .    The  polygon  ^1.42^3  .  .  .  is  the 


line  of  resistance  of  the  frame,  and  is  called  the  funicular  polygon 
of  the  forces  P,,  Pj,  P3,  .  .  .  with  respect  to  the  pole  0. 

The  two  polygons  are  said  to  be  reciprocal,  and,  in  general,  two 
fiRuros  ;.i  grr.phiral  statics  are  said  to  be  .xiprocal  when  the  sides 
in  the  one  figure  are  parallel  or  perpendicular  to  corresponding 
sides  in  the  other. 

A  triangle  or  polygon  is  also  saitl  to  be  the  rf'ciprocal  of  a  point 
when  its  sides  are  parallel  or  perpendicular  to  corresponding  lines 
radiating  from  the  point.  Thus  the  triangle  OseSi  is  the  reciprocal 
of  the  point  Ai,  and  the  polygon  ^41.42^3  ...  is  the  reciprocal 
of  the  point  O. 

If  more  than  two  members  meet  at  a  joint,  or  if  the  joint  is 
Mil)j(T(c(l  to  more  than  one  load,  the  resulting  force  diiigram  will  be 
a  (iiiadrilateral,  pentagon,  hexagon,  .  .  .  according  as  the  nuniber 
<il  in(>rnbers  is  3,  4,  5,  .  .  .  or  the  numlier  of  loads 2,  3,  4,  '.  .  . 

In  practice  it  is  usually  required  to  determine  the  stresses  in  a 
niimhcr  of  members  radiating  from  a  joint  in  a  framed  structure. 
If  the  reciprocal  of  the  joint  can  be  drawn,  its  sides  will  repre- 
wiil  in  ilirection  and  magnitude  the  stresses  in  the  corresponding 
iiiciiiIkts. 


4  THEORY  OF  STRUCTURES. 

The  converse  of  the  preceding  is  evitlently  true.  For  if  a  sys- 
tem of  forces  is  in  equilibrium,  the  polygon  of  forces  S1.92.s3  .  .  . 
must  close,  and  therefore  the  polygon  which  ha.s  its  sides  respect- 
ively parallel  to  the  radii  from  a  jwlc  0  to  the  angles  Si,  S2,  .S3,  .  .  . 
and  which  has  its  angles  upon  the  lines  of  action  of  the  forces,  must 
also  close. 

Ex.  1.  Let  0  Ix;  a  joint  in  a  framed  structure,  and  let  0*-,,  Os„  Os„  . . . 
be  the  axes  of  the  members  radiating;  from  it.  The  polygon  .4,-12.4,  ...  is 
the  reciproeal  of  0,  tlie  side  -l.-lj  representing  the  stress  along  Osi,  the  side 
.4j.43  that  along  ()»„  etc.,  Fig.  .5. 

Ex.  '2.  Let  the  resultant  forces  at  the  joints  be  parallel.  The  polygon 
of  forces  becomes  the  straight  liiie  s,.'.,,  which  is  often  termed  the  line  of  loads. 
Thus,  the  forces  P„  P,,  . . .  Pj  are  represented  by  the  sides  SiS,,  SjS,,  . . .  S4S,, 


FiQ.  7. 

which  are  in  one  straight  line  cloned  by  .<,.s,  and  .SyS,,  representing  the  remain- 
ing forces  /',  and  P,,,  while  the  triangles  Os,.s,,  0.<,x„  .  .  .  are  the  reciprocals  of 
the  points  -4j,  .4, .  .  .  Draw  OH  jierpo.idicular  to  .«,.'!,.  The  projection  of 
each  of  the  lines  O.v,,  0.<„,  Ox,.  .  .  .  peri)ondicula  .0  ,s,s,  is  the  .same  and  equal 
to  OH,  which  therefore  represents  in  magnitude  and  direction  the  stress  which 
is  the  same  for  each  member  of  the  fiame. 

Let  «,,  "2,  «3,  . . .  Ije   the   inclinations   of  the    members  ^,^4,,  A^„  ... 
respectively,  to  the  line  of  loads.    Then 

OH  "Ha,  tan  n  "Hs^  tan  a^; 

.:    OH   (cot   n,  +  COt    0,)=/^,4   //,S',,-.Vs-/'.+  /'3+P4+P.i-^.+^«. 

and  OH,  in  direction  and  magnitude,  is  equal  to  the  stress  common  to  each 
member.     Ainu,  tlie  >ir(S,s  in  any  mcmi)cr,  e.g.,  A^.\.^,^0!<^-OH  vmvx  a,. 

Let  the  resultant  forces  at  the  joints  -1,,.4,  be  inclined  to  the  common, 
direction  of  the  remaining  forces,  and  act  in  the  direction  shown  by  the  dotted 


FRAMES  LOADED  AT  THE  JOINTS. 


5 


lines.  Let  P,',  P,'  be  the  magnitudes  of  the  new  forces;  draw  «,«,'  parallel 
to  the  direction  of  P,'  so  as  to  meet  Os,  in  s,';  join  s,."!,'-  Since  there  is  equi- 
librium, s„'s,  must  be  parallel  to  the  line  of  action  of  P,'.  Thus 
.■••e'.VsV,  is  the  force  polygon. 

E.\.  3.    The  forces,  or  loads,  Pj,  P3, . . .  Pj  are  generally  ver- 
tiral,  while  P,,  P^  are  the  vertical  reactions  of  the  two  supports. 

.Suppose,  e.g.,  that  AiA^ ...  ^,  is  a  rope  or  chain  suspended 
from  the  points  .4,,  A,,  in  a  horizontal  plane  and  loaded  at  A^A^ .  . . 
with  weights  P„  P3,  .  . .  The  chain  will  hang  in  a  form  dependent 
upon  the  magnitude  of  these  weights.  The  points  H  and  s,  will 
(liiicide,  and  OH  will  represent  the  horizontal  tension  of  the 
iliuiii. 

Let  the  polygon  vl,A, . . .  j4,  be  inverted,  and  let  the  rope  be 
rpjilaced  by  rigid  bars,  .4,i4„  /lj.4, . . .  The  diagram  of  forces 
will  remain  the  same,  and  the  frame  will  be  in  equilibrium  under 
the  given  loads.  The  equilibrium,  however,  is  unstable,  as  the 
cliain,  and  consequently  the  inverted  frame,  will  change  form  if  the  weighta 
vary.    Braces  must  then  be  introduced  to  prevent  distortion. 


Fig.  9. 


Fig.  10. 


Take  the  case  of  a  frame  DCBA  . . .  symmetrical  with  respect  to  a  verti- 

riil  through  .4,  and  let  the  weights  at  A,  B,C,  .  .  .he  IF,,  W„  W ,  respect- 

iwly. 

Drawing  the  stress  diagram  in  the  usual  manner,  OH  represents  the  hori- 
zontal thrust  of  the  frame. 

Iho  ])<)rtions  ,<i,Sj,  SjS„  ...  of  the  line  of  loads  give  a  definite  relation  between 
till'  wciiihts  for  which  the  truss  will  bo  .stable.  The  result  may  be  expressed 
analytically,  as  follows: 

Let  rt,,  (T.,  «„  .  . .  he  the  inchnations  of  AB,  BC,  CD, . . . ,  respectively, 
to  the  horizontal. 

Let  the  horizontal  thrust  OH=H.    Then 

ir  /w  \  /It'  \ 

//__•  eot  «.-(-y'+ir,jcot  a.-(^^'+Tr.+  Tr,jcot  «,-... 

If  W,-W,-W,-..., 

cot  "i  --  3  cot  a,  -  5  cot  «,  —  ... 


6 


THEORY  OF  STRUCTURES. 


If  there  are  two  bars  only,  viz.,  AB,  BC,  on  each  sido  of  the  vertical  rentre 
line,  the  frame  will  have  a  double  slope,  and  in  this  form  is  employed  to  support 
a  Mansard  roof. 

If  there  are  a  number  of  bars  on  each  side  of  .1,  and  if  o,,  «,+,  are  the 
inrlinations  to  the  horizon,  of  the  wth  and  (/i+l)th  bars,  respectively,  count- 
ing from  A, 

2«  - 1      ^ 
cot  <*»-ti  =,,  — j  eot  «,. 

If  such  a  frame  as  the  above  is  inverted,  the  stresses  in  the  members  are 
reversed  in  kind,  but  remain  of  the  same  magnitude. 

4.  Non-closing  Polygons,— Lot  a  number  of  forces  Pi,  P2,  P.i,  .  .  . 
act  upon  a  structure,  and  let  these  forces,  taken  in  order,  be  represented 
in  direction  and  magnitude  by  the  sides  of  the  unclosed  figure 


Fig.  11. 


Fio.  12. 


MXPQ.  .  .  .  This  figure  is  the  unclosed  polygon  of  forces,  and 
its  clo.«irig  line  TM  represents  in  direction  and  magnitude  the  result- 
ant of  the  forces  Pi,  P2,  P3,  .  .  . 

For  PM  is  the  resultant  of  Pi  and  P2,  and  may  replace  them; 
QM  may  replace  PM  and  P3,  i.e.,  Pi,  P2,  and  P3;   and  so  on. 

Take  any  point  0  and  join  OM,  OX,  OP,  .  . 

Draw  a  line  AB  parallel  to  OM  and  intersecting  the  line  of  action 
of  Pi  in  atn/  point  B.  Through  B  draw  BC  parallel  to  ON  and  cut- 
ting the  line  of  action  of  P2  in  C.  Similarly,  draw  CD  parallel  to 
OP,  DE  to  OQ,  EF  to  OR,...    The  figure  ABCD ...  is  called  the 


NON-CLOSING  POLYGONS.  7 

fitnictilar  polygon  of  the  given  forces  with  respect  to  the  j)ole  0. 
The  position  of  the  jwle  0  is  arbitrary,  and  therefore  an  infinite 
miiiiber  of  funicular  jjolygons  may  be  drawn  with  different  poles. 
Also  the  iKJsition  of  the  point  B  in  the  line  of  action  of  Pi  is 
aihitrary,  and  hence  an  infinite  number  of  funicular  polygons  with 
tlicir  corresponding  sides  parallel,  i.e.,  an  infinite  number  of  similar 
funicular  polygons,  may  be  drawn  with  the  same  pole. 


Fio.  13. 


Fro.  14. 


5.  To  show  that  the  Intersection  of  the  First  and  Last  Sides  of 
the  Funicular  Polygon  (i.e.,  the  Point  G)  is  a  Point  on  the  Acttial 
Resultant  of  the  System  of  Forces  Pi,  Pn,  P3,  .  .  .—First  consider 
two  forces  Pi,  P-z,  MNP  being  the   force  and  ABCD  the  funicular 

polygon. 

Let  AB,  DC,  the  first  and  last  sides  of  the  latter,  be  produced 
to  meet  in  gi]  also  let  DC  produced  meet  the  line  of  action  of  Pi 
in  //. 

Produce  OP  and  MN  to  meet  in  K. 

Lot  the  lines  of  action  of  Pi  and  P2  meet  in  L. 

By  .similar  triangles, 


Hence 


or 


KP     HC     KN 
KN  ~HL'    KO 

HB      KO     Hgi 
'HC    KM    HB' 

KPKN   KO 
KN  KO  KM 

HC  HB  Hgi 
'HLHCHB' 

KP 
KM 

Hg, 
"HL' 

^^  THEORY  OF  STRUCTURES. 

anil  therefore,  since  the  anple  //  is  equal  to  the  angle  A',  the  line 
i*M  is  parallel  to  the  line  Lgi. 

But  I'M  represents  in  magnitude  the  resultant  of  the  forces 
Pi,  Pj.  and  is  parallel  to  it  in  direction. 

Therefore  L(ji  is  also  jwirallel  to  the  direction  of  the  resultant. 

But  L  is  evidently  a  point  on  the  actual  resultant  of  Pi,  P^. 
Hence  71  must  Ix?  a  jjoint  on  this  resultant. 

Xcxt,  let  there  l)e  three  forces,  Pi,  Po,  P3. 

Replace  Pi,  P.  by  their  resultant  X  acting  in  the  direction  Lgi. 
The  force  and  funicular  polygons  for  the  forc(!s  A'  anil  P^  are  e\  i 
ilently  MPQ  and  AijxDE,  respectively;   and  g^,  the  point  of  inter- 
section of  Afjx  and  ED  produced,  is,  as  already  proved,  a  point  on 
the  actual  resultant  of  A'  and  P3,  i.e.,  of  P,,  Po,  and  P3. 

Hence  the  first  and  last  sides,  AD,  ED,  of  the  funicular  jjolygon 
ABODE  of  the  forces  Pi,  Pj,  P3,  with  resfKTt  to  the  pole  0,  inter- 
sect in  a  point  which  is  on  the  actual  resultant  of  the  given  forces. 

The  proof  may  be  similarly  extended  to  four,  five,  and  anv 
numlier  of  forces. 

If  the  forces  are  all  i)arallel,  the  force  polygon  of  the  two  forces 
Pi,  P2  becomes  a  straight  line,  MXQ.     Draw  the  funicular  poly- 


Fir..   l.V 


Fi(i.   Kl. 


pon  AliCD  as  before,  and  through  y/i,  tin-  intersection  of  the  first 
and  In^t  sides,  draw  ^,)'  parallel  to  Mlj,  and  cutting  liC  in  5'. 
By  similar  triangles, 

/',      MX     r/il'         ,     Pi     QX    .7,1' 

av  "  o.\-  '^  jiv  ''"' '   o.v  ^  ox  " c'F • 
Pj   ry 

•  Pj^IiV 

Hence  Ygu  whieli  is  p.-uaiiti  I<>  liic  liiirction  of  the  forces  P,,  Pj, 


FUMCULAR  CURVE. 


9 


divides  the  distance  between  their  lines  of  action  into  segments 
whic'li  arc  inx-ersely  proi)ortional  to  the  forces,  and  must  therefore 
1)("  tlio  line  of  action  of  their  resultant.  The  proof  may  be  extended 
to  tiny  number  of  forces,  as  in  the  preceding. 

Funicular  Curve.— Let  the  weights  ujwn  a  beam  AB  become 
iiiHiiite  in  number,  and  let  the  distances  between  the  weights  diminish 
iiii  {('finitely. 

The  load  th^n  becomes  continuou-s,  and  the  funicular  polygon 
is  '>.  curve,  called  the  funicular  curve. 

The  ecjuation  to  this  curve  may  be  found  as  follows: 

Let  the  tangents  at  two  consecutive  points  P  and  Q  meet  in  R. 
Tliis  i)oint  is  on  the  vertical  through  the  centre  of  gravity  of  the 
load  upon  the  portion  MN  of  the  beam. 


M.^*     ^     *<^ftW 


B 

T 


Fio.  17. 


Pia.  18. 


Let  In  l)e  the  line  of  loads,  and  let  OS,  OT  be  the  radial  lines 
from  n,  the  ])o\p,  jiarallel  to  the  tangents  at  P  and  Q.  Take  A 
as  the  origin,  and  let  w  be  the  intensity  of  the  load. 

Let  0  l)e  the  inclination  of  the  tangent  at  P  to  the  beam,  and 
let  the  polar  distance  OV  =p. 

The  load  upfm  the  i)ortion  MX  is  wdx.    Then 

wdx  " SY  - SV  -  TV  '  pUiuO-p  tan  {O+dff) 
-- /)f/'?,  approximately. 


Tlierefore 

Integrating  twice, 


cPy 


dO 
"rfx 


since    0 


'dx' 


py--  ffxvdx^+ciX+Ca, 
fi  and  Ci  being  constants  of  integration. 


W  THEORY'  OF  STRUCTURES 

If  w  is  constant, 

and  the  cu-  o  is  a  parabola. 

6.  Centres  of  Gravity.— Let  it  be  required  to  determine  the 
cent IV  of  gravity  of  any  piano  area  synunetrical  with  respect  to 
an  axis  A'A'.  I)i\i(le  the  area  into  suitable  elementary  areas  ai, 
02,  03, .  .  .  having  known  centres  of  gravity. 


Fio.  19. 


Fio    20. 


Draw  the  force  (the  line  In)  and  funicular  polygons  correspond- 
ing to  thcsp  areas,  and  let  g  Iw  the  \mmt  in  which  the  first  and  last 
sides  of  tiie  funicular  polygon  meet.  The  line  drawn  through  g 
parallel  to  l/(  mi.  pass  through  the  centre  of  gravity  of  all  the 
elementary  areas  and,  thcTcfore,  of  the  whole  an-a.  Hence  it  is 
the  point  (/  in  which  this  line  intersects  the  axis  A'A'. 

Hail  and  similar  sections  may  Ih'  divided  into  elementary  area.s 
l)y  ilrawing  a  number  of  parallel  lines  at  right  angles  to  the  axis 
of  .synnnefry,   and  at   si  oh  distance's  aj)art   that 
each    elementary    figure    may,    without    sensil)le 
error,  be  considenvl  a  rectangle  of  an  area  eiiual 
to  tlie  product  of  its  breadth  by  its  mean  height. 
In   the   case   of  a   very   irregular  section,    an 
accurate    temi)late   of    the   .section    may    iw    cut 
out   of   cardboard   or   thin   metal.     If   the   tem- 
plate is  then  su.siH'iided  from  a  pin   through  a 
IK)int  near  the  edge,  the  centre  of  gravity  of  the  section  will  lie  in 
the  vertical  through  the  pin.     By  changing  the  |)oint  of  susiwusion, 


Fui    21. 


MOMENT  OF  INERTIA. 


11 


a  new  line  in  which  the  centre  of  gravity  lies  may  be  found.  The 
iiitersi'ction  of  the  t^ro  lines  must,  therefore,  be  the  centre  of  gravity 
rcriuired.  Another  method  of  finding  the  centre  of  gravity  is  to 
carefully  balance  the  template  upon  a  needle-point. 

The  area  of  such  a  section  may  be  determined  either  by  means 
of  a  planimetcr  or  by  balancing  the  template  against  a  rectangle 
cut  out  of  the  same  material,  the  area  of  the  rectangle  being  evi- 
dently the  same  as  that  of  the  section. 

7.  Moment  of  Inei..a  of  a  Plane  Area. — Let  any  two  consecu- 
tive sides,  C2C3,  C3C4,  of  the  funicular  polygon,  Fig.  19,  meet  the 
line  (jG  in  the  points  ma,  713. 

Let  xi,  X2,  X3, . . .  be  the  lengths  of  the  perpendiculars  from  the 
centres  of  gravity  of  ai,  02,  as, ... ,  respectivelj',  upon  gG. 

Draw  the  line  OH  perpendicular  to  the  line  of  loads,  and  let 

OH -p. 

By  the  similar  triangles  C3m3n3  and  034, 


nnd 


X3 


03X3- 


34 
'  P 


as 


■Ts 


or    maWa  -  ■ 


03X3 


—  —  -wisna-y  =  area  of  triangle  Caffians. 


p    2 

But  the  total  area  A  bounded  by  the  funicular  polygon 
r/'/'a . . .  and  the  lines  gCi,  gk  is  the  sum  of  all  the  triangular 
anas  Cif/mi,  dmarii,  Ctntiiii,  . . . ,  described  in  the  same  manner  as 
('39713113.      Therefore 

^'p   2^p   2'^  ■•'  ~    2p    • 


The  sum  J(ax*)  is  the  moment  of  inertia,  /,  of  the  plane  area 
with  resjx?ct  to  gO.    Hence 

A-;r-,    or    I -2Ap. 


The  moment  of  inertia  /„  of  the  area,  with  respect  to  a  parallel 
nxis  ut  distance  j/i  from  gG,  is  given  by  the  equation 

where  S -ill +ilt+ .. . 


12 


THEORY  OF  SritUCTURES. 


I^t   the  new  axis   intersect  C^g  and  kg  in  the  points  a  and  r 
bince  the  triangles  qgr  and  0\n  are  similar, 

and  therefore  the  area  A'  of  the  triangle  qgr 

Fi  ~  2p  • 


Hence 


A'ote.— If  p  be  made  =  ^-- 


Iu=2pA  +2pA'  =2;;(.4  +A'). 

A 
2' 


and  therefore 


7=^2    and    %,2-^^', 

/v=.4(.4+.4'). 


The  angle  Wn  is  also  evidently  a  right  angle. 
8.  Bow's    Method.- A-   examination   of   the  frame  and  stress 
diagrams,  Figs.  4  and  f),  s..o\vs: 

(1)  That  if  the  lines  representing  external  forces  on  parts  of 
the  frame  meet  in  a  point,  the  corresponding  line«  in  the  stress 
diagram  form  a  closed  polygon. 

(2)  That  if  the  lines  representing  parts  of  the  frame  diagram 
form  a  closed  polygon,  the  corresponding  lines  in  the  stress  diagran. 
meet  in  a  point.  * 

(3)  That  if  lines  meet  in  a  point  in  the  stress  diagram,  the  corre- 
sponding lines  in  the  frame  diagram  are  contiguous  and  form  either 
closing  or  non-closing  polygons. 

(4)  That  of  all  the  stress  diagrams  which  can  be  dra«Ti  for  a 
g.^o„  frame  under  given  loads,  there  is  only  one  which  satisfies 
the  throe  recii)rocaI   relationships  expr.>ssed   in   CI)    (o)    and   (3^ 
and  It  is  call.'d  the  recii)rocal  diagram  of  stress.     '        '  '' 

As  a  coasequence  of  these  r(>lationsl,ips,  Row  devise<l  a  sv.tem 
of  notation  which  greatly  facilitates  tl„.  construction  of  the  "stress 
diagram. 

Lines  are  drawn  to  represent  the  .xt,.,.„n!  force?  acting  at  the 
joints.    A  letter  is  then  assigned  to  each  enclo.sed  area  of  the  frame 


TYPES  OF  TRUSS. 


13 


and  also  to  each  space  between  the  lines  of  action  of  the  external 
forces.  Thus  each  line  in  the  frame  diagram  is  defined  by  the  two 
lottors  in  the  two  spaces  separated  by  the  line  in  question.  The 
corresponding  line  in  the  reciprocal  diagram  is  parallel  to  this  line 
and  is  similarly  named. 

Cioncrally  speaking,  this  method  will  be  adopted  in  dealing  with 
tlie  strps.ses  developed  in  framed  structures. 

Q.  Roof-trusses.  —  A  roof  consists  of  a  covering  and  of  the 
trusses  (or  frames)  upon  which  it  is  supported.  The  covering  is 
giiicrally  laid  upon  a  number  of  common  rafters  which  rest  upon 
horizontal  beams  (or  jmrhns),  the  latter  being  carried  by  trusses 
spaml  at  inter\'als  varying  with  the  type  of  constniction,  but  aver- 
aging about  10  feet.  The  truss-rafters  are  called  jrrincipal  rafters, 
and  the  trusses  themselves  are  often  designated  as  principals. 

In  roofs  of  small  span  the  trusses  and  purlins  are  sometiinea 
liisponsed  with. 

Types  of  Truss.— A  roof-truss  may  be  constructed  of  timber, 
of  iron  or  steel,  or  of  these  materials  combined.  Timlxr  is  almost 
inxariably  employed  for'  small  spans,  but  in  the  longer  si)aus  it 
hn<  been  largely  superseded  by  iron,  in  consequence  of  the  com- 
1  lined  lightnes.«,  strength,  and  durability  of  the  latter. 

Attempts  have  been  n.ade  to  classify  roofs  according  to  the 
mode  of  construction,  but  the  variety  of  form  is  so  great  as  to  render 
it  impracticable  to  make  any  further  distinction  tiian  that  which 
iiiiiy  lie  drawn  between  tho.se  in  which  the  reactions  of  the  supports 
are  vertical  and  those  in  which  they  arc  inclined. 

I'ig.  122  is  a  simple  form  of  truss  for  spans  of  less  than  30  ft. 

I'ig.  L'3  is  a  suiH-rior  framing  for  spans  of  from  30  to  40  ft.;  it 
may  he  still  further  strengthened  by  the  introduction  of  struts 
Fi^'s.  24  and  2o,  and  with  such  modification  has  Ixen  employed 
to  span  oiH-nings  of  00  ft.  It  is  safer,  however,  to  limit  the  use 
of  tli(>  type  .shown  by  Fig.  24  to  spans  of  less  than  (K)  ft.  Figs.  2G, 
27.  2S,  20,  and  30  are  forms  of  truss  suitable  for  spans  of  from  00  to 
HH)  ft.  anil  upwards. 

Arched  ro<jfs,  Figs.  20  and  30,  admit  of  a  great  variety  of  treat- 
nirnt.  They  have  a  pleasing  apl^arance,  and  cover  wide  spans 
without  intermediate  supports.  The  (hitness  of  the  arch  is  limited 
by  the  re(]uirement  of  a  minimum  thrust  at  the  jtbutments.    The 


9# 


14 


THEORY  OF  STRUCTURES. 


thrust  may  bo  resisted  either  by  thickening  the  abutments  or  by 
introducing  a  tie.  If  the  only  load  upon  a  roof-truss  were  its  own 
weight,  an  arch  in  the  form  of  an  inverted  catonarv,  with  a  shallow 
rib,   might  be  Uocd.    But  the  action  of  the  wind  induces  oblique 


Fig.  22. 


Fio.  23. 


Fig.  24. 


Fio.  27. 


Fio.  28. 


Fio.  29. 


Fio.  30. 


and  transverse  stresses,  so  that  a  considerable  depth  of  rib  is  gen- 
erally needed.  If  the  depth  exceed  12  in.,  it  is  better  to  connect 
the  two  flanges  by  braces  than  by  a  solid  web.  Roofs  of  wide  span 
are  occasionally  carried  by  ordinari-  lattice  girders. 

Princijxils,  Purlins,  etc.— The  principal  rafters  in  Figs.  22  to  28 
are  straight,  abut  against  each  other  at  the  peak,  and  are  prevented 
by  tie-rods  from  spreading  at  the  heels.  W\\on  made  of  steel,  tee  (T), 
rail,  and  cliaimel  (Ix)th  .single  —  and  double  I)  bars,  bnlb-tee  (J) 
and  rolled  (I)  btvims,  are  all  excellent  forms. 

Timlwr  rafters  are  rectangular  in  section,  and  for  the  sake  of 
economy  ami  apiK-arance  are  often  made  to  tajx'r  uniformly  from 
heel  to  ix'uk. 

The  Ihm'I  is  fitted  into  a  .'suitable  cast-iron  skew-back,  or  is  fixed 
between  wrought-inm  angle-brackets  and  rests  either  directly  upon 
the  wall  or  iiimii  a  wall-piate. 

When  the  span  exceeds  m  ft.,  allowance  .should  l>e  made  for 
alterations  of  length  due  to  changes  of  temiK'rature.  This  may  Iwt 
efTcctcd  by  interposing  a  set  of  rollers  Ix'tween  the  skew-back  and 
wall-plate  at  one  heel,  or  hy  fixing  one  liecl  to  the  wall  and  allow- 
ing the  opposite  .skew-hack  to  .slide  freely  over  a  wall-plate. 

The  jiuiclion  at  the  peak  is  made  by  means  of  a  ca-sting  or  by 
steel  plates. 


ROOF  iVEIGHTS. 


15 


Light  iron  and  timber  beams  as  well  as  angle-irons  are  employed 
as  purlins.  They  are  fixed  to  the  top  or  sides  of  the  rafters  by 
brackets,  or  lie  between  them  in  cast-iron  shoes,  and  are  usually 
held  in  place  by  rows  of  tie-rods,  spaced  at  6  or  8  ft.  intervals 
between  peak  and  heel,  running  the  whole  length  of  the  roof. 

The  sheathing-boards  and  final  metal  or  slate  covering  are 
fastened  upon  the  purlins.  The  nature  of  the  covering  regulates 
the  spacing  of  the  purlins,  and  the  size  of  the  purlins  is  governed 
by  the  distance  between  the  main  rafters,  which  may  vary  from 
4  ft.  to  upwards  of  25  ft.  When  the  interval  between  the  rafters 
is  so  great  as  to  cause  an  undue  deflection  of  the  purlins,  the  latter 
should  be  trussed.  Each  purlin  may  be  trussed,  or  a  light  beam 
r..ay  be  placed  midway  between  the  main  rafters  so  as  to  form  a 
supplementary  rafter. 

Struts  are  made  of  timber  or  iron.  Timber  struts  are  rectangular 
in  section.  Steel  struts  may  consist  of  L  bars,  T  bars,  or  light  col- 
umns, while  cas'  iron  may  be  employed  for  work  of  a  more  orna- 
mental character.  The  strut  heads  are  attached  to  the  rafters  by 
means  of  cast  caps,  steel  straps,  brackets,  etc.,  and  the  strut  feet  are 
easily  designed  both  for  pin  and  screw  connections. 

Ties  may  be  of  flat  or  round  bars  attached  either  by  eyes  and 
j)ins  or  by  screw-ends,  and  occasionally  by  rivets.  The  greatest 
care  is  necessary  in  properly  proportioning  the  dimensions  of  the 
eyes  and  pins  to  the  stresses  that  come  upon  them. 

To  obtain  greater  security,  each  of  the  end  panels  of  a  roof  may 
be  jirovided  with  lateral  braces,  and  wrind-ties  are  often  made  to 
run  the  whole  length  of  the  structure  through  the  feet  of  the  main 
.'Struts. 

Due  allowance  must  be  made  in  all  cases  for  changes  of  tem- 
perature. 

10.  Roof-Weights.— Tn  calculating  the  stres-ses  in  the  different 
iiiciiihers  of  a  roof-truss  two  kinds  of  load  have  to  be  dealt  with, 
tlio  (iiio  jx'rmnnent  (or  dead)  and  the  other  accidental  (or  live).  The 
].criiianetit  load  consists  of  the  covering,  the  framing,  and  accumti- 
Idtians  of  snov. 

Tallies  at  the  end  of  the  chapter  show  the  weights  of  various 
coverinp  and  framings. 

The  weight  of  freshly  fallen  snow  may  vary  from  5  to  20  lbs. 


16 


THEORY  OF  STRUCTURES. 


per  cubic  foot.  English  and  European  engineers  consider  an  allow- 
ance of  6  lbs.  per  square  foot  sufficient  for  snow,  but  in  cold  climates, 
similar  to  that  of  North  America,  it  is  probably  unsafe  to  estimate 
this  weight  at  less  than  12  lbs.  fjer  square  foot. 

The  accidental  or  live  load  upon  a  roof  is  the  wind  pressure, 
the  maximum  force  of  which  has  been  estimated  to  vary  from 
40  to  'i)  lbs.  per  square  foot  of  surface  perpendicular  to  the  direction 
oj  blow.  Ordinary-  gales  blow  with  a  force  of  from  20  to  25  lbs., 
which  may  sometimes  rise  to  34  or  35  lbs.,  and  even  to  upwards 
of  50  lbs.  during  storms  of  great  severity.  Pressures  much  greater 
than  ')0  lbs.  have  been  recorded,  but  are  wholly  untrustworthy. 
I'l)  to  the  present  time,  indeed,  all  wind-pressure  data  are  most 
unreliable,  and  to  this  fact  may  be  attributed  the  frequent  wide 
tli\-ergence  of  opinion  as  to  the  necessarj'  wind  allowance  in  any 
l)articuiar  case.  The  great  differences  that  exist  in  all  recorded 
wind  pressures  are  primarily  due  to  the  unphi'osophic,  unscientific, 
and  unpractical  character  of  the  anemometers  which  give  no  cor- 
rect information  eiiher  as  to  j)ressure  or  velocity.  The  inertia  of 
the  movhig  i)art.«,  the  transformation  of  velocities  into  pressures, 
and  the  injudicious  placing  of  tlic  •nemometer,  which  renders  it 
subject  to  local  currents,  all  tend  to  vitiate  the  results. 

It  would  be  practically  absurd  to  base  calculations  upon  the 
violence  of  a  wind-gust,  a  tornat'o,  or  other  similar  phenomena,  as 
it  i.s  almost  absolutely  certain  that  a  structure  would  not  lie  within 
its  range.  In  fact,  it  may  be  jussumed  that  a  wind  pressure  of  40  lbs. 
per  square  ioot  upon  a  surface  i)erpen(licular  to  the  direction  of 
blow  is  an  amjjle  and  perfectly  .safe  allowance,  es[X'cially  when  it 
is  remembered  that  a  greatt .  pressure  than  this  wouhl  cause  the 
overthrow  of  nearly  all  existing  towers,  chimneys,  etc. 

II.  Wind  Pressure  upon  Inclined  Surfaces.— llie  pr(>ssure  upon 
an  inclined  surface  may  be  obtained  from  the  following  formula, 
which  was  e.xixTiinentally  deduced  by  Hutton,  viz.: 


Pn=psma 


1.84co»n-l. 


(A) 


/)  being  the  intensity  of  the  nind  pres.sure  in  pounds  per  square  foot 
upon  a  surface  iMTiK'udieular  to  ?!ie  direction  of  blow,  and  /)„  being 
the  normal  iiitensity  upon  a  surface  iQclincd  at  an  angle  a  to  the 
direction  of  blow. 


DISTRIBVTIOS  OF  LOADS. 


17 


Tx^t  /)/,,  p„  be  the  components  of  p„,  parallel  and  perpendicular, 
rcs'lK-ctively,  to  the  direction  of  blow.    Then 

Ph  =  Vu  sin  a    and    p„  =  /)„  cos  a. 

Honce,  if  the  inclined  surface  is  a  roof,  and  if  the  wind  blows  hori- 
zontally, «  is  the  roof's  pitch. 

Again,  let  v  be  the  velocity  of  a  fluid  current  in  feet  per  second, 
iiiid  1m^  that  due  to  a  head  of  h  feet. 

Lot  tr  be  the  weight  of  the  fluid  in  pounds  per  cubic  foot. 

I^t  p  be  the  pressure  of  the  current  in  pounds'  per  square  fcot 
ii|i()M  a  .surface  perp)endicular  to  its  direction. 

If  the  fluid,  after  striking  the  surface,  is  free  to  escape  at  right 
angles  to  its  original  direction, 

9 
Hence  for  ordinary  atmospheric  air,  since  to -.08  lb.,  approximately, 


(B) 


When  the  wind  impinges  upon  a  surface  oblique  to  its  direction, 

1    •         •        e   1  .    /r  sin  /9\2  ,        ,     , 

the  intensity  of  the  pressure  is  J  ~i^7r~ }  ,  i'  btung  the  absolute  im- 
pinging velocity,  and  /?  being  the  angle  between  the  direction  of 
blow  and  the  surface  impinged  upon. 

Tables  prepared  from  formula;  A  and  B  are  given  at  the  end 
(if  the  chapter. 

12.  Distribution  of  Loads. — Engineers  have  been  accustomed 
to  Mssume  that  the  live  load  is  uniformly  distributed  over  the  whole 
(tf  I  ho  roof,  and  that  it  varies  from  30  to  35  lbs.  per  square  foot  of 
ciivcrod  surface  for  short  spans,  and  from  35  *o  -10  lbs.  for  spans 
of  more  than  60  ft.  But  the  wind  may  blow  on  one  side  only,  and 
;.lili(nigh  its  direction  is  usually  horizontal,  it  may  occasionally  be 
inclined  at  a  considerable  angh>,  and  be  even  normal  to  a  roof  of 
liiirli  pl'"h.  It  is  therefore  evident  that  the  horizontal  compo- 
ni  ut  (pit)  of  the  normal  pressure  (p„)  should  not  be  neglected,  and 
ii  niiiy  cause  a  complete  reiertml  of  stress  in  members  uf  the  truss, 
osiiorially  if  it  is  of  the  arched  or  braced  type. 


18 


THEORY  OF  STRUCTURES. 


If  Pn  is  the  total  normal  wind  pressure  on  the  side  of  a  roof  of 
pitch  a,  its  horizontal  eomponent  P„  sin  a  wll  tend  to  push  the 
roof  horizontaLy  over  its  supports.  This  tendency  must  be  resisted 
by  the  reactions  at  the  supports. 

In  roofs  of  small  span  the  foot  of  each  rafter  is  usually  fixed 
to  Its  support,  and  it  may  be  assumed  that  each  support  exerts 

the  same  reaction,  which  should  therefore  be  equal  to  ^"  ^'" ". 
In  roofs  of  large  span  the  foot  of  one  rafter  is  fixed,  while  that  of 
the  other  rests  upon  rollers.  The  latter  is  not  suited  to  withstand 
a  horizontal  force,  and  the  whole  of  the  horizontal  component  of 
the  wind  pressure  must  be  borne  at  the  fixed  end,  where  the  reaction 
should  be  assumed  to  be  equal  to  P„  sin  a. 

In  designing  a  roof-truss  it  is  assumed  that  the  wind  blows  on 
one  side  only,  and  that  the  total  load  is  concentrated  at  the  joints 
(or  points  of  support)  of  the  principal  rafters. 

For  example,  let  the  rafters  AB,  AC  of  a  truss  be  each  supported 
at  two  intermediate  points  (or  joints),  D,  E  and  F,  G,  re-spectivelj-, 
and  let  the  wind  blow  on  the  side  AB. 

Take  BD=CF=h,  DE^FG^h, 
EA=GA=h;  and  let  l^+l^  +  h^l; 
BC^2l  cos  a,  a  being  the  angle  ABC. 
I^t  W  be  the  permanent  (or 
dead)  load  {wr  square  foot  of  roof- 
surface.  Le  p„  be  the  normal  wind 
pressure  per  square  foot  of  roof  surface.  Let  d  be  the  horizontal 
dustance  m  feet  from  centre  to  centre  of  trusses. 


The   total  normal    live   load 


The     total 


at  E^pnd    -;^-;   at  A=p„d  ., 


concentrated  at  B=p„d-^;   at  Z> 


h 


=  («•(/ 


vertical 
at  E  and  G 


dead     load 
./a  +  Zs 

=  U"d — :r — : 


concentrated 
at  -1  =  H'rt/3. 


at    D   and   F 


Let  Rx,  /?2  bo  the  resultant  vertical  reactions  at  B  and  C,  respect- 
ively (i.e.,  the  total  vertical  reactions  le.ss  the  dead  weights,  ml''-, 
concentrated  at  these  pointsj. 


CRANES 


19 


Take  moments  about  C.    Then 

Ri2l  cos  o-simi  of  moments  of  live  loads  about  C+sum  ot 
moments  of  dead  loads  about  C 
-moment  of  resultant  wind  pressure  about  C+ mo- 
ment of  resultant  dead  load  about  C 


'Pnldl^+l  cos  2a]  +wd{li  +2/2+2^3)^  cos  a, 


I 


where  ^  +?  cos  2a  is  the  perpendicular  from  C  upon  the  line  of  action 

of  the  resultant  wind  pressure  which  bisects  AB  normally. 

The  moment  of  the  horizontal  reaction  at  fi  or  C  about  C  is 
eviilcntly  nil. 

R2  may  be  found  by  taking  moments  about  B. 

The  rafters  AB,  AC  have  one  end,  B,  fixed  and  the  other, 
r,  on  rollers,  the  reaction  R2  at  C 
being  vertical.  A  force  P,  inclined  at 
o  to  the  vertical,  acts  upon  AB  at  D, 
and  Q  is  the  resultant  vertical  load  on 
tlic  rafters.  I^t  P  and  Q  meet  in  E. 
Taking  ED=P,  and  the  vertical  DF  =  Q, 
then,  on  the  same  scale,  FE  is  the  result- 
ant /?3  of  P  and  Q.  Let  it  make  an 
angle  0  with  the  vertical.    Then 


Ri*- 


Fig.  32. 


8ma-d  =  jT  sm0  =  ^ 
"  Ra 


sma. 


Ix^t  R3  and  R2  meet  in  0.  Then  OB  is  the  direction  of  the 
nstiltant  reaction  Ri  at  B.  Taking  the  vertical  BG^R-i,  then,  on 
saiiK'  scale,  OG  =  Ra  and  OB=Ri,  ORG  being  a  triangle  of  forces. 
l.ct  Ri  make  an  angle  ^  with  the  vertical.    Then 


sm  (4>-d) 


R2  .    „    R2  .    ^ 
■K-  sm  p=^-  sm  6. 


If  BG=l  and  if  p  and  q  are  the  perpendicilars  from  B  upon 

Ril~Pp  +  Qq. 

13.  Simple  Frames  of  Various  Types. — (a)   Jib-crane. — Fig.    33 
ro]  .resents  an  ordinary  jib-crane.    OX  is  the  post  fixed  in  the  ground 


20 


THEORY  OF  STRUCTURES. 


at  O.  BC  IS  tho  jib.  and  AC  the  tie.  The  jib-tio  and  gearing  arc  so 
separated  from  tho  post  as  to  admit  of  a  free  rotation  round  its  axis. 
Fig.  34  IS  the  stress  diagram  when  tiie  crane  supfwrts  a  weight  W 
as  inilicated. 


Fig.  3.3. 


Fig.  34. 


The  weight  Tr,  howoAor,  is  not  suspended  directly  from  B  but 
IS  niLsed  or  lowered  by  means  of  a  chain  passing  over  pullevs  to 
a  eham-barrel  which  is  usually  fixed  to  the  j^ost.  Disregarding 
pullex-  friction,  the  tensio.i  in  the  chain  is  ^^,  n  being  the  nun.ber 
of  jalh.     Let  the  dotted  line,  Fig.  35,  indicate  tho  direction  in  which 


Fig.  35. 


Fig.  36. 


the  chain  pa,sses  to  the  chain-barrel.  The  loads  on  the  apex  are 
now  W,  acting  vertically,  and  ~,  acting  in  the  direction  of  the  chain. 
Thus  Fig.  36  i.s  tho  stress  diagram,  and  AC  represents  in  direction 
and  magnitude  the  resultant  force  at  the  apex.    AD  m  the  tension 


CRANES. 


21 


in  the  tie,  and  CD  the  compression  in  the  jib.  The  dotted  lines  in 
Fig.  36  also  show  that  generally  the  effect  of  chain  tension  is  to 
increase  the  thrust  on  the  jib  and  to  diminish  the  tension  in  the  tie. 
Draw  the  horizontals  AE  and  CO  to  meet  the  verticals  at  D  in  £ 
luul  (J  respectively. 

Tlicn  DG=vert.  component  of  thrust  in  jib  =  pressure  in  post  at  0, 
and     DE  =  vert,  component  of  tension  in  tie  =  tension  along  post. 

Therefore  ZX?-/)£  =  Tr  +  FC=  resultant    pressure  in  post  at  0. 

Also,  /I £  =  horizontal  component  of  tension  AD  in  tie,  and  the 
nioiiipnt  of  this  forct  with  respect  to  0=-AExXO.  This  moment 
fends  to  upset  the  crane. 

OH,  the  horizontal  projection  of  the  jib  (or  tie),  is  the  radius 
or  thrme  of  the  crane. 

If  the  post  revolves  about  its  axis  (as  in  7n<-cranes),  the  jib  and 
gearing  are  bolted  to  it,  and  the  whole  turns  on  a  pi'/ot  at  the  toe  K. 
In  this  case,  the  frame,  as  a  whole,  is  kept  in  equilibrium  by  the 
weight  ir,  the  horizontal  reaction  H  of  the  web  plate  at  0,  and 
tlie  reaction  R  at  G.  The  first  two  forces  meet  in  H  and  therefore 
tlie  reaction  at  K  must  also  pass  through  H. 

Hence,  since  OHK  may  be  taken  to  represent  the  triangle  of 
forces, 


H-=TV 


OH 

Ok 


and    R  =  ]\ 


HK 
OK- 


In  a  portable  crane  the  tendency  to  upset  is  counteracted  by 
means  of  a  weight  placed  u{K)n  a  horizontal  platform  OM  attached 
to  the  post  and  supported  by  the  tie  XM. 

The  horizontal  pull  at  X=AE,  Fig.  35,  and  if  OM  is  taken  to 
rei)resent,i4£,  then  on  the  same  scale  OX  is  the  counterweight  at  M. 

(h)  Derrick-crane. — The  figure  shows  a  combination  of  a  der- 
rick and  crane,  called  a  derrick-crane.  It  is  distinguished  from 
the  jil>-crane  by  having  two  backstays,  .47),  AE.  One  end  of 
tlie  jib  is  hinged  at  or  near  the  foot  of  the  post,  and  the  other  is , 
held  by  a  chain  which  passes  over  pulleys  to  a  winch  on  the  post, 
so  that  the  jib  inay  Iw  raised  or  lowered  a<!  required. 

The  derrick-crane  is  generally  made  of  wood,  is  simple  in  con- 
stiuction,  is  easily  erected,  has  a  vertical  as  well  as  a  lateral  motion. 


22 


THEORY  OF  STRUCTURES. 


Fig.  37. 


and  a  range  equal  to  a  circle  of  froin  10  to  &)  ff>et  radius.    It  is  there- 
fore useful  for  temporary  works,  setting  inasonry,  etc. 

The  stresses  in  the  jib  and  tie  are  calculated  as  in  the  jih-crane. 

and    tho-^e    in    the    backstays   and 
post  may  he  obtained  as  follows: 

Let    the   j)]aiie   of   the   tie   and 

jib  intersect  the  plane  DAE  of  the 

two  backstays  in  the  line  AF,  and 

suppo.se     the     backstays     rfpi;ice<I 

by  a  single  tie  .1/'.     Ta'  c  OF   to 

represent  the  horizontal  j  ,ill  nt  A. 

The  pull  on  the  "iinaginH"  "    ^tay 

AFis  then  represented  by  AF  and  is 

evidently  the  resuUant  pull  on   the 

two    backstays.      Completing    the 

parallelogram  FGAII,  AH  will  repre- 

sent  the  pull  on  the  backstay  .If", 

and  AC,  that  upon  AD,  their  horizontal  comi^nents  being  OK  OL 

resiHTtively.     The  figure  OKFL  is  also  a  parallelogram.  '       ' 

If  the   backstays  li;-  in  planes  at   right   angles   to   each   other, 

0/.=0Fcos«  =  7'sin«c..s«.  and  is  a  maxiMmm  when  0 -Q° 
and  ' 

OK^OFsmfi^Tsma^md,  ami  is  a  maximum  when  <?  =  nO°, 

0  being  tt...  anglr.  y'O/.,  an.j  n  th.-  inchnation  of  the  ti..  to  the  vertical 

Hence  the  stress  in  a  l,ackstay  is  a  n.aximum  wh.-n  th,.  plane 
ot  the  backstay  and  p..st  coincides  witli  that  of  the  jib  and  tie 

Again,  l..t  ,,?  b.-  tl...  inclination  of  th..  backstays  to  the  vertical 
The  vertical  comiM)nents  of  tlie  backstay  stresses  are 

7' sin  a  cos  <?  cot ,?    and     T  sin  «  .sn  fl  cot  y9;     • 
and,  therefor*',  the  corresponding  stress  along  tlu;  i)o.st  is 

r  sin  «  cot ,?  (cos  0  +  sin  0), 
which  i.<?  a  maximum  wlicn  (9-4.*)°. 

r.  Shenr-!<'f,.s  (or  Shear.)  ami  Tn/Hni,  (or  Gins)  are  often  employed 
when  heavy  w,  igfits  are  tn  Ik-  lifted.  The  former  consists  of  tuo 
strut.s,  AD,  AE,  united  ai  .1  and  supjKjrted  by  u  tie  AC,  which  may 


BRIDGE-  AXD  ROOF-TRUSSES. 


23 


bo  made  ailjustable  so  as  to  atlmit  of  being  lengthened  or  shortened 
The  weight  is  suspended  from  A,  and  the  legs  are  capable  of  revolving 


Fio.  38. 

around  DE  as  an  axis.  Let  the  plane  of  the  tie  and  weight  intersect 
till'  plane  of  the  legs  in  AF,  and  suppose  the  two  legs  replaced  by 
a  single  strut  AF.  The  thrust  along  AF  can  now  be  easily  obtained, 
and  licnco  its  components  along  the  two  logs. 

Ill  tripods  one  of  the  three  logs  is  asually  longer  than  the  others. 
Thoy  ;iro  united  at  the  top,  to  which  point  the  tackle  is  also  attached. 

(/.  liridqe-  and  Roof-trusses  of  Small  Span.— A  single  girder  is 
tlic  simplest  kind  of  bridge,  but  is  only  suitable  for  very  short  spans. 
For  longer  spans  the  miiUlle  point  of  the  girder  may  be  supported 


Fki   39. 


Fio.  40. 


:  >\  struts,  Fig.  30,  through  which  a  j)ortion  of  the  weight  is  transmitted 
1(1  tlie  abutments.     I.et  P  \yp  the  iM)rti()n  of  the  weight  thus  trans- 
niittod,   as  in  Fig.  40.      Fig.  41,  in  which 
Mi     P   is    the  line   of   loads,  is    the  stress 
ilia>;ram,  and  the  thrast  along  each  strut  is 

p 
.ir(  */iC) -7,- cosec  a.     Drawing  the  hori- 

■.)ntal  CO,  AOi'-BO)  and  CO  nre  the  com- 

I'oiicnts  of  the  thrust  at  the  foot  of  a  strut. 

llius   the   vertical  and  horizontal  pn'ssures 

on  till   ma.sonry  at  the  foot  of  a  strut  are 

P        ,  P      ,  .      ,  Fio.  41. 

-    and  .J  cot  a  res|x>ctively. 

In  Fig.  42  a  stratnififj-nU  is  lntm<iuce!l,  .and  the  girdft  in  ."iupportef! 


-'-  THEORY  UF  STRICTURES. 

at  two  intonnediate  points.     Let  />  be  the  portion  of  the  weight  at 
each  ot  these  points  which  is  transmitted  through  the  struts  and  the 


^'"^  «•  Fig.  43. 

straining-cill,  Fig.  43.    Then  Fig.  44   is  the  stress  diagram,  ABC 
being  the  Inie  of  h.ads.     .1/;.  CD  are  dra«-n  parallel  to   the  corre- 
sponding struts,  an.l  AB^BC^-P.     Draw  I)B  horizontally. 
The  thrust  along  the  straining-cill  =/>)/?  =  P  cot  a. 

„'        "  ''    '•»"^''"t  =  AD  (or  CD) -Pcosec  a. 

By  means  of  stiaining-cills  the  girders  may  be  supported  at 


Fio.  44 


Fio.  4.'). 


several  pomts,  1,2 an.l  the  weight  concentrated  at  .-ach  inav 

be  assume<l  to  U^  one  half  of  the  load  betwtvn  the  two  adjacent 
IKHnts  of  supiH.rt.     The  ,aIeuIati<M.s  for  the  ..tres.sea  in  the  struts 
etc.,  are  made  precisely  as  above.  ' 

If  the  .stmts  are  very  long,  flu-y  are  liable  to  lM>nd,  and  counter- 
braces,  .1.1/,  /i.V,  are  added  to  counteract  this  tendency. 

e.  The  tria„nle  is  the  only  geometrical  (ig.m-  of  which  the  form 
cannoL  he  change.l  witlu.ut  varying  the  lengths  of  the  sides  For 
this  H'ason,  ail  compound  trusses  for  bridges,  roofs,  etc..  are  made 
up  of  triangular  frames. 

Fig.  4(i  n'presetaa  the  simplest  form  of  roof  tru.ss  AC  BC  are 
rafters  „(  e.,ual  letmth  Inclined  )o  the  horizontal  at  an  angle  «  and 
each  carries  a  uniformly  distrihuted  loatl   11'. 

The  rafters  react  horizontally  u|)on  each  other  at  C  and  their 
feel  are  kept  in  p.^^.tion  by  tfie  tic-Uutu  AB.    Consider  the  rafter  AC. 


LOADED  TRIANGULAR   TRUSSES. 


25 


The  resultant  of  the  load  upon  AC,  I  e.,  IF,  acts  through  the 
middle  point  D. 

Let  it  meet  the  horizontal  thrust  //  of  BC  upon  AC  in  F.  For 
equilibrium  the  resultant  thrust  at  A  must  also  pass  through  F. 

Also,  if  ;-  is  the  angle  FAE, 

,       AE     AE     1 

AFE  is  evidently  a  triangle  of  forces  for  W,  H,  and  R.    Therefore 
H  =  W  cot  Y^  =  g'Cot  a, 


R  -  W  cosec  ;• 


■"•s/l+2^", 


R  cos  Y  =  H  =  tension  in  tie, 

R  sin  ^  =  ir  =  pressure  on  support. 

It  must  of  course  be  remembered  that  when  the  horizontal  mem- 
b<r  acts  as  a  tie  the  reactions  at  the  two  supports,  due  to  the  external 
loads  on  the  truss,  must  '-ecessarily  be  vertical. 


FiQ.  46. 


Fio.  47. 


If  the  rafters  AC,  BC  are  unequal,  let  a,,  az  l)e  their  inclinations 
to  .1,  B,  «'S|HTtively. 

T^-t  ir,  be  the  uniformly  distributed  load  upon  AC,  W2  that 
upfin  BC. 

\r{  the  direction  of  the  mutual  thmst  P  at  C  make  an  angle  ^? 
with  the  vertical,  so  that  if  CO  is  drawn  perjx'ndicular  to  FC,  the 
.inRlo  COB-ii;   the  angle  ACF -^-ACO-fXf-i^-ax). 

Draw  AM  perjjpndicular  to  the  direction  of  P,  and  consider 
ilir  rafter  AC.  As  before,  the  thrust  Ri&iA.  the  resultant  weight  H*, 
at  the  midiile  point  of  AC,  and  the  thrust  P  at  C  meet  in  the  point  F. 


mw 


•26  THEORY  OF  STRUCTVRES. 

Take  moments  about  -4.    Then 

P  AM^WiAE. 
But  AM  =AC  sin  ACM  =AC  cos  O^-ai), 


and 


AE=    ,  -  cos  ai.     Therefore 


2    cos  G'''-ai)' 
Similarly,  by  considering  the  rafter  BC, 

p  _  IE? ^^^  «2  Tl'z     cos  aj 


Ilcnce 

and  therefore 


2   sin(,3+«2-90°J  2  cosC?+a2)* 

Hi      <•"««! ^  n  _  _  Hj  _^oa  a2 

2cosOJ-aij°     ~       2  cosOS+aJ' 


tan  ^  ■ 


Tr,  +  H'a 


M'l  tana2-irj  tanai' 


The  horizontal  thrust  of  each  rafter  =  P  sin  ,?. 
The  vertical  thrust  upon  the  supjxirt  .1  =  ll',  -P  cos  ^. 
The  vertical  thrust  upon  the  support  fl  =  n\, -i- P  eos  ,9. 
Sometimes  it  is  exjiedient  to  support  the  centre  of  the  tie-beam 
upon  a  colunm  or  wall,  the  king-post  iKMng  a  pillar  against  which 

the  heads  of  the  rafters  rest  in  such 
a   maimer    that    the    reaction    upon 
AC  at  C  is  at  right  angles  to  AC. 
Consider  tl«<  rafter  .If. 
The   normal    reaction   W   of  CO 
upon   AC,   the  resultant  weight    11' 
at    the    middle    poitit    J),    and    the 
thrust  R  at  .4  m«<et  in  tiie  iM)int  F. 
Taking  moments  about  .1, 

If  AC-WAE,     or    /e'=^'c(»a. 


Iki.    W. 


Thus     the    total    thrust    trans- 


mitted through  CO  to  the  sup|X)rt  atO-H'cos^a 


KING-POST  TRUSS. 


27 


w  w 

The  horizontal  thrust  on  each  rafter  =-r  cos  a  sin  a  =-rsin  2  a. 

2  4 

14.  King-post    Truss. — The    simple    triangular    truss    may   be 

modified  by  the  introduction  of  a  king-post,  Fig.  49,  which  carries 

a  portion  P  of  the  weight  on  the  tie-beam,  and  transfers  it  through 


Fio.  50. 


tlic  rafters  to  the  end  of  the  tie,  where  it  is  again  resolved  into  hori- 
zontal and  verticol  com|K)nents,  the  former  straining  the  tie  in 
tension,  and  the  latter  cau-sing  a  pressure  on  the  supports. 

( )f  the  weight  W  on  each  rafter,  it  is  ju^sumed  that  one-half  is  car- 
ried on  a  support,  and  the  other  half  on  the  ridge.  Thus  the  loads 
on  the  frame  are  11'  at  the  ridge,  P  at  the  foot  of  the  post,  and  the 
t\v;)  vertical  reactions,  each  =  h{W  +  P),  at  the  supports.  Fig.  50 
is  cvidentlv  the  stress  diagram,  AB  being  the  line  of  loads.      Then 

...     ,.„     W  +  P 
reaction  at  each  support  =Aa  -  1  ij= — :: — 


.i/y  =  ir; 

Therefore 


YX~P. 


AC  =  thrust  along  rafter  =  ^(\V+P)  cosec  a, 
CA'  -/>>1'  -tension  in  tiv^{]\'+P)  cot  a, 
DC  -  I'A'  -  tension  in  post  ■=  P. 


Fio.  51. 

In  Fig.  ')1  the   \Mj*i   divides  the   tie-beam   into   two  segnients 
a  and  h.      If  W  is   the  load  on   the  ridge,  a!»auni«Hl  equal  to  one 


28 


THEORY  OF  STRUCTURES 


half  of  the  load  on   the   two  rafters,  and  if  P  is  the  load  concen- 
trated at  the  foot  of  the  post, 

the  reaction  at   the  left     support  =  ~(ir  +  P)  and 


«■ 


right 


=  ^6("'  +  ^)- 


Fig.  52  is  evitlently  the  stress  diagram,  AB  being  the  line  of 
loads. 

Then  AB  =  \y;    reaction   at  left  support  =X^  =11— „  at  right 

support  =  YB  =  ir-4i..     Therefore 
a  +  0 

AC  =  thrust  along  one  rafter  =  JI' cosec  a  • 

n  -X-rt  f 


a  +  b 


BD^ 


"     the  other  rafter  =  11'-%  cosec  5 : 

a  +  b  '^' 


CX  =/>}'  =  tension  in   tie  =  11'^  cot  a  =11"^  cot  fi; 
DC  =  I'A'  =  tension  in  post  =  P. 

The  tension  in  the  tie  diminishes  as  a  increa.ses,  and  is  nothing 
when  a  =90°,  i.e.,  when  the  rafter  is  vertical. 

15.  Incomplete  Frames.— The  frames  discussed  in  the  preced- 
ing articles  will  support,  wilhout  charuje  of  form,  any  load  consistent 
with  strength,  and  the  stre.sses  in  the  .sev(>ral  memlM'rs  can  be  found 
in  terms  of  the  load.  It  .sometimes  hapix>ns,  lu.wever,  that  a  frame 
\s  incomplete,  ^o  that  it  tends  to  change  form  under  every  distribution 
of  load.     An  example  of  this  class  is  the  simple  trajx^zoidal  truss, 


Fio.  M. 


consisting  of   the  two  horizontal   members  AB,  DE,  and  the  two 
equal  inclined  meml»er.«i  AD,  BR,  Fig.  53. 

Firxt,  let  there  be  a  weight  W  at  each  of  the  points  /),  E. 


INCOMPLETE  FRAMES. 


29 


Fig.  54  is  the  stress  diagram  in  which  12  =23=  IF. 
lOvidontly  also  XI  =A'3  =  If  =  vertical  reaction  at  each  support. 
14  (  =  34)  =ir  cosec  a  =  thrust  along  each  sloping  member. 
Tension  in  .45=24  =  11'  cot  a  =  thrust  along  DE. 
Scxt  let  there  be  a  weight  Wi  at  D  and  a  weight  W2{<W\) 
at  E. 


Fio.  55. 


Fig.  56. 


Let  123  be  the  line  of  loads,  where  12-Tri  and  23  =  Tr2. 

The  stress  diagram  for  the  joint  E  is  the  triangle  243,  and 
if  \n  is  dra^Tj  parallel  to  AD,  the  point  n  evidently  falls  outside  4 
and  the  stress  diagram  does  not  close.  The  frame  will  therefore 
he  distorted  unless  a  brace  is  introduced  connecting  B  and  D  or 
.1  and  E.  In  the  former  case  the  stress  diagram  becomes  12345, 
tlic  lino  45  being  drawn  parallel  to  BD. 

Drawing  the  horizontal  5A',  the  vertical  reaction  at  yl  is  XI,  and 
It  H  is  A'3. 

In  general,  if  Fig.  57  represents  any  frame  of  four  members, 
resting  upon  supports  at  A  and  B,  and  if  IF  is  a  weight  concen- 


Fio.  67. 


Fig,  68. 


trated  at  D,  then  124  ia  the  stress  diagram  for  the  joint  at  D. 
]>!'iwing  43  parallel  to  DE.  it  i-i  evident  that  23  i?  the  only  weight 
\iiich  can  be  supported  at  E  without  producing  distortion. 


30 


THEORY  OF  STRUCTURES. 


i6.  Queen-post  Truss.— It  is  a  common  practice  to  modify  the 
meoinpLte  frame  represented  by  Fig.  53  by  introducing  two  ver- 
tical queen-poKts  (queen-rods  or  queens)  DF  and  EG,  Fig.  59,  through 
which  the  loads  are  transmitted  to  D  and  E.  The  frame  thus  modi- 
fied IS  still  incomplete,  and  if  there  are  no  diagonal  braces  DG,  EF, 


Fio.  59. 


Fio.  60. 


the  distortion  of  the  frame  under  an  unevenly  distributed  load  can 
only  be  prevented  by  the  friction  at  the  joints,  the  stiffness  of  the 
members,  and  by  the  queens  being  rigidly  fixed  to  AB  at  F  and  G. 

Let  Wi  be  the  load  at  F  transmitted  through  the  queen  FD  to  D. 

Ix-t  ll'a  (<ir,)  be  the  load  at  G  transmitted  through  the  queen 
GE  to  E. 

//  the  frame  is  rigid,  the  reactions  ff,  at  A  and  R2  at  B,  which 
will  balance  these  weights,  can  easily  be  found  by  taking  moments 
about  B  and  .4  successively.    Thus, 


and 


where  .1^=/  and  FG  =c. 

In  the  line  of  loads,  Fig.  60,  12-11',  and  23-TF2. 

Take   A'I=/?,    and    X3^R.,.    Then   A'14    and    12M1    are   the 
reciprocals  of   th?  joints  at  .4  and  D  respectively,  so  that  there  is 

1st.  A  thrust  =25  =A'4  =/e,  cot  a  in  DE,  and 


2d.  A  thrust  =  45  =A'2^ 


n'l-ir^.'-r 


mDF 


•2  I 

'downward  pressure  at  F  on  the  tie  .4B. 


QUEEX-POST  TRUSS. 


31 


Again,  A'36  and  236n2  are  the  reciprocals  of  the  joints  at  B 
anil  E  respectively.  For  equilibrium  this  last  should  close  with 
tlie  reciprocal  of  the  joint  D,  i.e.,  n  should  coincide  with  5.  But 
it  appears  that  there  is 

1st.  A  thrust  =n2  =6X  =^2  cot  a  in  DE,  and 


2d.  A  tension  =6n  -X2 


W1-W2I-C 


in  EG 


2  I 

=  an  upward  pull  at  G  on  the  tie  AB. 

Thus  in  DE  there  is  an  unbalanced  thrust 

■=  (fti - ^2)  cot  a-iWi- W2)j  cot  a  -^^'~     '  ^-^  j, 

d  being  the  depth  of  the  truss. 

The  tie  is  also  acted  upon  at  F  and  G  by  two  forces  of  equal 
magnitude  but  acting  in  opposite  directions,  thus  forming  a  couple 

of  moment  — ^-^ — -  —j~'^>  which  tends  to  cause  a  rotation  of  the 

tie-beam. 

To  neutralize  the  tendency  to  rotation  and  to  take  up  the  unbal- 
anced force  in  DE,  a  brace  may  be  introduced  from  D  to  G,  Fig.  61  > 
or  from  E  to  F.  In  the  former  case  the  brace  will  be  in  compression, 
and  in  the  latter  case  in  tension. 


T-r 


Fio.  61. 


Fio.  02. 


Fig.  62  is  evidently  the  stress  diagram  for  the  modified  truss, 
and  the  stress  developed  in  DG  by  the  unbalanced  force  is 

„     W1-W2I-C        ,    W1-W2I-CS 
5G 2 r"""^ 2 Fd' 

s  being  the  length  of  the  diagonal. 


urn 


*»P' 


32 


THEORY  OF  STRUCTURES. 


With  a  single  collar-beam  DE,  Fig.  63,  and  a  load  2W  uniformly 

II' 

distributed  over  the  rafters,  it  may  be  assumed  that  -^  is  cancentrated 

at  each  of  the  joints  D,  C,  and  E.    Then  Fig.  64  is  the  stress  diagram, 
123-1  being  the  line  of  loads. 


Fio.  63.  Fig.  64. 

The  stress  in  the  tie-beam       =X5  =  Jir  cot  a, 
'      "    "    collar-beam  =56  =JTrcota, 
"     "AD  {or  BE)       -15  =Jir  coseca, 
'      "CD  (or  CE)       =  26  =  i  ir  cosec  a. 


With  a  collar-beam  DE,  two  queen-posts  DF,  EG,  Fig.  65,  and 
a  uniformly  distributed  load  of  211'  over  the  rafters,  the  stresses  in 
the  members  at  the  joints  D  and  E  become  indeterminate.  A  further 
condition  is  therefore  required,  and  it  is  sometime  s  assumed  that  the 

component  of  each  of  the  weights  y  at  .D  and  E,  normal  to  the 


Pro.  6.5. 


Fio.  66. 


rafter  on  which  it  is  concentrated,  is  taken  up  by  the  collar-beam  and 
the  queen. 

Fig.  66  is  evidently  the  stress  diagram,  1234  being  the  line  of 
loads. 


INCOMPLETE  FRAMES. 


38 


Tr  w 

56  ( =68)  =  Y  pos  «  =  component  of  -^  normal  to  rafter, 


II 


57  ( =78)  =-y  cos^  a  =  thrust  along  each  queen. 


W 


The  effect  upon  the  tie-beam  is  the  same  as  if  a  load  —  cos^  a  is 

concentrated  at  each  of  the  joints  F  and  C  the  loads  being  directly 
borne  bv  the  supports  at  A  and  B. 

W  W 

67  =—  cos^  a  tan  a  =-t-  sin  2a  =  thrust  along  collar-beam, 

A'5  =  Yl  =ZS  =  tension  in  tie-beam, 

A'l  =Z4  =|]r-A'Z  =f Tr-58  =  Jir-TT cos2  a  =vertical reac- 
tion at  each  support, 

W  3 

26  ( =36)  =  J  cosec  a;  15  (  =48)  =jW  cosec a. 

This  frame  belongs  to  the  incomplete  class,  and  if  it  has  to  sup- 
port an  xinequalhj  distributed  load,  braces  must  be  introduced  from 
U  to  a  and  from  E  to  F. 

The  truss  ABC,  Fig.  67,  having  the  rafters  supported  at  two  inter- 
mediate points,  may  be  employed  for  spans  of  from  30  to  50  feet. 


Fio.  67. 


Fio.  68. 


Suppose  that  these  intermediate  joints  of  support  trisect  the  rafters, 

Mv\  lot  each  rafter  carry  a  uniformly  distributed  load  W. 

W 
It  may  be  assumed  that  -r-  is  concentrated  at  each  of  the  joints 

//,  T),  C,  E,  K.    I^et  P  be  the  load  borne  directly  at  each  of  the 
joints  F  and  G. 


34 


THEORY  OF  STRUCTURES. 


Fig.  68  is  evidontiy  the  stress  diagram,  16  being  the  line  of  loads^ 

and  12  =23  =34  =45  =  56  =Y- 

Also,  A'l  =  I ir  +  P=Z6=  vertical  reaction  at  each  support; 

ZY=P  =  YX. 


Then 


17  =  (Hr  +  P)coseca=6-ll, 

78=-rCOseca  =  1011, 
6  ' 

28  =  (Jir  +  /')coseca=5  10, 

A'7  =  (|ir  +  P)cota  =  F-6, 

ri2  =  (jir  +  P)cota, 


»...=(f..) 


cot  a. 


17-  King-post  Roof-truss.     (Fig.  69.)— This  truss  is  a  simple  and 

economical  frame  for  spans  of 
not  more  than  30  feet.  To  elimi- 
nate excessive  bendin^;  and  to  di- 
minish the  danger  of  transverse 
failure,  the  middle  pomts  of  the 
comparatively  long  rafters  are 
supported  by  struts  OD  and  OE.  A  portion  of  the  weight  on  the  rafters 
is  then  transmitted  through  these  struts  to  the  vertical  tie  (king-post 


Fio.  69. 


Fio.  70. 


'..x 

1 

X.      ^» 

Fi<!.  71. 


or  rod)  CO,  which  again  transmit.'?  it  through  the  rafters  to  act  partly 
a.s  a  vortical  pro.«sure  upon  th(>  supjjort.s  and  partly  as  a  tension 
on  the  tic-beam.  The  main  purpose,  indeed,  of  struts  and  ties  is 
to  transform  tran.sverse  into  longitudinal  stresses. 

If  W  is  the  uniformly  distributed  load  on  each  rafter,  it  may  be 


BEST  CRASE. 


35 


assuinod  that  one-half  of  the  load  upon  AD  and  upon  BE  is  borne 
(lirpctly  by  the  supports  at  -4  and  B,  and  that,  of  the  remainder  of 

I!' 

the  load,  -  is  concentrated  at  each  of  the  joints  D,  C,  and  E. 

Lot  P  Ijo  the  load  concentrated  at  0. 

Then  Fig.  71   is  evidently  the   stress  diagram,  1234  being  the 

ir 

line  of  loads  in  which  12  =23  =34  =y. 

3         P 
Also,  A'l  =jTr  +  2-  =  l'4  =  vertical  reaction  at  each  support. 


(3         P\ 
j.U'+^l  cot  a  =  78, 

"       "       "  king-post  =67  =Y+^. 

DA  (or  BE)  =  15  =  (|  W  +|)  cosec  a, 
CZ)(orCi?)=26  =  /jir+^) 


"  thrust 


cosec  a. 


W 


"    2X)(or£0)=56=YCOseca 


i8.  Bent  Crane. — Fig.  72  shows  a  convenient  form  of  crane  when 
niucli  headroom  is  required  near  the  post.  The  crane  is  merely  a 
girder  with  one  end  fixed,  and  the  sections  of  the  members  and  other 
details  of  construction  are  governed  by  the  lifting  power  required. 


Fio.   72. 


Fio    73. 


For  light  loads,  say  not   exceeding  10   tons,  the    flanges   may  be 
braced  together  as  shown  in  Fig.  72.    The  flanges  may  be  kept  at  the 


36 


THEORY  OF  STRUCTURES 


same  distance  apart  throughout,  or  the  distance  between  them  may 
be  gradually  diminished  fro'ii  the  base  to  the  peak. 

Let  ir  be  the  weight  concentrated  at  the  jieak. 

Fig.  73  is  the  stress  tliagram,  and  the  magnitudes  of  the  stresses 
obtained  from  this  diagram  may  be  checketl  by  the  method  of 
moments. 

Take  IP  =  10  tons,  and  consider  the  equilibrium  of  the  portion  of 
the  crane  above  the  section  M\. 

Let  X,  y  he  the  length  of  the  p^'rpendiculars  from  0  upon  Yf 
and  the  line  of  action  of  11'  resix«ctively.    Then 


r/X(/  =  10Xj-,  and    therefore   Yf 


=  10^  =  1^  =  22 J    tons, 


vhich  is  the  value  obtained  for  Yf  from  the  stress  diagram. 

19.  Roof-trusses  of  Considerable  Span,— (a)  I<:ach  of  the   joints 
in  the  principal  rafters  of  the  roof-truss  represented  by  Fig.  74  is 

t 


Fio.  TS. 


loaded  with  a  weight  11'.  The  line  of  loads  is  Xr\  Fig.  7^,  in  which 
AO-OP-ir- ...  -TV,  and  Fig.  75  is  the  diagram  giving  the 
strrssTR  in  all  the  nieniber.'^  of  ihe  truss. 


ROOF-TRUSSES. 


37 


(6)  Next,  let  the  truss  be  subjected  to  loads  which  act  on  the 
joints  of  one  of  the  principal  rafters  and  which  are  inclined  to  the 
vertical. 

{c)  Again,  let  the  frame  Fig.  76  be  subjected  to  inclined  loads 


Fio.  77. 


nt  th(>  joints.  The  reactions  due  to  these  loads  may  be  found  as 
follows: 

Produce  the  lines  of  action  of  the  loads  to  meet  the  springing- 
liiK'. 

I{<s()l\»«  the  loads  XO,  OP,  PQ,  QH,  and  US  into  their  vortical 
!iiii|  liorizontal  com|)onentH,  the  former  Ix'ing  represented  by  N'O', 
<n>\  l»Q',  Q'lr,  and  R'S,  Fig.  77. 

rak.«  any  fjole  O"  and  join  it  witii  O  ,  P',  Q'  R',  S. 

Construct  the  funicular  fwlygon  1234.')6  and  draw  0"Y  par- 
uUfl  lo  the  cUwiiig  ill-..'  IG.    Then  <Si'  and  i'.V  are  t\w  actual  reac- 


38 


THEORY  OF  STRUCTURES. 


tions  at  the  support.-s,  and  the  reciprocal  diagram  Fig.  77  can  be 
easily  constructed    as  already  described. 

In  this  figure  the  i)oiiits  //,  /.  J,  K,  L,  and  M  coincide,  and  there 
can  be  no  stre.ss  in  ///.  /J,  JK,  KL,  or  LM. 

id)  Lork-jnint  Trusses.— In  certain  truss(>s  the  stres.ses  at  two  or 
more  joints,  called  hx'k-joints,  are  indeterminate,  and  the  reciprocal 
figure  cannot  be  dir(>ctly  drawii.  A  further  condition  is  required,  and 
it  is  sometimes  as.sunied  that  the  stres.ses  in  two  of  the  nunilx-rs 
meeting  at  a  lock-joint  are  e<]ual.  The  stresses  are  also  determinate 
if  the  relative  yield  of  the  bars  is  known.  A  simple  and  independent 
method  may  be  illustrated  by  means  of  the  truss  shown  in  Fig.  78. 


Taking  any  pole  O",  Fig.  79   is  the  somi-funicular  polygon  of  the 
hmU  on  the  r(K»f,  and  the  l)en(iing  moment  at  any  |M)int  i.x  moas- 


ROOF-TRUSSES. 


39 


unil  by  the  intercept  between  the  closing-line  19'  and   the  line 
ll.i .  .  .0  (Chapter  II). 

Let  Fi  be  the  force  in  01'; 

X  be  the  vertical  distance  between  OY  and  the  tnise  apex. 

Then    Fix  =the  bending  moment  at  the  apex 

=  99'xO"y. 
Take  0"r=x; 

then  Fi  =99', 

and  therefore  99'  is  the  stress  in  the  member  OY. 
Let  Fs  be  the  force  in  GY. 


Then  Fj^r  -bending  moment  at  apex  of  space  O 

=  55' xo"r 

=  55' XX. 
Therefore      Fa  =  2x55', 

and  thus  the  stress  in  GY  is  twice  the  intercept  55'. 

In  Fig.  SO  take  1'0=99  and  1'G-2X55'.  The  reciprocal  dia- 
unmi  can  now  Ije  ea.'^ily  constructed. 

Consider  the  truss  Kig.  81  with  two  lock-joints.  In  this  truss 
tWc  rise  is  about  one  ttfth  of  the  span;   the  struts  FG  and  ST  are 


Fig.  81. 


Fio.  S2. 


normal  to  the  rafters,  and  about  one  tenth  of  the  span  in  length. 
Till'  members  AH  and  liC,  CD  and  DE, . . .  ,JK  and  KL  liieet 


40  THEORY  OF  STRUCTURES. 

Oil  11  circular  arc  drawn  through  the  inner  end  of  the  strut  FG  and 
through  the  ends  of  the  rafter. 

Let  J  be  the  distance  between  the  meniber  MZ  and  the  apex 
of  the  truss. 

Let  tiie  truss  be  fixed  at  the  left  support  and  rest  on  rollers  at 
the  right  support. 

Consider  the  effect  of  a  normal  wind  pressure. 

At  the  points  of  intersection  of  the  springing-line  with  the  lines 
of  action  of  the  loads  resolve  the  loads  into  vertical  and  horizontal 
components,  and  represent  the  former  by  a'b',  b'c',  c'd',  . . .  i'j. 
Take  the  pole  0"  at  a  distance  from  the  load-line  a'j  equal  to  the 
vertical  distance  between  the  ridge  and  MZ.    Draw  the  funicular 

. la 

''V ^b' 

^  ^  ^  ^  >    ,— -♦*' "^v'"o. 

\fi' 

Fio.  8a 


pf>lygnn  123  .  .  .  r.<  for  the  vertical  components  of  the  loads,  and 
draw  ()"Z  parallel  to  the  closing-line  nI.  Then  jZ  is  the  reaction 
!if  the  right  supiK)rt,  and  is  of  course  vertical.  Hence  if  the  vertical 
tlirniiuli  the  truss  a|K>x  intersects  the  closing-line  In  in  m  and  xr 
prn.iuccd  in  n.  nni  is  the  total  stress  in  MZ,  and  thus  the  position 
of  I  lie  point  .1/  ill  Fig.  s;j  is  defiric(l.  The  diagram  can  now  l)e  com- 
plrti'il  in  thi'  usual  iiiaiiiirT. 

(>)  The  principal  rafters  of  the  truss  repre.sente(l  by  I'ig.  .S4 
arc  each  /  ft.  long  and  arc  spaced  r/  ft.  centre  to  centre.  They 
carrv  !i  (jeail  weiirjii  nf  )/  lbs.  j)er  square  f(>ot  of  rfjof-coviiring,  nnd 
uj)on  the  rafter  Mi  tliere  is  a  normal  wind  i)ressure  of  ;>,  lbs.  per 


ROOF-TRUSSES. 


41 


s(iuar('  foot.    The  end  B  is  fixed,  and  rollers  are  placed  under  the 
foot  of  AC. 


Fio.  84. 


Fio.  85. 


Resolve  the  reaction  at  B  into  the  horizontal  component  H  and 
the  vertical  component  R. 

Then    H  =  total  horizontal  component  of  the  wind  pressure 

=  />„Wsin  a. 
Also,  taking  moments  about  C, 

Ri  21  cos  a  -wldl  cos  a  +  p„ldl-  +1  cos  a ), 

,  ,,       ,  r.      w''^        pJd /I      cos  2a  \ 

imd  therefore  ^i  - -rr- + -^-^  I  t  +    -pi — ■• 

2         cos  a  \4  2     / 

AKiiin,  since  there  are  rollers  at  C,  the  reaction  at  C  is  vertical^ 
aiiil  taking  moments  a)M)ut  B, 

Ri2l  cos  a  - «7d  •  Z  cos  a  +  pnW  •  ^, 


(ir 


2       4  cos  a 


Tlion    Fig.  85  is   the  stress  diagram,  1234   being  the  "lin«  of 
lads,"  in  which 


Also, 


L/3-uW;     .T2-p„--21. 
Ri'lY,     YX'H,    and    A'i-ZZg. 


42 


THEORY  OF  STRUCTURES. 


The  dotted  lines  show  how  tlie  stress  diagram  is  modified  when 
rollers  are  placed  under  B.  The  stresses  in  the  members  are  dimin- 
ished, and  hence,  in  designing  such  a  roof-truss,  the  wind  pressure 
should  be  assumed  to  act  upon  the  side  on  which  the  foot  of  the 
l>rincipal  is  fixeil  to  its  supjKjrt. 

(/)  A  roof-truss  for  a  larger  span  may  have  the  middle  points 
of  Its  principal  rafters  supported  by  struts,  as  in  Fig.  86.  The  ver- 
tical dead  load  at  each  of  the  points  F.  A,  G'  =— . 

The  normal  wind  pressure  o\er  AB 

P'>^d   ,  ,       ,       „ 
=  ^—  at  A  and  at  B, 

PM 
=  r,~  at  F. 


and 


The  horizontal  component  of  the  wind  pressure 
=pjd  sin  a  =  horizontal  reaction  at  B. 


Fia.  8a 


Fia.  87. 


The  vertical  reactions  /?,  an.i  R,  at  B  and  C  can  be  found  at 
once  by  takmg  moments  about  C  and  B  successively,  and  are 


^A 


Ri'iwld  +  ,^^'"'''  (cos2a+jV 
J  cos  a\  '/' 

^  4  cos  a 


ROOF-TRUSSES. 


43 


Fig.  87  is  the  stress  diagram,  1234567  being  the  "line  of  loads" 


I  J 

in  which  12=45=^,  23  ■ 


pjd 


wld 

T 


5^,  34=^=56=67. 


Also,  ir=ffi,     YX=H,    X7=«2. 

Th(>  clottod  Hnes  show  how  the  stresses  are  modified  when  the 
rollers  are  under  B,  and  also  show  that  the  straining  is  greatest 
wiicii  the  end  of  the  rafter  on  which  the  wind  blows  is  fixed. 

iff)  The  principal  rafters  of  the  truss  Fig.  88  are  supported  at 
two  intermediate  points. 

A 


Fio.  88. 


/)o?n.— Pitch  =30";  AD 'BD"AE''CE~2S  ft.;  trusses  =  13  ft. 
centre  to  centre;  dead  weight  =8  lbs.  per  square  foot  of  roof-sur- 
face; wind  pressure  on  the  side  AB  normal  to  roof-surf  ace  =  28  lbs. 
ixr  s(iuare  foot;  DF  ==DH 'EG 'EK;  DF  and  EG  are  vertical; 
rollers  under  the  end  C;  span -79  ft.;  AF^BH  =21  ft.;  FH -3^  ft. 

The  wind  pressure  =4459  lbs.   ( — ^13 -28]    at   each  of    the 

points  F,  H, 
and -3822  lbs.  (-y13-28J    at    each    of    the 

points,  A,  B. 
Tliedearfload  -1274  lbs.  (-  — •13-8J    at   each    of    the 

points  F,  H,  K,  G, 
and -2184  lbs.  (-21 -13 -8)  at  the  point  A. 
The  resultant  reaction  at  B 

1  16.'i62 

-  :j  (4  X 1274  +  2184) +--;=-- 13201.8  lbs. 

The  huri2ontal  reaction  at  B 

-16562  sin  30P -8281  lbs. 


44 


THEORY  OF  STRUCTURES. 


Fig.  89  IS  the  stress  diagram  123  .  .  .  10,  being  the  "line  of 
loads"  in  which  12=67=3822  lbs.;  23=45=4459  lbs-  34  = 
1274  =  56  =  89  =  9 •  10 ;  78  =  2184  lbs.  Also,  1 F  =  13  202  fe  •  FY 
=82811bs.  '  ■' 


Fig.  S9. 


The  dotted  lines  show  the  modified  stresses  when  rollers  are 
under  B,  and  again  show  that  the  straining  is  greatest  when  B  is 
fixed. 

(/))  Bon-stritig  Roof-trms.— In  the  truss  represented  by  Fig   90 
the  joints  of  the  upper  and  lower  chords  lie  on  curves  which  are 
either  circular  or  i)aralx)lic  arcs,  and  it  is  assumed  that  the  por. 
tion  of  a  chord  k'tween  any  two  consecutive  joints  is    straiglit 
I  nder  a  uniformly  distributed  load  the  reciprocals  of  the  joints 
at  the  feet  of  the  verticals  do  not  close,  and  the  truss  evidently  belongs 
to  the  class  of  incomplete  frames.     E.,uilii,rium  is  impossible  and 
deformation  is  only  prevented  by  the  stiffness  of  the  truss  antl  fric- 
tion of  the  joints.      To  render  the  stresses  determinate,  the  trusK 
IS  modifie.1  by  the  introduction  of  the  sloping  members  shown  in 
Fig.  01.  an.i  the  stress  diagram  for  the  uniformly  distributed  load 
IS   F:g.  92,    I234r)()  being  the    line  of   loads,  in  which  12  =  ir=';j 
-=34=45=r>G  =  load  concentrate<l  at  each  of  the  upper-chord  joint's 
Next  let  there  be  a  given  win<!  pressure  on  the  kft  half  of  the 
truss,  and  let  it  be  assumed  that  the  truss  is  fi.xed  at  the  left  sup- 


ROOF-TRUSSES. 


45 


]iort  and  rests  upon  rollers  at  the  right  support.  At  the  latter 
t!io  reaction  is  vertical  and  can  be  found  by  taking  moments  about 
tlic  loft  support.  The  reaction  at  the  left  support  may  be  resolved 
into  a  vertical  and  a  horizontal  component,  and  the  latter  must 


Fio.  90. 


Fio.  91. 


Fro.  92. 


Fio.  93. 


Pio.  94. 


iKccssarily  be  equal  in  magnitude  to  the  horizontal  component 
of  tho  total  wind  pres.sure.  The  vertical  component  is  now  easily 
ni.tuiiK'd  by  taking  moments  about  the  left  .support. 

\  reversal  of  stress  in  any  of  the  sloping  members  or  verticals 
nil  tlio  right  half  of  the  truss  can  be  prevented  by  introducing  the 
(oimter-braces  shown  in  Fig.  93.    The  stress  diagram  for  the  wind 
I'llVct  is  Fig.  94,  1234r)67  being  the  "line  of  load,"  in  which 
ir  =  vertical  reaction  at  left  support, 
FA'  =  horizontal  reaction  at  left  sup|)ort,         * 
A'7  -vertical  reaction  at  right  sup|K)rt, 
and  also        12  =i  normal  pressure  on  2a  =23 
34-i      "  "         "  46=45 


56-i 


6d-67. 


46 


THEORY  OF  STRLCTLRES. 


21.  Bridge-trusses.— A  bridge-truss  proper  consists  of  an  upper 
chord  (or  flange),  a  lower  chonl  (or  flange),  and  an  intermediate 
portion,  called  the  web,  connecting  the  two  chords.  Its  depth  is 
made  as  small  as  po.ssible  con.sistor.  vith  economy,  strength,  and 
stiffness.  Its  purpose  is  to  carry  a  liistributeil  load  which,  as  in 
roof-trusses,  is  iussumed  to  be  concentrated  at  the  joints,  or  panel- 
pointf,  of  the  upper  and  lower  chords. 

(a)  Fig.  <)7  is  evidently  the  stress  diagram  for  the  bridge-truss 
represented  by  Fig.  95,  1234  being  the  line  of  loads,  in  which 

12  =  ir,.    23  =  ]r2,    34  =  11-3, 

the  loads  concentrated  at  the  panel-points  FKG  respectively. 


Fi(i.  90 


P'iG.  97 


Drawing  OX  parallel  to  the  closing  line  MX  of  the  funicular 
polygon,  then  A'l  and  A'4  are  the  vertical  reactions  at  B  and  C 
resp  '•tively. 

-1..  J  by  the  method  of  moments  it  follows  at  once  that 


and 


Ai  =  jir,  +  .ui\.^iir3 
A4  =  iir,  +  ur,^jir3. 


This  truss  inverted  is  sometimes  used  for  bridge  purpo.ses,  and 
may  be  constructed  entirely  of  timber.  The  stres.scs  remain  the 
same  in  iniiir!iitude.  l)iit  are,  of  course,  reversed  in  clinnictcr. 

(h)  V\ii.  'tS  represents  a  through-})ridge  tru.ss  of  the  A\'arren  tjlJC, 
and  is  coinposcd  of  a  number  of  equilateral  triangles. 

Let  1 2345C  Ik?  the  line  of  loads,  in  wliieh 

12  =  U'l,     23  =  Wj,     3 1  -  ir...     i:,  ^  :|-,.     an,  I  -,G  =  ]\\. 


BRIDGE-TRUSSES. 


47 


With  any  pole  0  describe  the  funicular  polygon  and  draw  OX 
])ar;ill('l  to  the  closing  line  QR.    Then  A'l  and  A'6  are  the  vertical 
reactions  at  S  and  T  respectively. 
Fio.  98. 


Fio.  99.  Fio.  100. 

By  the  method  of  moments 

If  the  truss  is  inverted,  the  loads  are  carried  on  the  upper  chord 
and  the  bridge  is  one  of  the  deck  type.  The  stresses  in  the  several 
incnihers  remain  the  same  in  magnitude  but  are  opposite  in  kind. 

iV)  The  Howe  truss,  Fig.  103 ,  is  suitable  for  bridges  of  the  through 
typi',  has  been  widely  used,  and  may  be  constructed  of  timber,  of 
in  III.  or  of  timber  and  iron  combined. 

I'luler  a  uniformly  distributed  load  composed  of  W  concentrated 
Mt  t  acli  panel-point,  Fig.  103  is  the  stres-s  diagram,  123  ...  8  being 
ilic  line  of  loads.  Also  A'l  =3^11' =A8,  OX  lieing  drawn  parallel  to 
till'  dosing  line  of  the  funicular  polygon.  Fig.  102. 

If  the  l<)ad  Is  vvevenly  distributed  the  stresses  in  certain  of  the 
tncmlMTs  I  ly  Im?  reversed.  For  example,  let  the  truss  carry  a  single 
liii.l  coiKcntratcd  at  the  panel-point  M,  FiS-  101. 

Take  any  pole  0  and  draw  the  funicular  jwlygon,  Fig.  104 
'',V  is  now  drawn  parallel  to  the  closing  line, 

A'l  =  reaction  at  left     support  =fP, 
find  A'8=      "         "  right      "       -iP. 


If 


48 


THEORY  OF  STRUCTURES. 


These  results  may  also  be  verified  by  the  method  of  moments.    The 
stress  diagram  is  Fig.  105,  and  it  is  evident  that  the  stresses  in  the 


Fio.  103. 


Fig.   104. 


Fig.  105. 


verticals  and  ties  on  the  right  of  M  are  reversed  in  kind.  Such  a 
reversal  may  be  provided  for  by  giving  sufficient  sectional  area  to 
the  members  thus  strained  or  by  the  introduction  of  counter-braces 
which  are  indicated  by  the  dotted  lines  in  Fig.  101. 

Under  a  live  load,  as  when  a  train  pa.sses  over  a  bridge,  counter- 
braces  may  be  introduced  on  each  side  of  the  centre,  and  although 
they  nia>-  not  be  required  in  every  panel,  they  give  increased  stiffness 
to  the  truss. 

(il)  The  Pratt  dock  truss.  Fig.  106,  is  merely  an  inverted  Howe 
truss,  and  the  stresses  under  the  same  uniformly  distributed  load 
are  the  same  in  magnitude  Lat  are  reversed  in  kind.     Thus  in  the 


Howe  truss  the  upper  chord   is  in    compression  and  the  lower  in 
tension,  the  verticals  are  ties  and  the  sloping  members  struts.     In 


bridoe-thusses. 


49 


tlir  Tnitt  dock  truss  the  upper  chord  is  in  conir->'«  -n  and  the  lower 
in  tension,  the  verticals  are  struts  and  the  sloping  members  ties. 
Tlic  (lotted  lines  are  counter-braces  introduced  to  provide  for  the 
,.ii(ct>  of  a  varying  or  lire  load. 

V„/(..— The  directions  of  the  lines  of  action  of  the  stresses  in  the 
.  \.  ra!  menilxTS  at  a  joint  in  any  framed  structure,  and  hence  also 
ii,(ir  (luiracter,  i.e.,  whether  they  are  ten.sions  or  compressions,  are 
, ;i-ils  determined  by  following  in  order  the  sides  of  the  reciprocals 

dl  the  joints. 

(()  Petit  Truss.— This  type  of  truss  is  suitable  for  a  bridge  of 
long  .sp;in.     Fig.  107  represents  the  truss  for  a  through  bridge  and 


Fio.   107. 


Fm.  in«. 


it<  str(>s.s  diagram  for  a  load  W  concentrated  at  each  panel-jwint  is 
1-ifr.  I  OS. 

in  the  stress  diagram  ABCD  ...  is  the  "line  of  loads"  and 
AB  =  ]V  =  HC=CD^  ... 

Also,  AM  =  vertical  reaction  at  the  left  support 

-our. 

In  drawing  the  reciprocals  of  tlic  joints  /<,</,  r  ...,  it  must  be 
n  niriiiluTcd  tliat  one  hall  "f  the  loads  concentrated  at  the  feet  of 
the  verticals  .">(),  910,  and  V.Ui  are  transinilteil  tln-oiigh  the  mein- 
I'l  IS  1.'),  S'.t,  and  12K?,  resiH'ctively. 


Fio.  109. 


Fi«.  110. 


22.  Bowstring   Tni88.--In  its  simplest  form  the  bowstring  truss 
i-  represented  by  Fig.  109  for  a  through  and  by  Fig.  110  for  a  deck 


fiO 


THEORY  OF  STRUCTURES. 


bridpe,  the   axis  of  the  curved  chord  being  either  the  arc  of  a  circle 
or  of  a  parabola. 

'       '  !  tice  the  fxirtion  of  the  curved  chord  between  consecutive 
j(j...t   .J  usually  straight  (Figs.  Ill,  112),  and,  under  a  uniformly  dis- 

HQFEDCBA 


Fio.   112. 


tributed  load  of  W  concentrated  at  each  panel-point  of  the  horizontal 
chord,  the  stress  diagram  for  one  half  the  truss  is  Fig.  113,  if  the 
joints  lie  in  a  circular  arc,  and  Fig.  114  if  they  lie  in  the  arc  of  a 


Fio.  113. 


Fid    114. 


parabola.  In  the  latter  case  the  diagonals  are  unstrained  and  might 
he  dis|H'nsed  with,  so  that  the  uniformly  distributed  loati  is  trans- 
niitti'd  to  the  bow  through  the  verticals  only. 

In  the  stress  diagram.^  .IW  =  M' =/ir -TV.) -/)£"- load  concen- 
trau'd  at  each  panel  {Htint  and  A'.l  -vertical  reaction  at  a  support 
=-.'}§  If. 

I'ndfT  a  moving  or  /iirl  load  the  stresses  in  web  m*ml)ers  may 
Im-  reversed  in  kind.  Sup|H)se,  for  «'xarnple,  that  a  load  /'  is  concen- 
trated ;it  till' |'iuiel-i>oint  i>.  Then  Fig.  1  Hi  is  the  stress  diagram, 
B("V.\lIi  anil  .\;!I.'».V  In-ing  the  recipriH-als  of  the  joints  p  ami  q. 
Following  thi-  .«iiles  of  Ihese  reei|ir<K';ils  in  order,  the  stri-sses  in  ,i\ 
and  1.')  due  to  /'  ;ire  I'ouihl  to  1m'  ;i  teii.-ion  and  c<(mpn>ssion  resp<'c- 
tively,  while  \uider  the  uniformly  di.-lrihuted  load  the  corres|K)nding 
Btrcssps  are  a  conipre.-..-;ion  aiid  a  i'  iimhh.     If  the  sln's.s's  due  to 


r^^.- 


BRIDGE-TRCSSES. 


51 


I'  exceed  the  latter  in  magnitude,  the  members  34  and  45  must  be 
Uesigned  to  bear  safely  stresses  which  may  be  alternately  tensile 


Fio.  115. 


COSFOH 

Fia.  116. 


and  compressive.  This  can  be  done  either  by  giving  each  member 
in  (piestion  a  sufficient  sectional  area  or  by  introducing  the  counter- 
braces  shown  by  the  dotted  lines  in  Figs.  117,  118. 


Fio. 

117. 

x.,JW-2^x 

x^ 

/ 

/ 
/ 

t 

t 
t 

M^ 

x/y 

1 

1 

i/t 

1 

n/ii 

/ 

%x 

•"■/a     b' 

p  c 

0 

t 

^n 

a     H 

Fio.  11«. 


CDtFaHUMII  •     T      t 


Fio.  119. 


The  stress  dinnram  for  the  latter  case,  with  the  load  P  concen- 
lr;it<(l  at  />.  is  Fig.  1  n>. 

It  will  of  course  l»c  noted  that  the  truiw  represented  by  Fig.  US 
is  Munlv  the  truss  Fig.  117  inverted.  Inder  tiie  same  l«.ml  the 
^tr.s,>..s  ill  c(»rres|Kmding  memln'rs  are  of  the  same  magnitude,  but 
arr  ris<rs<Ml  in  kind.  Thus  the  curve<l  chord  is  in  compression  in 
1  i>:    117  and  in  tension  in  Fig.  11«. 


77/ /.y  ».'.•  1"  OF  S  TU I  C  rUUKS 


Tlip  truss  Fip.   1-0  is  n  hnwstriii}:  truss  with    isoscolos    bracing, 
the  upixT  chord  lH'iii>:  straif^ht  iM'twcfii  coiiscciitivc  joints. 


► 


Kii.   \:i. 


Vu:     122. 


rii.l.r  any  ;:ivcn  loa.ls  .1/^  li(\  CI),  />;;.  HF.  FC.  an.l  CU,  oon- 
cciitral'i!  at  thf  |iaiu  l-iniinis,  V\\i.  !_*_',  is  the  .-Irrss  tliaj?rani,  AM 
ami  XII  iM'iiift  the  n-actimi-i  at  thr  sii|i|Mirls.  Tlic  tncnilnTs  12  and 
ll.l"i  an'  always  in  tiiisiun  l)Mt.  undtr  a  lixc  load,  every  other  woli 
iiietnlMT  may  lie  placid  in  letision  nr  in  eonip'—ion  and  must  thcrc- 
i'nre  lie  de>ij;ned  fnr  a  rexcrsal  of  stress. 

23.  Piers.  V\)i.  \S-\  nprisciiis  unc  df  the  liraced  piers  for  su{v 
portinj:  a  deck  hridne. 

Ileijtht  of  jiier    .Vl  It. :  ot"  trn-s    ;{(lfi. 

Width  of  pier  at  ti.p     17  It  ;   at  l.ottoni     X\\  ft. 

The  liridjie  when  most  hea\  ily  Inatled  throws  a  weight  of  KM)  tons 
on  each  of  the  ]Hiints  .1  ;iii,|  /{. 

Weiijlit  of  half  jiier     ,'!(>  ti>n«. 

The  increased  weiuhi  ;ii  lach  of  the  |>oints  (\  I)  and  K.  F  from 
the  |Mirti<ins  Ah  and  (F  if  the  pier     .'» tons. 

Hesiijiiuit  !iuri2oii(ul  wind  preK-iure  <>n  triiin-40  ions  at  H7J  (pft 


Ill 


pii::rs. 


53 


r.csiiltant  horizontal  wind  i)ro.ssure  on  truss  =20  tons  at  65  feet 

;llili\('  l)aso. 

Ki'sultaiit  horizontal  wind  pressure  on  pier=2i  tons  at  each  of 
thr  points  f  and  K. 

With  the  wind  jm-ssure  acting  as  in  the  figure,  the  diagonals 
CH,  El),  and  GF  are  reciuired.     When  the  wind  l)!<)ws  on  the  other 


Fio    123.  Fio.  J24. 

side,  the  di;ig(inals  I)  to  A,  F  to  C,  and  //  to  F  are  brought  into 
play.    The  moment  of  the  eouple  tending  to  overturn  the  pier 
-  40  X  H71  + 1»()  X  li.*)  +  4  X  25  =  4«KK)  ton-feet . 

[)\r  moment  of  stability  -  (2(M» +  ;«))  X  .,-  -3S71i  ft.-tons. 

I'liiH  the  dilTen-nee,  -  |<KM)-:<S71i{  -  I02S.\  ft. -tons,  must  Ix*  pro- 
viI'mI  tor  in  the  anehorage.     The  pull  on  a  vertical  anchorage-tie  at 

\il'.\\u,  if  //  lie  the  Ijorizontal  force  U|k)P  .he  pier  at  A  due  to 

■Ailid  pressun', 

//  X  .iO  "  40  X  .H7  i  +  20  X  (M  -  4S00 ; 

II -mums. 


;«l,Vi  tons. 


54 


THEORY  OF  STUUCTCRE.S 


Fig.  124    is  the   stross  diagram,  12345(5   hoiiig   the   load   lines, 
in  which  12  =  2  tons  =23;   34=%  tons;   45  =  KM)  tons -56. 
Project  126  upon  the  vortical  /6.      Tlicn 


and 


/G  =  vertical  pressure  at  //, 
12,/=()ut\vard  horizontal  thrust  at  //. 


Project  1,11,12  upon  the  vertical  64/,  and  upon  the  horizontal  lo4 
Then 

/4  =  uplifting  force  at  (i, 
and  I3- 14-/,  12=  horizontal  pull  at  G. 

In  computing  the  stresses  in  '  leeward  ix)sts  of  a  braced  pier 
it  is  common  practice  to  jissume  that  the  maximum  load  is  upon 
the  bridge  and  tliat  the  wind  exerts  a  pressure  of  'M  lbs.  fjer  square 
fiHjt  upon  the  surfaces  of  the  train  and  structure,  or  a  pressure  of  50 
lbs.  jier  s(|uare  f(K)t  u})on  the  surface  of  the  structure  when  unloaded. 
The  negative  for  n-versed;  stresses  in  the  windward  jx)sts  of  the  pier 
are  determined  when  the  mininmm  load  is  on  the  pier,  the  wind 
pressure  remaining  the  same. 

24.  Fink  Truss.— In  tiie  tru.><s  represente.l  in  the  accompanying 
figure  the  length  of  the  iK-am  AH  is  so  great  that  the  .single  triangu- 
lar truss  U'li  with  a  single  central  .strut  CO  is  an  insuniei.'iit  supiwt. 
Th.>  two  halves  are  therefor.'  strengthened  by  the  .simple  triangular 
trasses  ACO  with  a  central  strut  <iF  and  lim  with  a  central  strut  /*.V. 

.\gain.  .-aeli  (|tiarter  l.-ngth.  viz..  AF.  FO,  ().\,  .\7i,  is  similarly 
tru.<sed.    The  .subdivisions  may,  if  necessary,  be  carried  still  farther. 

D  F<;h  n  I  U  n  g 


Fuj.   12.') 


This  truss  in  fn„r.  ei,,ht.  slxlrrn.  .  .  .  .livisions  or  poneh  is  kno«-n  as 
the  Kmk  truss,  and  has  In-en  widely  employed  in  .\merica,  the  numlxr 
of  panels  usually  Iwing  eight  or  sixteen. 

The  Mienibers  sli.,vvn  by  the  dotted  lines  mav  \h-  inlnnluced  for 
stitTness.  and  the*  pli.tform  may  Ik.  eitli.-r  at  the  top  „r  inittom  'fV 
weight  dir,H!!y  iH.nu.  !>y  ■*  ^tnit  i.  u^iaiiv  det.  ,  ii.ined  fro.n  the 
loads  uiKm  the  two  adjacent  i>an..|s  by  a.ssuming  the  corres|M.nding 


BOLLMAX  TRUSS. 


55 


portions  of  the  beam  to  be  independent  beams  supported  at  the  ends. 
riiiis  if  there  be  a  weight  11'  at  the  point  S  in  the  panel  FH,  the  portion 

CI/ 

„r  ir  borne  by  the  strut  OF  at  F  is  11'  ^,  and  the  portion  borne  by 

FS 
llio  strut  A7/at  H  is  11  ^jj. 

Let  Wi,  ir,,  H'a,  H'*,  H's,  U'e,  H't  bo  the  weights  upon  the  struts 
(or  posts)  I)E,  F(l,  HK,  OC,  LM,  AT,  QR,  respectively; 

L<>t  1\,  I'i,  Pz,  I\,  J  &,  Pf.,  ^7  be  the  compressions  to  which 
tlicsi'  jKJsts  are  severally  subjected; 

Let  a.  ,9,  y  be  the  inclinations  to  the  vertical  of  AE,  AG,  AC, 

rt'six'ctively; 

Let  T\,  Ti,  Ta,  .  .  .  l>e  the  tensions  in  the  ties,  as  in  Fig.  125. 
Tlu'  tensions  in  the  ties  meeting  at  the  foot  of  a  post  are  evi- 
dently ('(lual. 

\M-h  triangular  truss  may  l)e  considered  separately. 

Friin  the  truss  AEF,     JTi  cos  o  «  Pi  ■»  Tl'i ; 

from  the  t ru-ss  AdO.     'IT 2  cos  ^  =  /',  =  U'a  +  (^i  +  Ta)  cos  a ; 

from  the  truss  FKO,    iTi  cas  a~l\  =  \W ; 

from  the  trass  ACB, 

2Tt  cos  r  =  /%  -  "'4  +  {T-i  +  Tn)  cos  ;5  +  {T3  +  Tj)  cos  a ; 

from  the  tniss  QMS,    '2Ti  cos  a  -  Ps  =  "'si 

from  the  trass  OPR,      27,,  cos  ,9  =  /'«  -  ir«  +  (Ti  +  Ti)  cos  a; 

from  the  truss  \RB,     2T7  cos  u-Pt  Wi- 


Hence 


T3  -  Y  "^c  a, 


T..l{»'..''^)-.',    r,-"-'*... 


<<  u 


56  THr.ORY  OF  STRICTURES 

Again,  the   thrust  aloiij;  .1/-'     r,  sin  a-  T.sin  3  +  7^  sin  r; 
''        "      :>l  ^'  r.  sin..? +  7^4  sin  ^; 

:il()i,;r  /.7;  .^  Ti  sin  ..?+  7^4  sin  y  +  T^  sin  a; 
at  O  -  Ti  sin  r; 
etc..  etc. 
If  the  truss  carries  a  UMil'ormiy  (iistrilnitod  loatl  TT'', 

''i  =  T's  =  Ts  =  ^7  =  —  s('c  a, 
lb 


If 


ir 


7^2  =  To  =  ^  st'c  ,?,     ^4  =  —  sec  y. 

If  thp  ahovo  diagram  is  inverted,  it  will  represent  another  type 
of  truss  in  whirli  the  obliques  are  struts  and  the  verticals  ties. 

25.  Bollman  Truss. -Fig.  12()  repn-s^-nts  a  In-ani  trus.sed  by  a 
number  of  iiulrix-iuteiit  triangular  trus.s(>s.  Ww  vertical  jHists  boing 
equidistant.    The  weight  concentrated  at  the  head  of  each  post 

B 


Fk;    IL'fi. 

may  h-  found  by  the  method  .l<'s,.nb.-d  in  Kx.  2.  which  in  fact  is 
genr-rally  a|.plicable  to  ail  bridge-  and  roof-tru.s.ses. 

L't  Ti,  7\.  1m'  the  tensions  in  AH,  HE,  res|)ectivelv 
"     II',   Im-  the  weight   iit    l>\ 

"    o„  <«..  Im-  the  inclinations  of  .1/^,  /y//    res,)ective!v,  to  the 
vertical. 

Then       7',    ir, .  '"'"-       r.  =  n-.    "'""' 

sin(a,w,,/     "     '".sin(«r+„7)- 
Similarlv,  the  stress  in  any  oth.'r  tie  niav  1h'  obtained 

Tt'-  <oinpr.  »inn  j,,  th.'  lop  ehord  .s  the  algebraic  sum  of  the 
h.,nzont;.l  romponents  of  all  th.'  .stres.s,.s  in  the  ties  which  meet 
at  one  end. 

The  verticals  are  always  struts  an.l  the  obli<iues  tiea. 


METHOD  OF  SECTIONS. 


57 


This  truss  has  been  usetl  for  bridges  of  considerable  span,  but  the 
tics  may  prove  inconveniently  long. 

26.  Method  of  Sections. — It  often  happens  that  the  stresses  in 
till'  iiicinbers  of  a  frame  may  he  easily  obtained  by  the  method  of 
sections.    This  method  depends  upon  the  following  principle: 

If  a  frame  is  dividetl  by  a  i)lane  section  into  two  parts,  and  if 
fiK  li  part  is  conSiilered  separately,  the  stresses  in  the  bars  (or  mem- 
Imts)  intersected  by  the  secant  plane  must  balance  the  external  forces 
uiMiii  the  part  in  question. 

Hence  the  algebraic  sums  of  the  horizontal  components,  i'(A'), 
(if  the  vertical  components,  l'(Y),  and  of  the  moments  of  the  forces 
with  resi^ect  to  any  point,  I{M),  are  severally  zero;  i.e.,  analytically, 

J(A')=0,     ^(r)=0,    and    I{M) -0. 

These  equations  are  solvable,  and  the  stresses  therefore  determinate, 
if  the  secant  plane  does  not  cut  more  than  three  members. 

Example  1.  ABC  is  a  roof-truss  of  60  ft.  span  and  30°  pitch.  The  strut 
1)1  -ail -h  a.;    the  angle  F/^/l-90^     Mao  AF-FB-AG~GC. 

The  vertical  reaction  at  B-5  tons.  The  weight  concentrated  at  D-4J 
tons. 

I^t  the  angle  ABF-a. 

Then  ilB  -  30  see  30°  -  20v'3,    cot  a  -  ^^—  -  2\/3. 

o 


■ma- 


VVS 


2v/3 


If  thp  portion  of  the  truss  on  the  right  of  a  secant  plane  M\  lio  removed, 
til"  forces  C",  T,.  T„  in  the  members  AD,  AF,  FC  must  balance  the  external 


W^NItw- 


FiQ    127. 


Fio.  128. 


Forres  T)  tons  and  4}  tons  in  order  that  the  equilibrium  of  the  remainder  of 
the  truss  may  be  preaer\'ed. 


<»  THEORY  OF  STRUCTURES. 

Hence,  resolving  horizontally  and  vertically, 

T,+  T,  cos  (a+30°)  -C  sin  GO'-O, 
r,  sin  (o+30°) -C  cos  60=+5-4i -0. 
Taking  moments  about  F, 

C  5  -  5BF  cos  (30^  -  n)+ 4iDF  sin  30°  -0. 

co8(a+30=) ^^.     8in(o+30°)— ?^,    COS  (30' -  a) -—^. 

lVl3  2v/l3  2^13 


But 


BF-BDseca-.ViS,    and    DF~ 5  ft. 


Therefore 


a 


V3 


2\/l3 

Co— 5x5N  13    — ,_  +  li  -.Jxi^O. 
2v'l3 

Hence  C  -  ISJ  tons,  7",  - O.Sft  tons,  and  7*,  - 6.3.')  tons. 

Kx.  2.     Fijj.  128  re[)rps«-nt8  ii  portion  of  a  bridge-truss  cut  off  by  a  piano 
•V.V  ami  supixirtcd  at  the  abutment  at  .1. 

Till-  vertical  ri-aetion  at  .1  =4(»<(,4«()  lbs. 

The  weight  at  «-41t,."KK)  lbs. 
"        "  t=3S,7(K)llw. 

.»«-/«  =24  ft.;   /i/>-24ft.;   CE-29Jft. 

The  forces  (",  I)',  T  in  the  memlK-rs  met  by  MN  must  balance  the  ex- 
ternal forces  at  .1,  Ii,  <'. 

Uesolving  horizontally  and  vertically, 

7'4  />(■<«  .i-r'e(«,? -0, 

/>'  sin  '>  4  r-  nil,  ?  -4(m4(KI*4!t.J(K)+ 38700-0, 

a  and  ,?  l)eing  th<'  iaclinations  to  the  horizon  of  KF,  I)E,  respectively. 
Taking  moments  ulx)ut  K, 


Dat 


-T  y  2<»l  +  400400  K  4.S  -  49.'i00  x  24  -0. 

tan  <»-.,.-—     and     fan,?--?-    . 
24       '.)  '      24     (> 


11  9 

Sm  a--  ,      -       COS  n 

\  202  v/-''»2 


/>...,'     "'"'''": /S-     •o"^- 


>/85' 


9 


Hence 


THREE-HINGED  BRACED  ARCH. 
7'-629,427i'ilb8., 

9814500   /—     1090500 


09 


C- 


D'- 


117 

1994600 
13 


V85-- 


13 


-\/85  lbs., 


\/202  lb8. 


The  results  in  the  two  preceding  examples  can  also  be  easily  verified  by 
drawing  the  stress  diagrams  of  the  portions  of  the  trusses  under  consideration. 

27.  Three-hinged  Braced  Arch. — For  station  roofs  and  for  sheds 
of  wider  spans  than  those  for  which  simple  trusses  are  found  to  be 


Fio.   130. 

oronomical,  it  is  a  common  practice  to  employ  two  braced  trus.ses,  as, 
•• «.,  Figs.  129,  IW,  connected  by  a  central  hinge  and  hinged  also 
:it  the  two  abutments.  The  use  of  these  hinges  makes  the  stresses 
ili'tcrininate  and  the  stress  diagram  for  any  given  loading  may  !« 
ilrawn  as  soon  as  the  resultant  ri'actions  at  the  hinges  are  known, 
riicy  can  !>«>  easily  calculated  by  the  "  method  of  sections."  Fig.  130 
itpresents  one  of  the  three-hinged  braced  arches  which  supjjort  the 
riiof  (if  ii  large  hall. 

A  load  /'  at  any  iK)int  /»  of  the  left  truss  developes  a  reaction  at  b, 
ami  the  direction  of  this  reaction  must  necessarily  pass  through  the 
centre  hinge  C.  If  the  direction  falls  above  or  below  C,  the  hinge  is 
Miltjccted  to  a  Ijending  moment  and  rotation  or  defonnation  can 
only  Im!  pn'vented  by  the  inherent  stifTness  of  the  trusses.  \jci  the 
lines  of  action  of  /*  and  the  reaction  along  BC  meet  in  the  point  r. 


60 


TIIWRY  OF  STia'CTURES. 


Thoii,  for  the  oquilihriun.  of  the  loft  truss,  the  resultant  reaction  at 
.1  must  also  pass  throuftli  the  point  r. 

So,  again,  a  loa(U,>  at  any  p..int  7  on  the  right  truss  developesa 


Fiu.  130. 


reaetion  along  AC.  an.l  if  the  ,lin>etion  of  the  reaction  meets  the  line 
of  action  of  (J  in  the  point  .v,  then  lis  is  the  direction  of  the  resultant 
reaction  at  li  due  to  (J. 


Via.  131. 

At  r  resolve  the  reaction  along  BC  due  to  P  into  its  horizontal 
and  vertical  components  //  and  V. 

At  r  re.M,Iv.-  the  r.-action  along  AC  due  to  Q  into  its  horizontal  and 
vertical  coni|Mjnents  //'  and  I''. 


TUREE-HISGED  BRIIKIF-TRUSSES. 


61 


Tlion,  ovulontly  //=//'    ami     V  +  T'  =0. 

Coiisiilcr  the  e<|uilibriuin  of  each  truss  soparatoly. 

Ill  rijr.  1.31,  considering  the  loft  truss,  tho  hingo  C  is  acted  upon 
;i\  I  he  horizontal  reaction  //  and  the  iij)\vard  vertical  force  (cr  shear) 
1  .  1  ac  n-sultant  reaction  at  -I  may  he  also  resolved  into  its  vertical 
.1  'nui/.ontai  coniixtnents,  viz.,  I'l  and  //i. 

As  //  and  II\  are  the  only  two  horizontal  forces  acting  upon  tiie 
i:',:s>  ill  this  case,  they  are  cfjual  in  magnitude,  hut  act  in  opposite 
ilinctioiis,  and  the  //,(=//)  at  .1  may  he  either  taken  up  by  the 
aliutiiicnts  or  by  a  tie-rod  connecting  .1  and  3,  which  may  often  be 
I  iiiixciiiciitly  placed  Ih'Iow  the  floor. 

Take  moments  about  .1 .     Then , 

H  X  53.99+  I  •  X  59.33  =  P  X  9.33 
or  //  +  rx  1.099  =  PX. 173. 

Similarly,  by  taking  moments  about  B  for  the  right  tnias, 

H  X  53.99  -  r  X  59.33  =  Q  X  27.67 
or  //-rx  1.099  =(?X. 513. 

Hence  ^  =  Px.087  +  yx.256 

ami  T'=Px  08-QX.24. 

Also,  Vi  +  V^P    and    V^-V -Q, 

I  ..  licing  the  vertical  comjH  iient  of  the  reaction  at  B. 

Therefore  V i  •=  P  X  .9236  +  (^  X  .229 

ami  I'a  =  Px. 0764 +(^X. 771. 

The  principle  of  the  three  hinges  is  also  sometimes,  but  less 
l'iv(|ui'iilly  adopted  in  bridijv  irork  (Fig.  132).  The  two  halves  are 
I  MiiiKcted  by  a  central  hinge  and  are  al.so  hinged  at  the  abutments. 
A  -<rious  <i|)jection  to  this  ty|x«  of  bridge  is  the  large  deflection  due  to 
'!!'!>!:!'<  of  teniiMTufure.  Th.e  eh^iiun-ifinn  of  the  hinge  at  <'  gnnitlv 
Iniiiiiishes  the  deflection  and  stifT«'ns  the  structure,  but  at  the  same 
tiiiir  develojx's  temiKTature  .stre«i<es  in  the  several  members. 


02 


THEORY  OF  STRUCTCRES. 


-M 


The  reactions  at  the  three  hinges  are  deterniined  in  precisely  the 
same  manner  as  in  the  preceding  ca-se. 

A  load  P  on  the  left  truss  necessarily  produces  a  reaction  along 
the  sloping  chord  CB  and  if  the  line  of  action  of  P  meets  BC  pro- 


liQ.  132. 

duced  in  s,  As  is  the  direction  of  the  resultant  reaction  at  A  due  to  P 
So  Bt  IS  the  direction  of  the  resultant  reaction  at  B  due  to  a  load  Q 
on  the  right  truss. 

If  the  resultant  reaction  passes  through  a  panel-point,  e.g.  a,  and 
mtersects  .4^/  in  V,  a  weight  placed  on  the  horizontal  chord  at  x 
produces  no  stress  in  the  member  Ym.  This  can  be  easily  proved  by 
consKlermg  the  equilibrium,  under  the  weight,  of  the  portion  of  the 


l-M( 


Fio.  133. 


Fio.  134. 


truss  on  the  nght  of  a  vertical  .««^ction  MX.  The  moments  of  the 
stresses  m  hn  an.l  «m.  and  of  the  resultant  reaction  at  B.  are  all  nil, 
as  he,r  h,.es  of  ac,i„„  pa.ss  through  the  ,K,i„t  9.  Hence  the  momeni 
of  the  stress  m  }  ,„  an«l  th.-refore  the  stress  it.self  is  aho  nil 

Let  iMgs.  l.T?  i:i4  repre«.nt  the  two  halves  of  a  3-hinged  bridge 
of  W   t^  span,  2  ft.  de,,,  at  the  centre  an.!  10  ft.  deep  at  the  end.sf 

Let  loads  of  4.  4.  2.  2,  4  tons  Ik,  conc,.ntrate<l  at  the  panel-pointa 
A,  4.  .»,  0.  and  10,  res|)ectively; 

Lot  //  !«.  l!„.  !if.rizf>n{al  naction  at  omh  hinge; 

I^H  r,,  r,  1k<  the  v.Ttical  reactions  at  .4  and' fl  respectively; 


THREE-niSGED  BRACED  TRUSSES. 


6a 


\jet  ±  T'  be  the  vertical  shear  at  the  central  hinge. 
Consider  the  equilibrium  of  each  half  separately. 


Fio.  135.  Fio.  138. 

Taking  moments  about  A, 

-V-32-i/  8  +  4(16+24)  +  2x32-0, 
or  H  +  ^V-28. 

Taking  moments  about  B, 

-T  -32  +  ^.8-4x8-0, 
or  H-4F-4. 

Therefore  H-16    and    V -Z. 

Also,  Vi-lO-V-Ttons    and    Va-6  +  V-9  tons. 

Then  Figs.  135,  136  are  the  stress  diagram  of  Figs.  133,  134,  re- 
sjM'Ctively. 

If  the  throe-hinged  braced  arch,  Fig.  132,  is  inverte<l,  the  bridge 
iK'fomps  one  of  the  suspension  type  and  the  load  is  carrie<l  on  the 
Ifiwcr  chord.  The  stres-ses  under  a  given  load  remain  the  same  in 
niagnitudo  but  are  reversed  in  kind. 

If  the  sloping  chord,  inst'-ad  of  beinp  straight,  has  its  joints 
lyiiiw  on  the  arc  of  a  parabola  with  its  vertex  at  the  central  hinge,  it 
(■■in  easily  be  shown,  either  graphically  or  analytically,  that,  under  a 
unifdrinly  distributed  load,  mo  stress  is  developed  in  the  horizontal 
('!'.' iril^. 

Although  it  is  the  usual  practice  to  connect  together  the  sloping 
chonls  by  means  of  the  central  hinge,  it  is  really  more  rational  to 


&4 


TiiEnnY  OF  sTurcTrh'Fs 


cuMiiict  the  liurixuiii.il  clionl^  lop'thcr.  ami  the  truss  thon  requires 
lilt  (U'pili  at  tlic  (•(■iitic. 


TmiI.K    <>K    W  til, Mix    UK    l{iMil--<l)VKl(l\iiS, 


Wrislil  .,f 

ill    Ito      |,«T 

*   ".MTI 

.\ri:i 

.1 

2  :.  1. 

.{ 

i.-i 

SK.  1 

J.-. 

IN'MTII-Il-.H  I  t   I   -iMTIIlK. 


15ii;ii(|liii;  I  J-iiich 

lio.'iiiliiit:  .'iMil  -liiii  iniii.  .  .  . 
('.•1>I  -irnli  |.'.ilr>  '  ,1-iliclll.   .     . 

(  'li|l|  IT   ,  

<  '(iiiiii::iliil  iron I  In  :<  7.'i 

I  I'll.  :i>lili:illri|    1 

IVIl  .iihI  nr.iM-I s  ti>  Id 

( i;ih  .iiii/ril  iritii    I  In  3  .") 

I. mil  aiiil  |>la.-.|i'r {Mo  III 

l';iniili>      Ill 

SIhtI  Ir.id .'i  111  S 

."*lii'<l  /inr                 \    '.'."i  III  1' 


Ilrail  Wi'iglii  •<!  liiHif  in  Iba.  p 
(1.  ..f  C'ii\<Te>l  Area. 


Mu'il  Irmi    ri>rriii::iUili. 


:<  I 


iV,  incli  lliirk  '    . 
"       If.  W  li  .;.i,.|l:,lh> 
ShliiL'li"'  ■  It'i-ilirlii 

ii'iii:' 
>lii:illiiiiii      I'itp  .    I  null    yillow. 
Nun  Ill-Ill  i 

>lii'iilliiiii;      I'lM  .    l-iiiili   viiliiw,; 

>iii|lllrnp  I 

Slir.iililin;   I  1  iiuli  I  111 -null    .mil 

lll.'tlilr'     .  i 

Slicjiililiii.'  'l-iiiili   ;i>li,  liii'kiin-,  i 

•  ink  i 

>lii',illiiiii!      I'riii-.-,  I  null  thick  i 

."•lali's  iiinlliiarv  > 1 


l.irurl 


H  without  lioanis  nnd  11  with  IxMniii 

fur  >|iali»  ii]!  to  7."i  ft 
J    I       •  Wi  wiihiiiii  lioaiils  and  l.'i  uitji  iHianl* 
fur  ^|•all^  Irmii  7")  lo  i.'M)  ft. 


a 

."» 
■» 

:< 
;i 
t 
I 

•I 

.")  tn  '.I 
!•  to  II 


10  on  liithM  for  )>|):tn)i  up  to  7.">  ft. 

1 1  1111  iath.s  fll^^|>alls  fmiii  7.'>  to  I.V)  ft 


>lali-i  .'iiiij  ii^iii  I.mIw 

.•"Iiiiiiu'li'*,  iijiii-    

.""Iicii  li'.dl 

Thaiiii ;.'■' 

Tilf>.  tl.n 

irriHiM-^.  ami  lillcls.  ... 
Till-  .iiiil  iimri  ir 
TiiiiliiriiiL'  "I  nil  .1  aiiii  .-Ian-  umiI. 

■ulililiiiiiab 

Tin 

Ziiir 


I.t  without  honnln  or  on  Imhi*  nnci  Id 

nil  I  1  ill    l>iiiiri|.'<  for  K|uiiiN  up  to 

7ri  It 
17  "itiiiiiii  lionrd.t  or  on  Jailin  niul  3) 

I'll   I  I -ill    liourda  fur  i>|iunit  from 

7.">  to  I.jO  ft. 


Ill 

•> 

.1  to 

H 

tl  .' 

l.-.to 

■.11 

7  111 

III 

■Jit  tn 

:«i 

"i   .'i  to 

•  i    "l 

7  to  1 

-'•■•l 

1  to 

:.' 

Miiow:  I^Mo'J.MIm   |N'r  linri/iiiital  >i|ii,irf  fiKA, 
Wind:  ;WI  to  .'ill    "       ••    MMKal 'i|iian'  foot. 


r.l  RLES. 


65 


Ilii'  Camogie  Stool  Company  gives  the  following  table  of  approxi- 
iiKit.  l(i;iils  (inciiuling  woiglit  of  truss)  jXT  sciuuru  foot  for  roofs  and 
.-j>:in>  uinli'r  75  ft.: 

l!<M(f  I'ovrn'd  with  oirruftatniJ  HhcetM,  unhnanied.  ...  8  Ihs. 

••       on  hoards 11 

"  "  "     tilafo,  on  latlw 

.S'lmo,  on  Ixiards  1 1  inch  thick 1« 

HiMif  covered  wilh  .-ihinKle!*,  on  lalhs K) 

Add  to  alxive,  if  |iliu*tered  U-low  riiftem It) 

i^niiw,  liKhl,  wi'ijjhs  [wr  cul>ic  fiM)t ."i  to  12 

For*|>iins  over  7.">  ft.  add  4  11>».  tt>  the  aliove  loads  per 

.s(|iiiire  f(K>t. 
It  is  eiiNtoiiiar>-  to  add  ',V)  11>h.  |xt  s<|iiarp  fiMit  to  the 

aliove  for  .tnow  and  wind  when  M-parale  cuicu- 

lat  ioii.s  are  not  niatie. 


Tm.I  K  op  TIIF.  VaUES  ok  /'■.  I'r,  Ph.  IV  PofXDH  TEll  SylAHB  FoOT  OF    SrHFACl 
MllKN-   /*~10,   AH  DeTKIIMINEI)   ItV  THE   FlIIIMI'i.A  I'n  -  P  aUi  <il.lM  Con  o  - 1. 


Ktch  or  HiH,t. 

f-n 

A", 

/•* 

r," 

r,  (» 

to 

0  4 

111" 

»  7 

!l  tl 

17 

211° 

IH    1 

17  0 

!'.  2 

mt" 

2».  » 

22  H 

l.t  2 

It)" 

:w  3 

2o  .5 

21.  t 

.Id" 

:w  1 

21  .'-. 

2W  2 

m" 

to  (1 

2tl  0 

3t  1) 

7(1" 

ti  1) 

ltd 

.■w  r, 

H(»° 

to  1 

7  () 

.'fit  H 

•Ml" 

•too 

0  1) 

to  II 

Taiii.k  riiKi-Aiirn  .fhkm  tiif  Kiiiimii.a  /»■ 


— - 

— - 

WliH'itHf  ill 

\>l.«  iiir«  in 

IVfMin-F  in 

frrt  |i««r  M>t>in<l. 

iitilm  |HT  liHiir. 

IIh..  iirr  ».|   (>. 

1(1 

I>  S 

0  2.5 

20 

l.<  li 

1  00 

to 

27  2 

4  00 

till 

to   N 

0  IN) 

70 

17  t» 

II  2.5 

NO 

M   1 

1)1  00 

mi 

•il  2 

20.2/1 

too 

tM  l> 

25  00 

110 

71  H 

:«i  2.5 

ILII 

Ml    li 

31)  00 

i:to 

KN    t 

•12  25 

l.'iO 

102  0 

.50  25 

& 


afti 


THEORY  OF  STRUCTURES. 


EXAMPLES. 


I.  A  numl)or  of  roplnnar  foncs  liavc  to  lx>  in  oquilihrium.  The  lines  of 
action  anil  inaniiiluilcs  of  all  thr  forces  with  the  txccption  of  twouro  known; 
and  of  these  two,  the  line  of  aition  of  one  of  thern  and  a  (Kiiiit  on  the  line  of 
action  of  the  other  are  also  known.  Show  how  you  would  |rruphicallv  de- 
terinino  the  two  unknown  forces. 

a.  Show  that  I  lie  locus  of  the  |K>les  of  the  funicular  ixilvRona,  of  which  the 
♦^rst  and  last  sides  pass  through  two  lixetl  |»oiiits  on  the  clooinf;  hne,  is  a 
Rtrai(;iit   line  parallel  to  the  closing  line. 

3.  The  first  and  l.i.it  sides  of  a  funicular  polypon  of  a  sy.stem  of  foreea  in- 
tersect the  closing  lino  in  two  fi\i-d  points.  Show  that  for  any  position  of 
the  jiole  each  side  of  the  iH)Iypon  \mII  pass  throuuh  a  fixed  |H>int  on  the  dos- 
ing  line. 

4.  If  the  |)ole  of  a  funicular  ixilygon  dcncriU'  a  straight  line,  show  that 
the  corn>s[M>ndinp  si.les  of  successive  fuincular  inilypms  with  resjHMt  to  suc- 
ressive  |H.sitions  of  the  |K>le  will  intersect  in  u  straight  :in(>  which  is  parallel 
to  the  IfH-us  of   the  |Kple. 

5  Forces  act  uiM.n  a  rectangli-  in  ihi«  rn.inner  shown  by  Fig.  137.  De- 
termine the  niMgnitudi-.  po.sition,  and  din-ctiun  of  the  resuhant. 

6.  Fig.  13S  represents  a  jointed  frame,  the  lengths  of  the  links  l)eing  M 
indicated.    Tbt   Iranie  is  acted  u|»on  by  six  forces,  of  which  three  an-  given, 


■*4^ 


Ik.    I:i7 


Fi..    i;»H 


ntulthe  lines  (,f  ,,.ti..n  of  all  the  fores  bis,.,t  r<s|«ctively  th<'  cngled  nmrkwl 
by  a  .{..it.d  ar  .  Iind  thi-  f,.rre»  .V  Y.  and  /  md  ilet.ri.iine  the  din-ciion» 
in  which  III.  V  a.  I  SUn  |„m|  the  f.,r.,.  aloiiif  radi  link,  distinguishing  be- 
twcn  -It:  111 .  and   tiiw. 


7    l>ra«  the  fniiiriilar  [«ilvgon  for  the  sit  fur-es  t>f  |,  2  :\,  i,:^  and  «  ton" 
acting  at  an^mlar  di«ian.es  „f  (Mr.  as  >n  1  lu'   f:{'i.      \|..,  deiernime  »rr.,!  I,,.  j.IIv 


EXAMPLES. 


07 


tli.    inomciit  of  ti-i-m-  loads  with  respect  to  any  given  point  on  the  line  of 

at  imii  of  the  6-ton  force. 


.^ 


/It 


^-•JV 


Vik» 


•r 


\tt 


Fio.   139 


^r^ 


Fics.   140 


8.  A  rijrid  frame  CAOR  is  hinged  at  C  and  ro«ti»  upon  rollers  at  A?.  Find 
•1m  mignitudp  and  direction  of  the  rec-dtant  miction  at  (',  and  alao  the  vcr- 
11  il   iKtion  at  H,  due  to  the  horizontal  force  //  at  .1.  Fijf    IM). 

C   Skitch  the  reciprocal  diagram  of  a  frame  congi^tini;  of  four  equal  rods 

Iihi.c!    ..  iiH  til  form  a  parallelogram  with  aiijfles  of  Mf.  and  iiriKiMi  hv  two 

•li:nr»iiMls  .roHsinn  (>a.h  other  without   joint,  lh<'  hkIs  l)einK  pinne<l  at   the 

nil,,  1..  of  the  figure  only.     The  frame  ix  fni-  from  external  forces,  but  in  Much 

•ii-iraint  that  the  ionp-r  diagonal  Ih  under  a  tension  of  2  toiis*. 

I".  Four  bars  of  e<jual  weifht  and  lenjtlh,  freely  articulattnl  at  the  extrem- 
!!.-..  fi.riii  a  s.|uan'  AHCI).  The  system  n-std  in  a  vertiral  plane,  the  joint  .t 
l»inii  lived,  and  the  form  of  the  »«|uan'  ih  prexervcl  hy  nieanx  of  a  horizontal 
*irii,i:  .tmiieciinj;  the  joints  H  and  />      If  11'  It.'  the  >veinht  <.f  each  Iwr,  hIjow 

'hit  the  stress  at  <'  is  horizt>nial  and---,  i^)  that  the  stresn  on  flf  at 
H     J     and  makes  with  the  vertical  an  .in»(le  tan     'J.  l*"'  'hat  thestretw 

f.T, 

'  I!  :i!  H  i:-  11'  ^  and  makes  with  the  vertical  an  anjtb"  t<in  '1.  {</  that 
•'      tf.  -is  uiMMi  Ali  at    t  is  iW.  (-     that  ilie  dnsioti  i>f  the  stnn)t  in  21*'. 

M     live  luir-i  of  equal   IcnKtIi  and   weiftht,   freely  articulated  at   th«'  ex- 
'I'      i'-.  form  a  nuular  |>entaKi.n  MUllF       The  system  rests  in  a  vertie.il 
jii  .     ilie  l.iir  (  h  Ixiiid  li\e<l  III  .1  li,.ri/..iii  ,1  |><>«iii.in.  and  the  form  of  tl< 
[•tiisi.Hi  'eiiiK  preserved  \<\    means  ..f  a  sirinit  runra'ctiiid  the  joints  H  .tlid 
'       Il    the   wemht   of  ea.h  l.ar  1m'  H'  .show  ihat   the  tension  of  th«'  Mtrinii  is 

■>r  ^.'J  tan  iH"),  and  liml  the  majmitudes  and  dinvtionnof  theiitr«Me« 

y  ■'  •    M.,iits. 

>i\  Iwni  of  .v|uhI  length  and  weiijhf  ■  -Mi,  frw-ly  articulated  at  tie 
■    itias,  iorm  a  rcKular  lie\a|fi>n  AlU  1/1:1'. 


68 


TIIEOHY  nf  STUICTIHES 


I'irti,  if  iho  system  hnne  in  n  vertical  plnnc.  ilip  li.ir  Mi  Wne  fl\r»d 
h(iri/c>iitiil  |>i.sitioM,  ami  the  foriii  of  thi-  hcvairoii  lx"iii(r  pri'scrvcil  l.v  nieati^  n\ 


m  ft 


a  sirinti.otiiic.titip  the  nii.lilh'  points  d'  Mi  ati.l  l)K.  show  tliat  uii  th 


of  tlu-  siriiifj  is  .'{ir,    /m  il 


h-  stros  at 


II 


I'  tension 


and  hiiri/otital,  (c)  tlii 


le  stress 


at  />  is  IIX'Tf  ami  tnakc.  with  thr  vrrtiral  an  MnL'lr  i-ot      '  L'v  3. 

Show  thai    till'  siri>s.>s  at  I'  ami  /'  remain  horizontal  when  the  hiirs  AF 
A'/:'.  /<r.  (7>  ;,ro  rcplarcd  l>y   any   olh.Ts   uhirh   ar.' all  iMiiiaily  in.  Iiiic<|  to  the 


S.ii'iiil.   if  I  he   s\st 


•  •ni  rcNt  in  a  vertical  plane,  the  bar  />/■,'  JM-inn  fixinl 


a  hon/ontal  posiiion.  an<l  the  form  of  the  hexajion  Uini: 


of  a  siiin;:  c-onniMiini:  the  joints  ( '  .in<i  /•',  -ho«  ih: 


in 


>reser\'e(|   liy  InealU 


the  t 


H\  :{.  I'o  th 


ensi.m  i.f  the  St 


le  >lle«s  at  (    is  U\    , ;,  and  inakev  with  lli 


nii(r 


(n  the  s|ies>  ;i|   /;  |«  ll\/v.,  and 


iiiaUis  vvii 


h  <  li 


naiiRlesin     \   ,|, 


Th.r.l.  d  ,1, 


an  an-ile  sni     \ 


the  form  'i|'  the  I 


e  <\<\f\u  haiiir  ill  a  vertieal  plane,  ill.'  j..int     I   lifiiig  lixed  ntiil 


with  the  joints   A,  h.  .ui.l  ( 


I. A.iii.in   l«!iiL'  |iriservi-.|  l.y  means  of  strii 


th:it   !,(<  the  t 


igs   ...nileitiiii; 


.I/;  and    K    IS  ir\ 


ensi.m  .if  I'a.li  .if  the  st 


the    Ihairnitll.le-i    .•tlid    .iire.tloMs    ..f    il 


th.'  1.  nsi..ii  of  til,.  >irin;:  .\  h  is  Ull',  and  dett 


niifp 


le    stresses    ;, 


t    tl 


ih.-  strings  are  eomie.'te.l  with  pins  distill,  t  from  tin-  I 


■e  j.>iiit8,  utwuiiiiiig  that 


3.    .V  -ysl.tii  o|   h.;iv\   Ik.t-.  fn.ly    irti.  iilateil.  is  siis|H'nd('d  from  two  fixed 


|»Miit';   .lei.Tmm.'  ili,'  in  miii  iid.  -   iml  ilire.tioi,s  ,,f  ih,.  stic.s,s< 


th.'  I. 


:irs   .'ii'.'  y,\\  iif 


at  the 


.|U.'ii  u.iL'hl  an. I  l.'n);tli.  ,h.i«   ih,it    tl 


Joints 


aiiiTiis    «hi,h    siii'.-.-s^iv>'    liars    mik.'    with    ih.'    hori/...iiiiil   an 


le   iailKelits  of  the 


pn  (jrres-.iori, 
I',       If 


ill   arithiiieth' 


th.  ir  ui'it'liu.  ,iii.|  if  //  |,  ill,    I 


in-  ih.'  -lo|..  s  ..f  th,-  ,th  an.!     r  •  1  ith  I 


«ars  and   Fl'.,  Il", 


//. 


II  ,  •  II 


I'-n/oiit-il  sire-*-*  iti  .'aeh  !»ar,  the  .stress  in  the  rth 


14.   In  .1  tn;iiis.'ir.|  r.mf  of  I  .' 


II.   n«.  ,  ihi 


n-.'    Ii.i-  it>  r.ifli 


!■<  oelin.'il  a!  iiil    i-.  i| 


upiMT  Iri.'iriKular  |M>rli«in  (nf  -I  ft. 


.'ll''     lll.hn.d    at     .ill       I,,    ih,'    \,.!||.';,|,         If    III 


ii'.n 


ridi:.'.  111.. I  ll 


I.'    l.ia.l    .ll    e,| 


Il  Int. 


ll.      I'll.'  rafters  of  the  lower  |M.r- 
lere  is  ,1  |..;,.|  ,,f  imNI  ||,s,  at   the 


th.    ihni-l  of  the  roof. 
\    !o,,,|  ..f  .'(KMI  III-. 


rm.'.li.it.'  j..iiii  lie.',  sHarv  f,>r  eijiijliliriiirn,  and 


a  lii.e-e  i-  iii^..rl.-.|  Ulue.  n  ih..^,-  j.iiiii«       | md  l| 


enli.ih'.l  ,11  ,■;,,  h  ,,f  ||„.  iiiterineiliate  joilitH  iili'l 


I"    .    IIMNI  II. 


ll-  s||i'v«  111  the  lir.'i.e, 

ini-'t     .-uMis  .1 11,^.;  ,i.u(x  ;ni»i 


15.    \  'h.iin  of  ,,,,ial  link-  i«  Mis|K.n.|.'d  ami  l.ia.l.d  as  Av 


w  joinN  he  III 


ii'iilar  ar.'  ..f  .HI  ft    r,i.| 


iwii  III  Fig.  HI. 


>  lim.l  ,il  :!ll    I.,  ih.'  verli.al      I) 
>tri'»'.s  III  tile  link'. 


i  IS       The  loads  at   E  .'iiid  h' 
termiiie  thi-  loails  ut  the  other  joint*  and  the 


16    A  'I 


i.'iiii  111   ."( 


1-  sii.<[H'ni|ii|  from   two  iimiii',  in   i| 


I  e.|iial  r.«|s    ea-h  of  wi  i^ht  ll'l,  linke.1  ,it   thi«  eri.^ 


tile    h< 


i'..mp..|ienl   ,.f   the   ^irer..   at    each  Joint 


le  h..ri/...iital    plane.      Prove  (,i)  itmt 


i>i  eonNtnnt,  ('>)  that  tlw 


m 


KXAXU'l.KS 


69 


\,"!;il  c(ini[Mincnt  at  the  junction  of  th(>  nth  and  (fi  +  lith  fn>cknne<l  from 
,,i:  .mil  t>i  Him  1  n),  ;uiil  K)  that  tlic  tatip'iit  of  the  iil'liiiatioii  of  the 
,ii.  i»l  lo  tlif  horizon  is  |iro|Mirti(>niil  to  m~ii. 

17    A  hirycli".  Fip.   14'_*.  with  'M)-'\u.  whii'lf"    carries  a   riilor    uciuhitiK  1"»0 
1;.-      I'Im'  inctiilxT  He  M  horizontal,  and  .1,  the  intcrsj'ctioii  of  Wand  Kl), 


Fig.  141 


1  Hi    1 1: 


!•;  virtically  atxivo  thi'  tf>ntip  of  the  frfint  wh)-*'!.  iVtcrniine  the  stroA-K's 
!M  till'  -ivrral  MK'MilN-rs  wii  when  all  ihr  x.i'inhl  is  larricd  on  thi-  sadiilc.  ih'i 
HJi.  t   ill.    wcijiht  is  rurrii'd  on  the  Ircaili*'. 

I.**  Tlu-  It-nKthx  of  thi'  ixrnt.  lii-riKl,  and  jib  of  a  iranc  an-  ir>,  JO,  and 
:t(ir  ti-.|K'(tivi'ly.  If  the  crane  lifts  a  ivcijrht  of  "1  toiw,  dctcrniinc  the  strt-sw-s 
ih  ill.  >i  Acral  tn<-nilN*rs  when  the  chain  passi'S  km  alonji  the  jil>,  i/i  aioriR  tlio 
"•  .t/i.'!.  (</),">:   |.(m:  s.;{»  Ions,     i/.t  S;  li.ti.'i;   i;{.4,s  tons. 

10  riif  |Kist  0.1  of  a  |M>rlal>le  hand  initH-  makes  an  anple  of  l.'r  with 
ill.  ill  "/<,  and  an  aiiftle  of  lL»ir  with  the  tie  Mi.  Ihe  liackstay  AC  makes 
;iii  liijilc  of  4."i '  with  the  hori/oiital  strut  Ol).  \  weijcht  of  10  tons  is  sus- 
|Hii.i.'i|  I'liiiii  the  end  /(  of  the  jili.  Kind  the  amount  and  kind  of  stnss  in 
I  .h  .1  th.  iiiernlKTs  tut,  AH,  and  A(  .  Also  detennine  the  counter  Imlaiice- 
n.  ..'Ir   iLiuiredat  <'.  Ii,.v.  :{;{..,     .'7.:«;    ;W..'>;    •.'•{. 7  tons. 

.■  >  Ihe  |M»sf  of  a  jib-crane  is  jO  ft.;  the  wcijiht  lifleil  -H':  tht-  jib  in  in- 
'li!'!  ;i'  .Ml  and  tlip  lie  at  tiO'  lo  the  vertical.  Kind  un  the  stroiiM>s  in  the 
Jil'      .'I  !M  .  .'iiid  also  the  K.  M.  at  the  foot  of  the  |kisI. 

II. .\\  .'.I  Hill  these  stres.S4's  Im'  inodllied  if  thi-  chain  has /.mr  falls,  .ind  if 
ii  !  ,.~.-  to  the  chain-barrel  in  a  direction  bisectiiiK  the  aiiftle  lietwiM-n  tho 

J^i.    ,h.|   lie,' 

Alt.'.     {It)  Sins.s  ill  ti  •  -  »■;  in  jib  -  H\  :{;  ll.M.  -  tr.".\  "a  ft. -tons. 

('-I         -.S7  H;  in  jib  -  I.S7  »'. 

.'I     \n  ordinary  jib-er.'iiie  is  re<|inrei|  to  lift  a  weiitht  of  10  ions  at    1  hor- 

ip    •   I  .Ii-itaiice  of  !(  ft.  fro-ii  the  axis  of  the  |mi«i.      Ihe  haii«;ini!  part  of  the 

it.     i    -  III  /.•(ir  falls;    the  jib  is  |.*(  ft.  lonft  and  the  t<i|i  of  the  jiohI   i"  Itil  ft. 

■  "Uiiij.      Find  the  stresses  in   the  jib  and  tie   when  the  .  ham   pii.sseM 

1:  the  Jib   i  >j  along  iIm  tie. 

I !.     |.o«i   turn-'  round  a  \.^iical  a\is      Find  the  ilirectioti  and  ina^iiitiid« 

i.f  Ml-  hiissiiie  at  the  i<K',  which  is  ;t  ft.  lielow  icrotnid. 

An.-i.  (A)  Stress  III  tie -0.1  tons,  in  jib -■  I  liJ  Ioiih. 

(2i       * -.Ititons     in  jib -".1^  toas. 

I'r.  •i^iire  on  t<K'"-10\    III  tons,  niclined  to  vertical  at  an  anyk'  tan    '  ,'J. 


70 


TllKOIiY  OF  .STRlCTUIillS 


M.  Fig.  143  is  a  <Tam>  for  liftiiii:  <  tmis,  the  chain  l)eitig  in  /"«r  falls  and 
passing  from  .4  to  E.     Draw  the  stress  diagram,  (n)  wh.'ti  a  member  BD  is 


D<-«-* 


Fio.  UA. 


intrcHlu.-,>.l.   (/,)    when  a  n.emU^r  rK  is  infro,lured.     Aho  find  the   vertical 
upiiii  at   r,. 

aj    Th.-  hra.inpnf  ,h..  .rane    Fie  U4,  mnsists  of  iso.soeles  trianRles  havine 
r.,ual  l.a.;«.s  upon  an  outer  an-  of  ;«»'  ra.ii.i.  sul.tendinK  an  Hn^le  of  .K)°  at  the 
•<-"'.•.      I  he   radnts  of  the  inner  are  i.  27  ft.  an.l  the  flanges  are  (1  ft    atmrl 
Mt    the  eround  surface.     iVternune  the  s,,......  i,.  „||   the  mendn-rs    vheTa 

wemh.  of  ,„  ,„,...  is  U.ing  l.f.ed.     Also  „nd  ,  he  .s. res...  i„  theanehlXrs 

24    The  ,H,st  o,  a  derriek-erane  is  .«)  ft.   high;    the  horizontal  tra.v8  of 
the  ,«,,  iKH^stays  are  at  right  angles  to  ...,1.  other,  and  are  V,  ft.  and   K, 
n  length.     Show  that  the  angle  l.tween  the  .horter  traee  and  ,h        a"n;     | 
.he     h  and  ,„■.  when  the  stress  in  th.  |„,s,   ,.  ...  „,,xi„Hu„.  i,  ^o  .-„v  ' 

.Mso  Imd  th..  grea.es.  .,re.s.s,.s  ,„  ,h„  .i„r..r..Mi  .....ndK-rs  ,.f  ,ho  erane  when 
...■  J.  .  «h.,  h  .s  .V,  ft.  |..ng  and  ,s  h.uged  at  the  ,.„.,  ..f  ,he  ,H,st.  C  Lhnod 
at  4.,"  to  the  vertical,  the  weight   l.f.cl  Ih.uk  KNN)  lb.s. 

Alts,  .stress  in  jib  •  (WMUi!!    Ilw  •    in    li,.      i-.iv  «    lu^  .    „  ,, 

P<«       Kl^tH...    II.,..    n.av  .stn.ss    on    long  l.a.  k.stav -^  7;{(H'.7    Iks      „„  «h«rt 
backstay  -  lOVJf)  llis.  '  '    """"«>" 

»5-   If  the  crane  repn-M-nted  bv  the  figure  AB-AC-^H  ft 

^-_^.f...«/>=:.,f,.., ..htl.fted%..ons,^.,.'J;;, 

4.,  ,    he  chair,  hangs  ,„  I.,,,  f,.l|.,„„i  ,,,,,..,  f„„„   ^  ,„  i,\ 
the  stres.s,.s  in  all  the  n„  ndn-rs  ami  ,1„.  upward  pull  at  It 

.4n.v.  Str.-.    ,n    Hf     S,.s,    .tr-48.2.);    AB-->H27 

Verti.ul  pull  at  li     .tl.l  i  ,i,h. 

-I         K.     Hh      ,U:    K(;^lir     ,j  ft.      I  md  tin-  s,„.ss..h 
in    1 1-'   nM-n.lHTs  ,„    ,he   ,.r,„.„   ^hen   ih..   weigh.    ,1    ,.„„    |,f„,l      'fx^f^ 


c 
Fin.  US. 


\B'BD 

c 


i..«,.l  .1;   :/,,  at  H.   -n  .,  /,.      v|.,,„,»  ,i,„,  „„.  ,,^„^^,^  -^^,- 

.here   h.  an   additional    ueight    of    J   ,.„,   „,   .,,..|,  o,    ,)...   ,„„,.,« 
If.  I'.  /• ,  and  (». 


I 


14«i 


EXAMPLES. 

7 

Stnm. 

Cm**. 

Cut  6. 

Cm*. 

CM*  A 

.  Aa-Bb 

2 

0 

0 

0 

Cd-De 

H 

If 

M 

« 

Eg 

w 

3i 

2i 

H 

Xa 

V5 

0 

0 

0 

Xc 

VV2 

VV2 

0 

5V2" 

XI 

t8\/2 

!Sv/2 

!SV2 

HV2 

ab 

0 

1 

0 

i 

de 

0 

0 

0 

i 

be 

!>/5 

iv's 

0 

HVs 

dc 

An/317 

i\n/317 

iVv317 

iVv'317 

ft 

tIi\/317 

ili\/317 

tI.v/317 

A'.Vsit 

l9 

.".^5 

•'.V5 

AV3 

lS>/5 

27.  The  post  AB  of  a  jib-orane,  Fig.  147,  is  20  ft.;  the  jih  AC  is  inclined 
;ii  :«r  and  the  tic  BC  at  4")°  to  the  vertical;  the  weight  lifted  is  .5  tons.  Find 
tli>  ^trt>ss(!s  in  the  jib  and  tie  when  the  rhain  passes  (a)  along  the  jib,  (b)  along 
ihf  tic,  (c)  horizontally  from  C  to  the  post. 


Te^ 


Fio.  147. 


M       M'     10' 


Fia.  148. 


5- 


1  III'  chain  ha«  two  falls. 

XUii  draw  the  stress  diagram  for  the  three  rases  when  the  dotted  members 

added. 

Caw  a.  CMC  b.  Cue  t. 

yln«.  Stn-ss  in  jib,     (i.16  'd.&\  5.40 

"      "  tie,     2. 59  W)  .35 

.'8.  Fin.  148  represents   a   2-ton  travelling  crane  running  upon  a  pair  of 
^      Draw  the  stress  diagram  and  find  the  n^wtions  on  the  rails. 


^0.  lii  the  rranc  ABC  the  vertical  ixwt  .tfl-l.V,  the  jib  AC -23',  and 
•!  iiidic  HAC-MV.  Find  (a)  the  stresws  in  the  jih  and  tie.  and  also  the 
I"  ti  litit;  moment  at  the  foot  of  the  |M*8t  when  the  crane  lifts  a  weight  of  4  tuns. 

ihc  throw  is  increased  by  adding  two  hnriiontat  mi'mlwrs  CK,  BI)  and  an 
1:1  lined  member  DE,  the  figure  BE  Ijeing  u  parallelogram  and  the  diagonal 
'  /'  <oinci<lent  in  direction  with  CA.  Find  {!>)  the  str«>8Hes  in  the  several 
nicinlfer?'  of  the  crane  as  thus  nuMliftiHl,  the  weight  lifted  lieing  the  same. 

In  the  latter  caw  show  (t)  how  the  stresses  in  the  members  an;  affected 


<-  rili:Oi<Y  OF  STRICTIRKS. 

when  the  chain,  wliich  is  in  jour  fulls,  passes  from  E  to  li  and  then  down  the 

I  lost. 

Ahx.  (a)    Tension  in  tie  =3^  tons;   thrust  in  jib-6^f  tona. 
(b)    Stress  in   CK  -<,t.;}4;  in  /v/>  =  10.10;  in  C«- 13.49; 
Ml  ( 7>  =(i.l-);  in  l).\  -  10.7;  in  Rl)  -  7  tona. 
(r)    Stress  in    (K^S.W;    \i\  KI)  =  \Q-;    \nCR-\2.9: 
in   (/>-.'>.,S;    in />.» -10.7;  in  B/)- 7.4  tons. 
JO.  The  horizontal    traces  of  ilie  two  haclvstays  of  a  derrick-cTanc  are  z 
and  1/  feet  in  length,  and  the  angle  Utween  thcin  i.s  ,y.     Show  that  the  stress 

in  the  [Kist  is  a  inaxiinuin  when  ""'  =-,  tf  being  the  anffle  between  the 

cos  0  If  " 

trace  y  and  the  plane  of  the  jit>  anil  .ie. 

31.  The  two  backstays  of  a  derrick-crane  are  each  .18'  long,  and  the  angle 
Ix'iw.rn  their  horizontal  traces  2  tan  '/j;  height  of  crane-post -32';  the 
length  of  tlu'  jib  40';  the  throw  of  the  crane  -.I)';  the  weight  lifted -4 
Ions.     Detcrnnne  the  -tresses  in  the  siveral  nietnlx-rs  and    the  upward  pull 

at  the  foot  of  each   back>tay  when  the  plai f  ihi-  jib  and    po.st  ia)  bisects 

the  angle  Ix'tween  the  hori,.onial  traics  of  tin-  backstays,  (/*)    passi-s  through 
a  backstay. 

Alls.  In  jib  -.">;  in  tie  »L.'..V.'  tons;  in  backstay  in  («)  -1.77,  in  (b)  -.3.267 
Ions. 

iJ.  liiid  the  stresses  in  the  rneinliers  of  the  crane  represented  by  Fig. 
14'.»:    ;ils(i  find  bal:ince-\veight  at  ('. 

.1'-.  Sticks  in  «/;•_'.-,;  />/•;  ,i'C.«(;  />/i"lS.S.->;  />.! -L'C.OK;  B.l-l.tJ7; 
lii       U\.V.\  tons;    counterweight  a'  ('  'll.(>7tons. 

3J.   In  the  crane  represented  by  Fig.  l.")0,  draw  the  stress  diagram  when  a 


i    K.      I  IM 


lo.iil  iif   ttNHi  Ml-..  I-  -ii-|«-iided  from  .1       If  the  ihiiin  pa.sses  along  /?.!  over  a 

j'llliiy  .It    .1  :ii|.|   i*    in   /   '(;■  fall-,   deteniiine    the   ^Ire— .es  ill  till'  -4V'r;il  IIielTil>«>r«. 

H-  \  I'MJr  of  -(i,.:ir-l.-g-i,  eieli  J",  ft.  Inng,  uih  the  jMiint  of  siiS|iens:on  'JO 
ft.  vi-rtujilly  .iIma,-  ill,  grniinil  Mirfre c  i»  siipporteil  bv  a  tie  UN)  fi.  \iin\t; 
dlstani'c  Ik  luviii  I'l  .  I  -l'  I...'-  10^  .",  fi .  1  nid  the  thni.sts  along  the  legs  and 
the  leli«iuii  111  ill      '•:>■  \\\i<  u  a   ni  ii;||t  oi   .'  tiilis  i-i  InMllg  iiftod. 

.1/'".    Iiliili.li   IN  tie       I  .l;t7  lolls;  ciiiijiresslini  ill  each  leg       l.lifttoim. 

J5.  MK  It  is  :i  1)11  nliihitend  tni-s  1//  ,.||„|  r/<  biiig  liori/onlid  and  l.'i 
and  ,M>  ft.  in  lenutli.  i. -|km  tuely.      11.     Im^rtli  nf  M     is  10  ft.,  and   ito  in- 


EXAMPLES. 


78 


,1, nation  to  the  vertical  is  00°.      A  weight  11',  is  plared  at  C,  and   W,  at 
/(      What   iiiusi  1h'  the  relation  betwwii  11',  and  ,,4, 

II  M.tluit  the  truss  may  not  be  deformed?  For 
;,;,\  .iihir  relation  Ix-tween  H',  and  W„  explain 
liovv  vou  would  incMlifv  the  truss  to  i)revent  defor- 
inaiu'ii,  and  find  the  stn'saes  in  all  the  members. 

Ahs.  \\\-\.3VtCt\V,. 

36.  Draw  the  stress  diagram  for  the  frame,  Fig. 
l')l,  under  a  load  of  1  ton  at  the  ai)ex.  Also  find 
I  In  Ncrtical  forces  in  the  anchorages. 

37.  In   the   frame,  Fig.  1.52,  AC -AB- 10  ft., 
a„,l    a    l<)a<l   of  10()0   Ibe.  is  concentrated    at   A.  Fio.  151. 
\\!i:it  load  at  B  will  prevent  distortion?      If  the 

tiM  iuIk  r  BC  is  introduced  and  a  load  of  3000  lbs.  is  concentrated  at  B,  draw 

tl,r  ^iri'ss  diagram. 

38    In  a  triangular  truss  of  the  dimeasions  shown  in  Fig.  153,  a  load  of 


/ 

\p 

w 

\* 

>'/* 

cS;«r 

■</ 

Xy 

-«^ > 

I    t(.M 


Fiu.  153 

is  (oTicentrated  at  .1  and  at  B.     What  must  l»e  the  magnitude  of  the 
/'  at  ('  so  that  tb-re  may  be  no  distorticm  of  the  frame? 

It)  a  fiuadrilateral  truss  ABVD,  AD  is  horizontal,  AB  and  BC  are 
(1  at  unplcN  of  (10°  and  :«)°  n-^iMw-tively  to  the  horizontal,  and  CD  is 
i|  nt  4.')°  to  the  horizontal.  What  weight  must  lie  concentrated  at  C 
iiilain  the  e(|uilibrium  of  the  frame  uniler  a  weight  U'  at  W? 
I  weight  II'  is  placed  at  ('  as  well  as  iii  /),  what  memU-r  must  Ik-  intro- 
t.i  prevent  distortion?  What  will  lie  the  stn-ss  in  that  im-mlKT? 
/,v.  1st,  l.:Ui<>  ir  ;•.>(!,  stress  in /</>-. 13411' ctwec  (0.0° +  «).  where  ni-B/)/l. 
.  M'K  is  a  triangidar  truss  siipix.rted  at  .1  and  K;  .1A'=30  ft.;.ir- 
<!■:  -  IS  ft.  FB  and  FD  are  two  struts  from  a  \nm\t  F  in  the  tie  AE 
allv  Ixlow  (',  B  and  D  U-iiig  the  mi«ldle  jM.ints  of  AC  and  CE  respec- 
V  liiid  the  load  which  must  U>  placrd  at  D  to  pn-vent  distortion  (a) 
ill.  r.  is  li  verti.al  load  of  J  ton  at  B  and  at  C,  {!>)  when  then-  is  11  nor- 
iimI  pn-ssun-  of  \  ton  at  B  and  a  load  of  \  ton  at  B  and  at  (',  tissuming 
>d  .1  to  U'  lixed  and  the  «iid  E  to  rest  uiH.n  rollers. 

Alls.  (n). 'Si  ton;  (M  .42.')  ton. 
41  III  the  frante.  Fig.  l.'>4,  the  I>ar8  are  each  of  the  same  length  mikI  sl.|)e 
ii.wii,  /rtlning  horizontal.  A  load  of  I  ton  is  pla<vd  at  A.  What  l<«id 
.  !..  .•..ncciitrated  at  B  and  at  C  without  prinlucing  distortion?  If  the 
I  l.r  HH  i-  iiitrfHlui-ed  and  a  load  of  1  ton  is  plai'e<l  at  .1  and  at  B,  what 
11  Iw  the  load  at  (.■  so  that  there  might  Ix'  no  distortion?    If  the  member 


■,r\un 

11..  llIM 

!"  Ilia 

If 

li'l'  ri| 

1 

4" 

•  1    M  , 

■  I  1  ir 

TIlF.ailV  OF  STRUCTURES. 


or  is  also  intrmliKTd.uraw  th.-  stress  diaKrani  for  a  load  of  1  ton  at  each  of 
the  joints  .1,  «,  and  ('. 


Th 


4J.  In   the   triaiiKiilar  truss,    Fig.    l.V),  .l/?-'j 


If  til'  He  is  h 


.•o',  .ir  =  i.v 


id  He 


>ri/.(.ntal  and  is  supjK.rti'd  at  H  and  ('.     The  middle  poiuu* 


I.".  I 


AR  and  AC  are  siipiHirted  by  struts  l)E  and  DF  from  a  |v>int  7) 


Ix-low  .1.      Loads  of  KX)  and  S()()ll 


\\  hat  load  must  !)«■  conccnlratt'd  at  /• 


>s.  jtrc  (•onccnt  rated  at  A"  and  .-1 


to  prcveiit  distortion? 


vertically 
respectively. 


43.  Three    l.ars,    frcelv 


The 

hori/iintal  line,  and 


irlicuiated,    form    an    cpiilaleral    triangle    ABC 


i.vstem  rests  in  a  vcrli.,,!  plane  u|Hin  sup|«.rts  at   H  and  ( 


eight   M'  is  sus|K-nded  from  .1.     Ivtt 


in  the  same 


rmine  the  stress 


in  IW,  neglecting  the  weight  of  the  l.ars. 


Ann. 


\V 


44'  Three  liars,  freel 


kepi  ill  eipiiiil.riuni  hy  three  forces  act 


y  articulated,  form  a  triangle  ARC ,  and  fh 


l.\^3 


le  svstem  is 


in  e.ich  l(;ir. 


ig  on  the  joints.     Determine  the  strt 


•ss 


fore 


What   rel;it ion  holds  iM-twcen  th 


■s  meet  (.n  in  ihe  leiilroid,  i// 


le  stresses  when   the   li 


nes  of  action  of  the 


I  111  the  orihocentre  O  of  the  triaiigl 


45.  A  triangular  truss  of  white  pi 
and  a  tic-Uani  RC;    \\xv  span  of  tl 
uniformly    distrilnited    load    up 


\us.  (,/i   Rl-  (  A:AR\   (I,)  OAOROC. 
lonsists  of  two  eipial  rafters  AR   AC 


lie  iru->  is  :«(  ft.  and  it> 


stresses  in  I  hi 


■evcral  iiiendNTs. 


ea.h    rafter    is    SHU)   Ihs.     Jh 


rise  is  7J  ft.:  tht 


termine   the 


46.  A  ro<)f-tru>>  of  _•()  ft. 


An.-..   .<<ircss  in  ///•-,S4(M»  ll>: 


a  horizontal  lie-rod  I 


sp.'iii  aiii 


Is  ft. 


ill  -l/i-  I.IN)^  .Vllw. 


fool 


Were  raised   I  ft. 


Hi  ween   I  he   feet.      Th 


ise  is  com|M>.sed  of  two  rafter,  and 


111.      liinl  ihe  pull  on  ihc  ti 


47.  The  r.iflers  AR,  AC  «( 


lie  load  upon  thi'  truss  -.•KIOll.s.  |M.r 
What  woul.l  th.'  pull  U'  if  the  rinl 
.1"v.   .{ !-'.•>  lbs.;    tlJ.V)lb8. 


llgl 


les  ,.,  ;<  to  ih,'  vertical:   the 


a  rooi"  irc  uiiet|ual  in  lenirth  and 


are   inclineil  at 


AC^\\\.     liiidthet 


uiiifninily  ilistributed  load  u|Nin  AR-]\\,  upon 


elision  on  the  tie-lx'a 


ir,  +  ir 


48.   Ill  the  prci'ding  example,  if  tl 


Kin  '»  sin  ,? 
sin  (a  +  ,?)" 


he   tension    on   the    „e,  the   rafters  U'lng  spa.ed  .'i   ft.  centr..  ( 


th( 


roof  Iliad  I 


H'iiig  •-'(»  lbs.  iHT  sipiare  f.K.t 


4Q    The  ei|Ual  rafters  AR.  A(    f 


'p.ni      10  ft ..  <r  -  m"  and  ,?  ~  V,°,  find 
•  I  iTiitre  and 
A/I.S.  i«»s  llw. 


spaced  J\  ft.  centre  to  centre;  th 


II  roof  of  10  fi.  s,,„„  ,„„|  ^,j  f^ 


le  weight   of  the  riMif 


ri.si'  are 
covering,  etc.  - 20  Ibg. 


EXAMPLES. 


75 


]■.  r  si|iian'  f<K)t.  Fiiul  the  vertical  presnure  and  outward  thrust  at  the  fi»ot 
.1  a  rafter. 

Anx.  Total  verticai  prriwure  -  12.'>\/ri  llw.  -horizontal  thrust. 

50.  The-  lonf;tbi  of  the  tio-lx'atn  and  two  rafters  of  a  riK)f-lrusM  arc  in  the 
riiiiK  (if  ,">:4:3.  Kind  tlir  strc-wes  in  the  several  nieniU'rs  when  the  load  uiM»n 
I  ,>\\  rafter  is  uniforndv  distributed  and  etpial  to  ItJt)  ll>s. 

.\n.i.  Stres.s  in  tie -4S  llw.;  in  one  rafter -ttO  llw.;  in  other -S()  lbs. 

51.  In  a  triangular  truss  the  rafters  each  slo|)e  at  'Mf;  the  loa<l  U|>on  the 
i!" A     KK)  ll)s.     Find  the  thrust  of  the  roof  and  the  stri'ss  in  eaih  rafter. 

Anx.    1(H)  ihs.:   Sti.tllhs. 
Si.   A  roof-truss  is  coni|M>s<<d  of  two  e(|ual  rafters  and  a  tie-U-am,  an<l  the 
.s;:iii     1  times  the  ris*-;    the  load  at  the  a|N<\~-l(MN)  llw.     Find  the  stresses 
,    ilii-  several  niendn'rs. 

Si'i'iindly,  if  a  man  of  l.'iO  llis.  stands  at  the  mitldle  of  a  rafter,  by  how 
iiiiicli  will  the  stress  in  the  tie-U>an;  In*  inceas4'd'.' 

Auk.  1.  Stn'ss  in  tie -40(10  11)8.;  in  each  rafter -'JOOOv''-^ Ihs. 
•-'.  7')  ll»s. 
53.  A  kinn-[>o8t  truss  for  a  roof  of  M)  ft.  span  and  "J  ft.  rise  is  composed 
•  Mil.  c(|iial  rafters  Ali,  AC,  the  horizontal  tie-lx-am  liC,  the  vertical  tie  .!/>, 
r  I  ilic  struts  PK,  HF  from  the  middle  jioint  />  of  the  tie-l)eam  to  the  middle 

I Is  of  the  rafters;    the  r<Mif  load -'.'()  Ilis.  |ier  Ni|uai'e  fiM)t,  of  nntf  surface 

iiihl  the  rafters  an-  spaced  10  ft.  i-entre  to  (M-ntre.  Find  the  strps.s«'s  in  the 
-.  \  r:il  incndM'rs.  .S>cond,  lin<l  the  altered  stres.ses  due  to  a  man  of  l.'iO  lbs. 
ui  iL'ht  standing  on  the  ridf(e.  Third,  find  the  stresses  due  to  a  weight  on  the 
iH  -iKam  of  12  lbs.  jwr  square  Uwi. 


IM. 

2.1. 

3cl. 

Am.  HE 

."l<)_>.'l 

7ov/.'» 

«M)v/5 

EA 

37.'iO 

7.')V'5 

{n.n)\/r, 

HI) 

•.'-.'.'.OV  5 

IJO 

1800 

L.E 

1S7.-) 

0 

0 

AD 

7MK^rt 

0 

ISOO 

S4.  The  triangular  tru.s.s  ABC,  Fig.  l.'iCt.  is  of  the  dimensions  and  loaded 
.i-i  Indicated.     l)raw  the  stress  diagram,  and  show  how  the  .stn-sses  are  iniKli- 
liiij  wlicn  the  strut    DE   has  U-f-n  remove<l.     The 
liMii/oiital  com|Mtnent   of  the  normal  load  on  .AH 
I-  <"  Ih'  divided  e<|ually  U'tween  the  two  sup|Nirts. 


ss.  A  <iue<'n-tru«s  for  a  rtsif  consihts  of  two 
>' n/diital  mendN-rs,  the  lower  4S  ft.  long,  the 
I'  •■•■r  Hi  ft.  long;   two  inclined  mcmUTS  .\H.  IH', 

I  two  qiKHMis  HE,CF,t'm-]\  S  ft.  long;  the  |>oint9 

/'  divide  .t/>into  three  etjual  segments;  the  load 

.    'lithe  memlx'rs  .l/{,  /W',(7>is  I'JO  lbs.  jht  lineal 

'  t      Fiiiil  (11)  the  stn-sses  in  the  several  memlN-rs. 

!■  »■  I'll  will  tliew  stresses  l)e  niiHlitie<l  if  struts  are  intnsluced  from  the  feet 

'  'lie  <|Ueens  to  the  middle  jsiints  (l,  H  of  the  inclined  mend)»'rs?      In  this 

I  r  case  also,  determine  (<)  the  stresses  due  to  a  witid  pressure  of  120  lbs. 


Fi(i    l.V), 


MICROCOfY    RESOLUTION    TEST   CHART 

lANSI  and  ISO  TEST  CHART  No    2l 


^ 


jd     APPLIED  IfVMGE 


'^•'    *8^'    -   0300       Priori 


70 


TUEOIIY  OF  STRUCTURES. 


IKT  lineal  foot  normal  to  .1/i  assiimir,    ih.,i  tu  .  u    •      .. 
diviclod  Ix-twoen  ,ho  two  sup;;":  it.,  D         "°"''^'  "'''•'""  "  "^"''"y 
^ns.  («)  Stress  in  Il,.s.  i„  .i e  =  4o,((j  -,;  ^  j,/,  ^ ^^  ^ ^^,. 

^n  «»        •    „• '  '^  ^  ■*•'"*''••"''  =""  ^  '^^ ■■  ^*^^'  =  <^  = '  '^'    ' 
(6)  .Sfrt'ss  m  lbs.  in  AE  ^r,V.i>).S4  =  Of- 

fir  =  4(Mi(i.5f,  =  .1F;  .4«  =  -,740.5,;  ./)//• 

B(r' = 4r,4f)..w = r// :  eg  =  1 200  =  fh  ■ 

f7>  =  .-,;}(;.fl4  =(/.'.  ' 

(r)  Additional  strc-ss  in, If?  =  l()4()v/T;  ««  =  fl.SQ  ^  v 

56';rr!i:s'"4'r'T(''';*'^;"'^''''""''  '••  f'^'-^-* '«-»-'"-) 

:-  Z^'':    "-  1 ^  ^  su  .;r    ,^^j^',^;y^;^""^^^  point  />  of  the 

- . ........  to  a  ,^::^zr7i^^-iz  CrJ:.  ^;z 

'^/is.  (1)  Stress  ill  «/;•- 111)1.      Fl=7in.       .  i.-     -.->, 
57.    1  he  r.-iftiTS  A/i     \('  'in.  uii...w.»«    I     .   .1 

....« .., ..-..  .w  <,„,:ir;  j'^„;:  ri;  lis;  "''  "■-  *""  -"^ 

'"*  -       ""  («)  (6) 

.I'.v.   /</.•  24<K)v  o  14iXK  .-r;       ft/;   JSOO  2800 

AM    IKXK  .-,  14(X)N  .^; 

A  I-  14(K)v/.-)  14(K)v/r7; 

/•T    I.S()()\/.-7  l.SOOv/.'T; 

.!/>  1400  4(K); 
58.  A  lrl,•ltl^'ular  frame   AJU\  in   which    4«-  ir  o„  1    »/--•.      . 

'■■■■• ' " ■■  -"■■• ' -H..I  "r  .i«.  -','„:.:;::„', c  Z 

m.n..my  of  mat.Tial  will  U-  pr.-al,.st  when  tan  .!/«•- 4.'  £.+  7' 


DC  3(iOO      _  3000 

i>f':  io(K)v/.T 

/>/•■   JOO-s/.T        400V'3 


50.     A    n>..f-tni.-:-.    ! -iiaf-.:   i,f    )^y„ 


•qual  ralier.s  .Ifl.  ,it  iii,|,„p,|  „t  (W)°  to 


EXAMPLES. 


77 


the  vertical,  of  a  horizontal  tie-beam  BC  of  length  I,  of  a  collar-beam  DE 

of  Icnpth  — ,  and  of  queen-posts  DF,  tlG  at  each  end  of  the  collar-beam;  the 

truss  is  loaded  with  a  weight  of  2000  lbs.  at  the  vertex,  a  weight  of  4000  lbs. 
at  one  collar-beam  joint,  a  weight  of  1200  lbs.  at  the  other,  and  a  weight  of 
l.')(K)  lbs.  at  the  foot  of  each  queen;  the  (hagonal  DG  is  inser*->d  to  provide 
for  the  unequal  distribution  of  load.     Find  the  stresses  in  all  members. 

Ans.  Stress  in  BD- II 73.3i;  Bi^  =  .580615 \/:T;  DF  =  1.tOO; 
DA  ~  2000 ;  DE  =■  303.3i\/3 ;  DG  - 1  S66§ ; 

r7C  =  4a33j\/3";  G£:-2433J;  CE -98061; 
.lK=2(K)0lbs. 
6o.  A  triangular  truss  consists  of  two  equal  rafters  AR,  AC  and  a  tie- 
iK'iim  BC,  all  of  white  pine;  the  centre  D  of  the  tie-beam  is  supported  from 
.1  by  a  wrought-iron  rod  AD;  the  uniformly  distributed  load  upon  each  rafter 
is  S4()0  lbs.,  and  ujKin  the  tie-beam  is  .36,000  lbs. :  determine  (a)  the  stresses 
ill  the  different  memlx>rs,  BC  being  40  ft.  And  AD  20  ft.  What  (b)  will  be  the 
effect  upon  the  several  members  if  the  centre  of  the  tie-licam  be  supjxtrted 
ufwm  a  wall,  and  if  for  the  rod  a  post  be  substituted  against  which  the  heads 
of  the  rafters  can  rest?  Assume  that  the  pressure  between  the  rafter  and 
post  acts  at  right  angles  to  the  rafter. 

Ans.  (n)  Stresses  in  BD- 1.3200;   .ID -18000;  AB-13200\/2  lbs. 
(6)       "  "       -  4200;     "    -  4200;    "    -   0.300v/2lb8. 

6i.  \  triangular  tru-s-s  of  white  pine  consists  of  a  rafter  .AC,  a  vertical  post 
Mi,  and  a  horizontal  tie-beam  BC;  the  load  upon  the  rafter  is  IlOO  lbs.  per 
liiK'jil  foot;  .1('-:W  ft.,  .\B-(\  ft.  Finil  the  resultant  pressure  at  C,  Ufjon 
AC  assuming  that  the  pres,sure  upon  .!('  at  .1  is  normal  to  .\C. 

Find  the  stresses  in  the  several  members  when  the  centre  D  of  the  rafter 
is  also  sui)jK)rted  by  a  strut  from  B. 

Ans.  4702  llw.;  stress  in  BC -4nO0\/<A;  CD-112.t0;  DB~U2jO; 
DA-0;  J«-22r)0lb8. 
6j.  a  white-pine  triangular  truss  consists  of  two  rafters  .\B,  .\C  of  un- 
(■i|iial  length  and  a  tie-lx-am  BC  A  vertical  wrought-iron  rod  from  .1,  10 
ft.  long.  sup|H)rts  the  tie-lx'ain  at  a  point  D,  iividing  its  length  into  the 
sitiiHiils  BI)~U)  ft.  and  CD^'M  ft.  The  load  u|)on  each  rafter  is  .3(M)  lbs. 
l«r  hiieal  ft.;  the  load  ujM)n  the  tie-lx>am  is  1S,(KK)  lbs.,  uniformly  dis- 
tributed.    IVtermine  the  stresses  in  the  several  niemlx-rs. 

.1m*.  In  A B - 'Mm\- J  Iba.;    /lC'-482.V-'i  lbs.;  BD - TD - !)ftr)0  IIm. ; 
.l/>-!»000  ll)s. 
6j.  A  frame  is  conqHised  of  a  horizontal  top-lx-am  40  ft.  long,  two  verti- 
lal  struts  3  ft.  long,   ;iid  thre<'  fie-ro<]«  of  which  the  middle  one  in  horizontal 
and   I.")  ft.  long.     Find  the  Htress«'s  pHxluced  in  the  several  memU'rs  when 
I  single  load  of  12,000  lbs.  is  concentrated  ut  the  head  of  each  strut. 

An«.  Stress  in  horizontal  memlters  - .'i(NN)()  Iba. 
"      "  sloping  "      -.^UIH  " 

"      "  struts  -12000  " 


'^  THEOnr  OF  STRICTURES. 

64.  The  rafters  AB,  AC  of  a  roof-truss  are  20  ft.  lone  and  are  9,.nnort»^ 
at  the  centres  by  the  struts  A  ,  DF-  the  centre  D  of  Xh^l'^h.^^ RcT't 

to   ^^i"^-  "''I  'i^  *'"'  ''^'''^  ^^^  "P*'"  ''>"  ■^^•••™'  "members  if  .IB  be  subjected 
to  a  horizontal  pressure  of  156  lbs.  pt-r  lineal  foot?  subjected 

Ans.  (a)    Stress  in /yD=4t)00v/.T;  BJ?=.9l>00-  ^^-ooqo- 
A^Z>-4000;   .4Z)-2000;   />>-1200^-     ' 
.4F-5200;  CF-6400;  Ci)-3200\/3' 
(/>)    Tens,  in  5^-520\/3;  .4fi-2(50v/.T';  compres.  in 

f:/)- 520^3;  .4C-520V/3;  Z)C- 780. 
Ao  stresses  in  BD,  AE,  and  DF 

h.„^lT"^.  *^?  '!rr'  "'  "*"  '^^  """'"^^  «^  the  truss,  assuming  the  tie- 
beam  to  be  also  loaded  with  a  weight  of  600  lbs.  per  lineal  foot 

Am.  Stress  in  AB  increased  by  3000^3  lbs.;  in  BCby  9000  lbs  • 
in  .40  by  6000^3  lbs. 
65.  A  horizontal  beam  of  length  /  is  crussed  and  supported  by  a  vertical 
s  rut  at  .ts  muldle  pomt.     If  a  weight  W  roll  across  the^am  show  thaUl^ 
r  1  ZTL:T'  '"^"^^  P™Po«-te,y  With  the  distant" ^'ol  Z 


Ans.   Stress   in   horizontal  tie 


.n- 


yxcotd;  in  sloping  tie  =  j-x  cosec  0 ; 


in  strut -2-x,  e  being  angle  between  the  horizontal  and  the  sloping  mem- 
bers. 

66.  If  a  wheel  loaded  with  12.000  lbs.  travel  over  the  top-beam  in  the 
kst  quosfon.  what  meml)ers  must  Ik-  introduced  to  prevent  dLrtion? 
What  afe  the  .nax.mum  stresses  to  which  these  members  will  be  subSected? 

67.  .V  Ixvam  of  30  ft    span  is  supported  by  „n  invertod'!iTiiK>l-tnf,l^'';he 

rrfh    "1"  ""    •  '   K-  '""'  ^r*  '^^  '""*'""  *>""-"^'*'  me,nl.7  0        long 
tm.l  the  stros.s,.s  m  the  .s,.vcral  memlHTs  due  to  a  wight  W  at  the  hJd  f^ 

a  „uocn    .ntrcHiucing  the  diagonal  required  to  pn-vent'tlLo  tit      Alsofind 
the  stresses  due  to  a  weight  W  at  centre  of  U-ain 

Am.  (\)  Stress  in  AB-\'W-EF;  .4A'-23>n'-  BE-iW- 
BF^him-.DF;  /«■-',•■»•;■  CT-0       ' 
(2)  Strcssin  .4/i-;ir-Af;    .1/v'-1.74l>'-/iA'-i'- 
BF-O.  ^ 

68.  The  platform  of  a  bridge  for  a  dear  span  of  .U)  ft.  i.,  e„rricd  bv  two 
qu.-„-,russ,.s  ,.W,.  .|..p;    the  up.KT  horizontal  mcmU.  of  Z  n^^X 
.   ong      he  load  upon  the  bridge  -.^K,  ll.s.  ,.r  s.lu.re  foot  of  p,a,f,.r,n.  wL"h 
Js  1_  ft.  «i,le.     l-,n,l  the  str.'ss,.s  in  the  «<.veral  m.-nilM-rs 

.4»(s.  Str..ss   i„   vertical -.(KMHI   lbs.;  in  each  sloping  ineniU'r  -  lOOtM)  lb«  ■ 
lu  oath  horuuntni  !!». ■!,!>.  r-«<X)Oibs.  'l«J<JO  lbs.; 


\ 


EXAMPLES. 


79 


"-^^P- 


A 

FiQ.  157. 


69.  If  a  single  load  of  3000  lbs.  pass  over  the  bridge  in  the  last  question, 
; ,1(1  if  its  effect  is  equally  divided  between  the  trusses,  find  (a)  the  greatest 
Mirss  in  the  members  of  the  truss,  and  also  (b)  in  the  members  which  must 
iM'  introduced  to  prevent  distortion.  Also  find  (c)  the  stresses  when  one 
half  the  bridge  carries  an  additional  load  of  50  lbs.  per  square  foot  of 
jilatform. 

Ana.  (a)  In  sloping  end  strut -3333  J  lbs.;  horizontal  tie - 

2666}  and  1333 J  lbs.;   horizontal  strut -2666J  lbs. 
(c)  In  sloping  end  strut  -62.50  lbs.;  horizontal  tie  - 
5000  lbs.,  horizontal  strut  -5000  and  3000  lbs. 
(h)  In  case  (o)- 1666 J  lbs.;   in  ca-^o  (c)-2500  lbs. 

70.  The  platform  of  a  bridge  for  a  clear  span  of  60  ft.  is  carried  by 
two  trusses  15  ft.  deep,  of  the  type  shown  by  the 
acconipaiiying  diagram;  the  load  upon  the  bridge 
is  ."rfj  lbs.  per  square  foot  of  platform,  which  is 
12  ft.  wide.  Find  the  stresses  in  the  several  mem- 
bers. 

Ana.  StK3a  in  BE -13500;    BG -e750\/5;    SG-4500; 

£/)- 13500;    CD-2250V5;    GA-4500v'5; 
/li)- 9000  lbs. 

71.  If  a  single  weight  of  2000  lbs.  pass  over  a  truss  similar  to  that  shown 
in  the  preceding  question,  find  the  stresses  in  the  several  members  when  the 
load  is  (1)  at  £,  (2)  at  D. 

Ans.  (1)  Stress  in  BG -1500^5;  BE -3000 -ED;  EG -2000; 

GD-lQOO\/l;  AG-5O0V5-AH-CH;  DH-O-FH; 
DF  - 1000 -FC  lbs. 

(2)  Stress  in  B/l-lOOOVs-C/l;  BD  -  2000 -Z)C- AD  Ibe.. 
Stresses  in  other  members  -0. 
^i.  The  feet  of  the  equal  roof-rafters  AB,  AC  are  tied  by  rods  BD,  CD 
which  meet  under  the  vertex  and  are  joined  to  it  by  a  rod  AD.  If  W„  W^ 
arc  the  uniformly  distributed  loads  in  pounds  upon  AB,  AC,  respectively, 
and  if  S  is  the  span  of  the  roof  in  feet,  find  the  weight  of  metal  (wrought  iron) 
ill  the  ties. 

Aus.  Q—^. — 'Scot,?,  /  being  inch-stress  in  pounds  and  ,3  the  angle 
MilK 

iu)  If  .IB -AC -20  ft.,  AD-a  ft.,  the  angle  BAD -60°,  find  the  stresses 
ill  tiif  si'verul  members  when  a  weight  of  3500  lbs.  is  concentrated  at  the  vertex. 

Ana.  7000  lbs.;  6309.8  lbs.;  3500  lbs. 

I'l  Tlic  roof  in  (a)  is  loaded  with  10  lbs.  per  square  foot  on  one  side  and 
.( !  IIh.  (ht  .s<iu.are  foot  on  the  other,  the  trusm-s  being  13  ft.  centre  to  centre. 
Kitcrniiiic  in)  the  stresses  in  the  several  members.  Examine  (h)  the  effect 
"f  I  horizontal  pressure  of  14  lbs.  |H>r  8()uare  fcmt  on  the  most  heavily  loaded 
Ml',  assuming  that  the  reaction  is  equally  divided  between  the  two  supports. 

Ana.  (a)  11180  lbs.;  10077.65  lbs. ;  5590  Iba. 


Fio.  158. 


*0  THEORY  OF  STRUCTURES. 

73.  In  the  truss  represented  in  the  accompanying  figure,  the  load  on 
AIS^W,,  on  AC^W,;  the  angle  ABD-^-  AD-BD 
-AE^CE.  Find  the  total  weight  of  metal  (wrought 
iron)  in  the  tie-rods. 

Am.  ~^- — ?  s  cot  ^,  S  being  the  span  and  /  the 
inch-stress, 
(a)  If  the  stress  in  Bl)  or  EC  is  equal  to  the  stress  in  DE.  show  that  fi- 
^°~J'  "  being  the  angle  ABC. 

(h)  The  trusses  are  12  ft.  centre  to  centre;   the  span  is  40  ft  •   the  hori 
zoiital  tie  IS  10  ft.  long,     the  rafters  are  inclined  at  So°  to  the  vert  cal     the" 
dead  weight  o    the  roof,  including  snow,  is  estimated  at  10  lbs.  ^r    q   ft   „ 
roof  surface      Determine  the  stress  in  each  member  when  a  wiml  bL     on 
th.V'^lt  T    "  '77  "'  ''  •'"  '"'•  ^^-  "•  "°^'"'^'  *«  *he  roof  surface,  assuming 

£iet^  irrcr  ""^  ^'^  '^  ''-^^'y '--  ^^  '■  (-^>  ^^  ^^-"y  divided 

Ans.  (1)  Stress  in  .4B-  8956.8  lbs.;  fiD  =  10015.2  lbs  -ffC- 
AD^  2503.8  lbs.  =^£;Z)£,'_8196  lbs.;    ' 
^C- 113.56.8  lbs. 
(2)  Stress  in  ^IB-   7756.8  lbs.;  BZ) -6840.9  lbs.  -EC- 
^      J     ,u    .  ^^'   1710  lbs.  =.4£;  ^C -10156.8  lbs.  ' 

■  4«  =  ir    unon  Tr'  T^'^'T'^  ^r'  '^"  «''^«'«P*»-Vi»g  figure,  the  load  upon 
1      c  "  u^  ^" ''   ^^'"^  ''>"^''''  -'^^^^  '^;   the  span  .  '^ 

B(  =.S;    the  ties  .t«,  BD,  AE,  CE  are  equal;   F  and  G  .^Vq 

ar..  the  ini.ldle  p,m„s  of  the  rafters.     Find  the  amount     -V=^%i^^^ 
of  metal  in  the  tie-rods  (wrought  iron)  '  *-' 

Fig.  159. 

Ans.  ^  •?  *Il±SKi±E^'^  '•of_l 
(o)  The  .struts /)F  and  ^(?  are  each  5  ft.;  the /ngle  4«rfi-L/H 

ycr:d  .11......  when  a  .irli::  ST  ^^^HTl  ^,^^,:i: 

SVcoMdiy,  determine  the  stresses  produce,!  in  the  members  of  the  truss  in 

'"'  r '^iT  3;:r 'i^ " "  's^  -'^;:  ~-  *«  --- '^^"m ;;: 

31 238.55 ;  1 9852.35 ;  1 2fi;}3.'6 ;  86i 3.5 ;  1 9999.09; 
"I";  fA;         FD-         CO;  a  A;         GE 

29620.44;  2868.5.10;  78.55.2:22420  44-  21485  10-  ir.-w'iK. 
(2)Stre.s8..sin      BD^        DA-^       DE.  EA-  'eJ.      ^^ 

375^39;   125V/39:   lOOOv/lJ^  875v'39;   112.5v'39- 
«/<•;     ^V;    FD;     CO;     GA;  oF 

(fc)  The  rafters  .\B,  AV  are  of  unequal  length  and  make  angles  of  60°  and 


l) 


EXAMPLES. 


31 


4)=  respertively  with  the  vertical;  the  strut  DF ~7^  ft.;  the  tie  DE  is  hori- 
zontal; the  dead  load  upon  each  rafter -100  lbs.  per  lineal  foot;  the  wind 
I)r(s.siire  normal  to  .4B  -300  lbs.  per  lineal  foot,  rollers  are  placed  at  C.  Find 
the  stresses  in  all  the  members.     The  rtifter  ^B  -45  ft. 

Show  by  dotted  lines  how  the  stress  diagram  will  be  modified: 

(1)  If  the  rollers  are  placed  at  B. 

(2)  If  the  strut  DF  is  omitted. 

(3)  If  a  single  weight  of  500  lbs.  is  concentrated  at  D. 

{(■)  If  it  is  assumed  that  the  horizontal  reaction  is  equally  divided  between 
B  and  (',  show  that  the  stress  in  DE  due  to  a  horizontal  wind  pressure  upon 
Ali  is  nil,  the  angle  ABC  being  30°. 

id)  In  a  given  roof  the  rafters  are  of  pitch-pine,  the  tie-rods  of  wrought 
inin;  the  span  is  60  ft.;  the  trusses  are  12  ft.  centre  to  centre;  DF-b  ft.- 
IJi;  the  angle  ABC -20°;  the  dead  weight  of  the  roof,  including  snow,  is 
0  11.S.  j)er  square  foot  of  roof  surface;  rollers  are  placed  at  C;  a  single  weight 
of  Mm  lbs.  is  suspended  from  F,  and  the  roof  is  also  designed  to  resist  a  nor- 
niiil  wind  pressure  of  26.4  lbs.  per  square  foot  of  roof  surface  on  one  side  AB. 
Determine  the  stresses  in  the  several  members. 

75-  In    the    truss    represented   in   the   accompanying  figure,  the  struts 
III'  1)11,  EG,  EK  are  equal,  and  the  ties  BD,   AD, 
i:.\,  i:C  are  also  equal;   the  load  upon  AB  ia  W„  and 
u|).m  AC  is  H\.    Find  the  weight  of  metal   (wrought 
iron)  in  the  ties. 

Ans.  1  ^  l!L±3(irj  + W^,)_co8M 
18  /  cos  /9  sin  ,3 

(h)  The  rafters  AB,AC  are  inclined  at  60°  to  the  vertical  and  are  each 

40  ft.  in  length.  The  foot  C  rests  on  rollers, 
and  the  foot  B  is  fixed.  The  strut  DF  is 
vertical,  is  10  ft.  long,  and  is  equal  to  the 
.strut  DE  in  length.  Also  AF-HF -10  ft. 
The  dead  load  carried  by  the  rafters  is 
120  lbs.  per  lineal  foot.  Provision  has 
also  to  be  made  for  a  normal  wind  pres- 
sure uixin  AB  of  300  lbs.  per  lineal  foot. 
Draw  the  stress  diagram,  and  show  how  it 
will  be  modified  if  the  strut  DF  is  re- 
Fio.  161.  moved. 

Vertical  reaction  at  B  - 10528  lbs.  both  before  and  after  DF  is  removed. 
Horizontal  reaction  at  B  -  6000  lbs.    The  dotted  lines  show  the  modi- 
fied stresses  for  one  half  of  the  truss. 
76.  The  l)oom  AB  of  the  accompanying  truss  is  supported  at  five  inter- 
niiiliatp    points    dividing    the  length  into  six 

sesrnieiits  each  10  ft.  long.  The  depth  of  the  ^^i  V^^V  ^S^ 
iiu-s  10  ft.  Draw  stress  diagrams  for  the  ^  ■>A2^'^»  ""N^!^^^ 
fallowing  cases:  Fro,   162. 

I ')  A  weight  of  100  lbs.  at  each  intermediate  point  of  support. 
(6)  Weights  of  100,  200,  300,  400,  500  lbs.  in  order  at  these  points. 


.l;/.s, 


82 


THEORY  OF  STRUCTURES. 


Ans.  (a)  Stress  in  a  =  375;  6=325;  c-375;  h-4oO; 
m  =  1 25\/l3 ;  n  =  oO\/5;  o  =  50 v/5 ; 
p  =  25V'l3lbs. 
(6)  Stress  in  a=875;  6  =  825;  c  =  925;  A  =  1350;  <i-1325; 
c  =  1125;  /^1375;  n -oOn/.T;  o  =  100n/5; 
P  =  141§\/13;  9  =  8J\2^i;  r  =  200\/5; 
s  =250^5;  <=458i\/l3;  to=291JvT3  lbs. 


Fig.  163.  •  Pio.  164. 

77-  Determine  the  stresses  in  every  member  of  the  frame  shown  in  Fig.  163. 

78.  Draw  the  stress  diagram  for  the  frame  Fig.  164,  loaded,  and  of  th» 
dimensions  as  shown. 

79.  Each  of  the  joints  A,  B,  C,  D,  E,  F,  G,  Fig.  165,  is  loaded  with  100 
bs.  Find  the  stresses  in  the  several  members  and  distinguish  between  struts 
and  ties. 

80.  The  dead  load  upon  a  roof-truss  of  type  Fig.  166  consists  of  lOOO 
lbs.  at  F,  1000  lbs.  at  A',  and  500  lbs.  at  G;  the  \vind  pressure  is  a  normal 


OLE 

Fio.  105.  Fio.  166. 

forre  of  .30  lbs.  per  square  foot  of  roof  surface  upon  AB;  the  span -90  ft.; 
the  rise  =  25  ft.;  the  tru.sses  are  25  ft.  centre  to  centre.  Find  the  stresses 
in  the  several  inenilj<>rs  when  rollers  are  (a)  at  (',  (b)  at  B. 

Ans.  (a)  Heucfioii    (vertical)     at    ('  =  12291  jj    lbs.;    vertical   reaction    at 
/i=2.3!»,j.S^  lbs.;    horizontal  reaction  at  7? -18750  lbs. 
Tension  in  rt/J  =48(125;    />/.  =.34475;   /.f;-21675;   £C-22125; 

l)H  =  7861 J ;  .1  /.  =  1 588Sr, ;  A7:;  =  250  lbs. 
Coiiif)ressi()n  in  /?F -3060 ijv^  106;  F// -27885^106; 
//.»•  =  \<)77Wm\;  .1 A  =232.->\/roTi; 
AV;  -  21()SJ\/T()0;  ac  =  24.-)8J>/r()6; 
W  =  1572-\/!Ofi;  T.fr-]M)r,\s/\Hi; 
LK  - 83} VTsi ;  Ed  - oOn/FoO  lbs. 


EXAMPLES 


83 


(6)  Only  alteration   in   stresses  is  tha  each  stress  in  the  different 
sections  of  the  horizontal  tie  is  diminished  by  18750  lbs.; 
all  the  remaining  stresses  are  unchanged. 
8i.  In  the  accompanying  roof -truss,  angle  ^BC'=30°;  the  span-90J  ft.; 
DF  -  EG -lOi  ft.;    each  rafter  is  divirled  into  four 
equal    segments    by    the    points    of    support;    the 
trusses  are  20  ft.  centre  to  centre;   the  weight  of  a 
|>ay  of  the  roof  =24,416  lbs.     Determine  the  stress 
Fia.  167.  in  each  member. 

Also  determine  the  stresses  due  to  a  wind  pressure  of  30  lbs.  per  square 
foot  of  roof  surface  acting  normally  to  AB,  when  rollers  are  under  (a)  C 
(6)  B.  ' 

82.  Determine  the  stresses  in  all  the  members  of  the  truss  loaded  and 
of  the  dimensions  xhown  in  Fig.  168. 


Fia.  168. 

83.  The  horizontal  boom  CD,  Fig.  169,  is  divided  into  eight  segments, 
each  S  ft.  long,  by  seven  intermediate  supports;  the  depth  of  the  truss  at 
each  end -16  ft.;  a  weight  of  1  ton  is  concentrated  at  C  and  at  D,  and  a 

Ci    .    .    ,  ^   >    .     .     ,D 

K      I-     O 


Fio.   169. 


weight  of  2  tons  at  each  of  the  points  of  division. 
ill  the  several  members. 


Fio.  170. 
Determine  the  stresses 


84.  Fig.  170  is  a  skeleton  diagram  of  a  roof-tru.ss  of  72  ft.  span  and  12 
ft.  deep;  G,  K,  L,  O,  //  are  rc9i»ctively  the  middle  points  of  AE,  EL  EF 
i.F  IH;  AE  =  EL-LF-FB-20  ft.;  the  tru.ssc-s  are  12  ft.  centre  to  centre'; 
the  ,\vmI  weight  of  the  roof -12  lbs.  per  square  foot;  the  normal  wind 
pn-<iire  uixin  AE  may  be  taken -30  lbs.  per  square  foot;  the  end  .4  is 
tixcl  iind  B  is  on  rollers.  Draw  a  stress  diagram.  Show  by  dotted  Hnes 
bt)w  the  stress  diagram  is  modified  with  rollers  under  A,  B  being  fixed. 

83.  The  platform  of  a  bridge  <>f  84  ft.  span  and  0  ft. 
'Ill  |i  is  carried  by  a  pair  of  trus8i>s  of  {he  tyjx'  shown 
II'  111,  iiffure.  If  the  load  lK)rno  by  each  truss  in  ;100 
ili^  IKT  lineal  fo«)t,  find  the  stresses  in  all  the  moml>ers. 


84 


THEORY  OF  STRUCTURES. 


Ans.  Stress  in  ^B-6000;  AC -\200'^73;  AD-SGOOvTi;  BC-4800' 
CD  - 14400;  D£;- 28800. 
Stress  in  horizontal  chord -48000;  in  each  vertical  -  3600  lbs. 
86.  The   inclined   bars  of  the  trapezoidal  truss   represented   by  Fig.  172 
make  angles  of  4.5°  with  the  vertical;    a  load  of  10  tons  is  applied  at  the 


Fio.  172. 


top  joint  of  the  left  rafter  in  a  direction  of  4.5°  with  the  vertical.  Assuming 
the  reaction  at  the  right  to  be  vertical,  find  the  stresses  in  all  the  pieces  of 
the  frame. 

Ans.  Vertical  reaction  at  D-—\/2;   stressinD£;  =  — v^;  Di5-6|; 

BE-65;  B.l=|v^;  AE^'^v%  AC-'f;  C£-|v/2ton8. 

87.  An  overhanging  roof,  Fig.  173,  is  supported  on  columns  and  loaded 
as  .shown.  Draw  the  stress  diagram  and  also  determine  the  forces  in  the 
vertical  columns  at  A  and  B. 


FiQ.  174. 


V    •  ^( 


Fig.  175. 


88.  A  factory-roof  of  32  ft.  span  is  loaded  as  shown  in  Fig.  174,  and  the 
angle  .1  is  90°.  Determine  the  stres-ses  in  all  the  members.  Also  draw  the 
stress  diagram  when  there  is  a  normal  wind  pressure  on  AB  equivalent  to 
J  ton  at  V  and  B,  and  i  ton  at  D  and  A,  rollers  being  placed,  Ist,  under  D 
and,  2d,  under  E. 

89.  .\  cantilever,  Fig.  175,  is  loaded  and  of  the  dimemsions  as  shown. 
Determine  the  stresses  in  all  the  members. 


Qo,  Draw  tho  ctrpo'  diap-am  for  the  truss  represented  by  Fig.  170,  the 
load  at  each  of  the  points  B  and  C  being  500  lbs. 


EXAMPLES. 


85 


Also,  if  the  rafter  AB  is  subjected  to  a  normal  wind  pressure  of  100 
lbs.  (KT  lineal  foot,  introduce  the  member  BF  to  prevent  distortion  and  state 
ill  'Miuiids  the  stress  it  should  be  designed 
to  bear.  Draw  the  stress  diagram  of  the 
modified  truss,  assuming  that  A  is  fixed, 
but  that  D  rests  upon  rollers. 

(AB -15' -AE;  BC-W;  angle  BAD 
-45°;  angle  £i4Z)- 30°.) 

01.  The  frame  Fig.  177  is  loaded  as  ^^°-  ^'^^■ 

indicated.    Find  the  stresses  in  the  members  and  the  horizontal  tension  in 
the  ciiord  at  the  base,  shown  by  the  dotted  line. 

92.  A  crescent-shaped  roof  of  the  dimensions  and  loaded  as  shown  in 
Fig.  178  is  fixed  at  the  right  supports  and  rests  upon  rollers  at  the  left  support. 


Fig.  178. 

The  joints  of  the  upper  members  lie  in  the  are  of  a  circle  of  24  ft.  radius,  and 
each  load  acts  towards  the  centre  of  this  circle.    Draw  the  stress  diagram. 

93.  A  station  roof  with  trusses  15  ft.  apart  is  supported  on  central  col- 
umns (Fig.  179)  and  the  weight  of  the  roof-covering  and  truss  may  be  taken 
at  30  lbs.  per  square  foot  of  covered  area.  A  horizontal  wind  pressure  exerts 
an  a.lditional  force  of  30  lbs.  per  square  foot  of  projected  vertical  area.    De- 


20 — ■* 


Ikj.  179. 


oik.ifoib.inon. 


MO  Ik... 


Flu.    180. 


Fio.  181. 


trrininc  the  stresses  in  the  several  members,  assuming  that  the  point  of  inter- 
'cction  of  the  total  resultant  force  with  the  vertical  post  is  a  virtual  hinge. 

94.  Draw  the  stress  diagrams  for  the  roof-truss  Fig.  180  when  subjected 
to  a  horizontal  wind  load  of  30  lbs.  per  square  foot  of  projected  vertical  area. 


86 


THEORY  OF  STRUCTCRES 


96.  lu    the   accompanying   roof-truss   .4B-.1C=30   ft.,   and   the   struts 

are  all  normal  to  the  rafters.  Find  the 
stresses  in  all  the  members,  the  load  at 
each  of  the  joints  in  thfc  rafters  being  o 
tons  (angle  ABC -SO"  and  angle  DBC 
-10°).  How  will  the  stresses  be  modi- 
fied if  there  is  a  force  of  2  tons  acting  at 
Jr.         .  ,  ''"'■'^  "^  ^^^   points   of  support   f)etwppn    4 

and  li  at  nght  angles  to  the  rafter,  an<l  a  force  of  1  ton  at Tisum  n!r  th'! 

*hc  end  B  is  fixed  and  that  C  rests  u,Km  .oilers?  '  ^  *''"' 

97.  Tile  load  upon  a  roof-truss  of  the  tvpc  Fig   18.3  is  1000  lh=    «f  ».„k 
Jo.nt;   the  .span  100  ft.;   the  rise=.>5  ft.     Fi^d  the  stTe CTth^'diLrn 
memlK>rs.     How  will  the  stres.ses  be  affected  by  an  additional  load  olS 
lbs.  at  each  of  the  jomts  between  the  foo^and  ridge  on  one  side? 
Ans.  Stress  in  BD-ooOO\/5;  DF -5000V  Z; 

FH  - 4.500^5;  ///.  =  4000^51  LN  =  3500v T" 

A'.1=300(V5;  D£  =  0;  FG  =  500;  /fA'-lOOO; 

LM  - 1.-)00;  ATO -2000;  .4P  =  .W00; 

BE^Uim  =  EG;  GA'  =  10000;  AM/ -9000; 

^/O  =  8000;  0P  =  7000;  TX;^-m\/b; 

FA'-1000\/2;  //.l/  =  500Vi3;  LO~lOQ0Vb; 

ArP=-500v/2'Jlb3. 


Fig.  183. 


Fio.  184. 


98.  The  frame  Fig.  184  supports  a  load  of  1  ton  at  A.  Find  the  stresses 
in  the  several  n.eml)ers,  and  .-.Iso  the  overturning  moment  at  the  Toot  of T 
post,wh.ch  .s  18ft.inlength.     .^C-l'.G.)     Also  s^^o.Lt i^: ^^^'Jlt 


EXAMPLES. 


87 


mndifi'jd  when  the  weight  hangs  by  a  chain  in  two  falls,  the  chain  passing 
over  a  pulley  at  A  along  a  horizontal  line  from  A  to  the  post. 

99.  A  braced  semi-arch  is  10  ft.  deep  at  th(>  wall  and  projects  40  ft.  The 
ufipiT  flang"  is  horizontal,  is  divided  into  four  equal  bays,  and  carries  a  uni- 
foniilv  di.stiibuted  load  of  40  tons.  The  lower  flange  forms  the  segment  of 
a  circle  of  104  ft.  radius.  The  bracing  consists  of  a  series  of  isosceles  triangles, 
of  wliich  the  bases  are  the  equal  bays  of  the  upper  flange.  Determine  the 
stresses  in  all  the  members. 

100.  A  bowstring  roof-truss,  with  vertical  and  diagonal  bracing,  of  50  ft. 
rise  and  five  panels  is  to  be  designed  to  resist  a  wind  blowing  horizontally 
mth  a  pressure  of  40  lbs.  per  square  foot.  The  depth  of  the  truss  at  the  centre 
is  10  ft.  Determine  graphically  the  stresses  in  the  several  members  of  the 
truss,  assuming  that  the  roof  rests  on  rollers  at  the  windward  support. 

101.  The  inner  flange  of  a  bent  crane  forms  a  quadrant  of  a  circle  of  20  ft. 
radius  and  is  divided  into  four  equal  bays.  The  outer  flange  forms  the  seg- 
ment of  a  circle  of  23  ft.  radius.  The  two  flanges  are  5  ft.  apart  at  the  foot, 
atid  are  struck  from  centres  in  the  same  horizontal  line.  The  bracing  con- 
sists of  a  series  of  isosceles  triangles,  of  which  the  bases  are  the  equal  bays 
of  the  inner  flange.  The  crane  is  required  to  lift  a  weight  of  10  tons.  De- 
termine the  stresses  in  all  the  members. 

102.  The  domed  roof  of  a  gaj-holder  for  a  clear  span  of  80  ft.  is  strengt>h- 
ened  by  secondary  and  primary  trussmg  as  in  the  figure.  The  points  B  and 
C  are  connected  by  the  tie  BPC  passing  beneath 
the  central  strut  AP,  which  is  15  ft.  long,  and 
is  also  common  to  all  the  primary  trusses;  the 
rise  of  A  above  the  horizi  ♦al  is  5  ft.;  the  second- 
ary truss  ABEF  consists  ot  the  equal  bays  AH, 
HG,  OR,  the  ties  BE,  EF,  FA,  of  which  BE  is  horizontal,  and  the  struts  GE, 
Fll,  which  are  each  2  ft.  6  in.  long  and  are  parallel  to  the  radius  to  the  centre 
of  (rH;  the  secondary  truss  ACLK  is  similar  to  ABEF;  when  the  holder  is 
emptv  the  weight  supported  by  the  truss  is  36,000  lbs.,  which  may  be  as- 
suni>>d  to  be  concentrated  at  G,  H,  A,  M,  N,  in  the  proportions  8000,  4000, 
Um,  4000,  and  8000  lbs.,  respectively.  Determine  the  stresses  in  the  difl 
fcretit  members  of  the  truss. 


Fro.  185. 


103.  The  top  beam  of  a  roof  for  a  clear  span  of  90  ft.  consists  of  six  bars, 
Ali.  BC,  CD,  DE,  EF,  FG,  equal  in  length  and  so  placed  that  A,  B,  C,  D, 
/;,  /',  G  are  on  circle  of  80  ft.  radius;  the  lower  boom  also  consists  of  six  equal 
rods,  AH,  HK,  KL,  LM,  MN,  NG,  the  points  H,  K,  L.  M,  and  AT  being  on 
a  (irdc  of  148  ft.  radius;  B  is  connected  with  Jl,  C  with  A',  D  with  L,  E  with 
\l,  :m<\  F  with  N\  the  opposite  corners  of  the  bays  are  connected  by  cross- 
braces;  the  end  A  is  fixed  to  its  support,  G  being  allowed  to  slide  freely  over 
a  smooth  bedplate.      Determine  graphically  the  stresses  in  the  various  mem- 


«^. 


88 


THEORY  OF  STRUCTURES. 


bcrs  when  there  is  .    normal  nnnd  pressure  per  lineal  foot  of  460  lbs.  upon 
A  IS,  340  lbs.  upon  BC,  and  60  lbs.  upon  CD. 

104.  A  lattice-work  arch  of  the  dimensions  and  loaded  as  in  Fig  186  is 
pin-jointcd  at  the  supports.  Find  the  line  of  resistance  which  wiU  pass 
tnrough  the  pm  joints  and  also  through  the  point  where  the  greatest  load 


Fio.   187. 

intrrsorts  the  upper  chord  of  the  arch.    Also  determine  the  horizontal  thrust 
of  the  iirch  under  the  given  loading. 

004  n';/'"'  'n    '■.^''?""*";\  *^«    «'^''''«   ("^  bowstring)  truss  f„r  «  roof  nf 
204  ft.  H,,M„.      1  ho  j,.,Mt.  (.f  the  upix>r  and  lower  Ix.ws  ho  on  tho  arcs  of  narab 
«a.s,  th..  n.o  of  ,l,o  i„wor  bow  iK-ing   17  ft.  nml  of  tho  upnor  bow  Ti  f t 
Dmw  tho  .iross  .liagnun  for  tho  .load  weight  of  the  truss,  which  is  very  aol 
proxmiatoly  o.iuivalont  to  IJ  tons  iht  rmnol.  i,o..  to  IJ  ton.,  ,.„nrpntrat«l  ^t 
each  cf  the  intermediate  joints  in  the  upjx-r  l)ow.    The   trusses  are  24  ft 


EXAMPLES. 


» 


apart,  and  the  load  on  each  truss,  including  snow  and  wind  pressure,  is  assumed 
to  be^  40  lbs.  per  square  foot  of  covered  area.  Draw  the  stress  diagram  when 
this  load  covers  (o)  one  half  of  the  truss,  (6)  three  fourths  of  the  truss.  Also 
show  how  the  diagrams  are  modified  by  the  introduction  of  the  dotted  members. 

1 06.  Each  rib  for  the  support  of  a  dome  of  320  ft.  diameter  consists  of 
two  concentric  booms  6.4  ft.  apart  and  connected  by  triangular  bracing  which 
divides  the  rib  into  15  bays  of  equal  length.  The  rib  is  pinned  at  the  centre 
and  at  the  two  supports  {A).    The  dead  load  on  each  of  the  exterior  jointa 


Fio.  188. 

1  to  15  is  8000  lbs.  Determine  the  horizontal  thrust  and  draw  a  stress  dia> 
pram  for  the  members  between  .4  and  B.  Also  determine  the  horizontal 
tlirusts  at  A  and  B  and  draw  the  stress  diagram  (a)  for  a  load  of  32  tons  at 
joint  So.  7,  and  (6)  for  loads  of  24,  28,  44,  and  48  tons,  at  the  joints  Nos.  5, 
0,  11,  and  14,  respectively. 


107.  In  the  preceding  example,  draw  the  stress  diagram  for  a  normal  load 
of  10  tons  at  the  joint  No.  8,  passing  through  the  centre  0  of  the  dome. 

108.  The  figure  represents  a  portion  of  a  Warren  girder  cut  off  by  the 
pl.mp  .U.V  and  sup|)orted  upon  the  abutment  at  A. 
The  reaction  at  .4-20  tons;    the  load  concentrated 
at  each  of  the  points  B-4  tons.     Find  the  stresses 
in  lach  of  the  members  met  by  MN. 

.I"i.  Stress  in  tension  _chord  -'J'v/T  tons;  in 
coniprcssion  chord  -  32V3  tons;  comprewion  in 
«im)?onal-l«V3  tons. 


FiQ.  189. 


90 


TIIEOR,    OF  STRUCTURES. 


109.  The  figure  is  a  portion  of  a  bridge-truss  rut  off  by  the  plane  AfN 
ff  and  supported  upon  tho  abutment  at  A-    AC-CE 

-'"  '•      thedopthBr=/)i'  =  i7jft.;   inthe'third 


=  14s'fft. 


M,Mo  ibi  P*"*"'  ♦*'«  compression  in  the  upi)er  chord  is  64,600 
lbs. ;    the  tension  in  the  lower  chord  is  53,800  lbs. 


.h«s.mibi. 


Fmd  the  reaction  at  A,  the  equal  weights  supported 
at  t  and  A,  and  the  diagonal  stress  T 

.4«,    Reaction  =3894.3    lbs.;   weight   at    C  and  at 
„«    TK    r  E^V2%A\h^.;   T  -  1 0.578  lbs. 

no  rhe  figure  repres.>nt8  a  bowstring  truss  of  80  ft.  span  cut  off  bv  tl. 
plane  1/A  a,.d  sup.K.rtcd  at  0.  The  upper  flange  P""' "'*  "^^  ^'>  *'' 
Ot  l)h  IS  an  arc  of  a  circle  of  Ho  ft.  radius;  OA^AB 
-etc.  =  10  ft.;  the  rise  of  the  truss  =  10  ft.;  a  load 
Of  lo  tons  IS  conccntn.t..d  at  each  of  the  jx^ints  .1  and 
^:  the  reaction  at  0-J.5  tons.  Find  the  stresses 
in  the  members  cut  bv  the  plane  MS 

.1..S.7',- 94.827  tons;  T,- 9.5.744  tons;  7>  =  .97tons. 
"1.  Ihe  figure  represents  a  portion  of  a  roof-tru.ss  cut  off  bv  a  plane  1/V 
and  supp<jrted  at  .4.  The  strut  DC  is  vertical; 
AD^Xi  ft.,  and  the  distance  of  D  from  .4f'-7J 
ft.;  the  angle  between  AC  and  the  horizontal 
-cos  'J;  the  vertical  reaction  at  .1-7  tons' 
the  horizontal  reaction  at  .4  =2J  tons;  at  each 
of  the  points  B  and  C  a  weight  of  4  tons  is  con- 
centrated. Find  the  stresses  in  the  memliers 
met  by  MN.  (AD  and  T,  make  equal  angles 
with  the  rafter  and  DB  -  DC.) 

„,    Fi„    .OQ  ■'*"■'■  '^>-'3-«"''s:  7'="2.1  tons;  r.-lo.Stons 

-pi^ried  at   f    -ZTl  "  '^"i.""  °^  "  "•-'"'"^'  ^^''^''-t ^"««  «i«h  one  end 
supported  at  .1.    The  load  at  each  panel-point  is  W  and  the  dimensions  are 

M 


192. 


«^indicated.     Find  the  stres^-s  in  the  members  ..ut  by  a  vertical  .-ction 

•4rM.  C- II. SOU'    T-TIW-  n     J -III 
113.   Find,  by  the   '  mcth,Hl  of  .s,.cti.ms  "  the  stress-,  in    he  m......  /  .V 

tru..8  Fig.  194  cut  by  a  vertical  section  MX  fZZ  nu  n  Ix-rs  of  the 

as  indicated  '  ''*^  '^""''"'"ons  and  l(»ds  bein^ 


EXAMPLES. 


91 


115.  The  cantilever  roof-tnisa  Fig.  195   is  supported   at   C  and  by  the 
stay  AB.    Find  the  position  of  the  load  which  will  produce  no  streaa  in  sr. 


Fio.  195. 


Fig.  190. 


Hence  also  find  the  maximum  tensile  and  compressive  stresses  in  the  mem- 
licrs  ptj,  qr,  rs,  sq. 

116.  The  roof-truFS  Fig.  196.  of  40  ft.  span  and  13  ft.  rise,  is  loaded  with 
4  tons  at  D  and  5  tons  at  C.  Find  the  load  P  at  £  which  will  make  the  stress 
diagram  close.    The  struts  are  each  4  ft.  in  length. 

117.  Fig.  197  is  the  truss  for  a  factory  roof  of  72  ft.  span  and  18  ft.  rise. 

It  St. 


"«V«" 


Fio.  197. 


Fio.  198. 


It  is  loaded  as  indicated.     Draw  the  stress  diagram. 

118.  Fig.  198  is  a  six-panel  truss  of  48  ft.  span,  the  dimensions  and  load- 
iug  being  as  indicated.     Determine  the  stresses  in  all  the  members. 

119.  A  ship's  gangway  is  loaded  and  of  the  dimensions  shown  in  Fig.  199. 

M 


Fia.  190. 

The  end  A  rests  upon  rollers,  and  B  is  hinged  to  the  wharf.    When  B  is  10 
ft.  vertically  abo^'e  ^1,  find  the  stresses  in  all  the  members. 

no.  The  end  A  of  the  cantilever  truss  Fig.  200  rests  upon  rolleni.  B 
H  hmged  and  the  bearing  at  C  produced  a  horiiontal  thrust.  Draw  the  streM 
diagram  for  the  given  loads. 


92 


THEORY  OF  STRUCTURES. 


"1.  Tie  cantilever  span  Fig.  201  i»  loaded  with  5500  lbs.  per  foot  run 


♦ iit! 


Fia.  201. 


J.. „^ ^ 

Fig.  202. 


123 


Fio.  203. 


•.  The  dead  weight  of  a  draw-span  having  equal  arms  of  100  ft    Fie 
B  202,  is  1200  lbs.    per  lineal  ft.' 

of  truss.  Determine  the  stresses 
in  all  the  members  when  the 
bridge  is  open. 

"3.  Fig.  203  is  the  skeleton 
diagram  of  a  cantilever  for  a 
viaduct  in  India.  Determine  the 
stresses  in  the  various  members 
under  the  indicated  loading. 

?l  'Z';'' J'-'  /^^  '•'"'^'^  °'  "^^h  panel- 
-ir',^  n  u^  ^.l^r^'"  *■"'«''*  «^  ™»tre  span) 
-ir.,800  lbs.;  the  width  of  roadway -15  ft  the 
oad  p,.r  square  foot  of  roadway -80  lbs.  Find 
the   stresses   in    all    the   members,   assuming   the  Fio.  204. 

reaction  at  the  pier  F  to  be  vertical 

Ans.  /,-<, -28000;  <. -36500;  <, -45000;  <. -53500- 

r. -55200;  ^-48400;  .,-41600-^.';  c.-S/sT- 
C.-7300V/34;  v,-10200;  r,-ir,300;  V.-20400. 

"5.  The  truss  Fig.  205  is  hinged  at  A  and  B  fh» 
reaction  at  B  being  horizontal.  The  loads  and  dlmln! 
sions  are  as  indicated.     Draw  th.  stress  diagram. 

n«n'f!"  '"*",•  "P'^'.  *""'"    °^    »'"'    ^"ti'-als   o^    an    8- 
panel  truss   le  ,n   the  arc  of  a  parabola  whose  param- 
I-  or  IS  12S  ft.,  the  depth  at    the  centre   being   3"  S 
Draw  stress  diagram,  (a)  for  a  load  of  4  tons  at  each 
Fio.  205.  pane -point;    (6)   for    an    additional    load    of    6    ton, 

1    .  '^   "'^''  Second,  and  third  DanelDninta     /Tk.„..  u 

Bingle-intersection  type.)  ^        ^     '•    ^Through 


EX  AMPLES. 


Am. 


Fio.  206. 

IJ7.  The  loads  upon  the  truss  of  a  swing-bridge  when  open  are  as  indi- 
cated in  Fig.  207,  C  being  the  counterweight  at  D  and  R  the  reaction  at  each 


Fig.  207. 


Fio.  208. 


of  the  bearing  points  A  and  B.  Find  R  and  C,  and  draw  a  stress  diagram 
for  the  truss  under  the  given  loading.  The  panels  are  each  6  ft.  in  length 
and  the  truss  is  8  ft.  deep  at  the  Ist  vertical  and  12  ft.  deep  over  the  turn- 
table. 


ij8.  a  /our-panel  A  truss,  Fig.  20S,  of  100  ft.  span  is  40  ft.  deep.  The 
panel  dead  load  is  18,7.50  lbs.  Determine  the  stresses  in  all  the  members. 
Also  determine  the  stresses  in  the  several  members  due  to  a  panel  live  load 
of  53,250  lbs.  concentrated  at  the  2d  and  3d  panel-points. 

lag.  A  Warren  girder  80  ft.  long  is  formed  of  five  equilateral  triangles. 
WVifthts  of  2,  3,  4,  5  tons  are  concentrated,  respectively,  at  the  1st,  2d,  3d, 
and  4th  apex  along  the  upper  chord.  Determine  the  stresses  in  all  the  mem- 
bers of  the  girder. 

Ans.  Compression  chord :  Stress  in  1st  bay-2v  3;  2d-5§v^3;  3d-7v/3 
4th-6J\  3;  5th -2§\  3.     _ 
Tension  chord:    Stress  in  Ist  bay-4v3;   2d-6|v3;   3d-7Jv^3 

4th -5J\  3. 
fyiagonaU:    Stress    in    Ist    and    2d-4v''3;    3d    and    4th-2§v'3 
5th  and  Gth-§v  3;  7th  and  8th-2v'3;  0th  and  10th 
-5i\  3. 
130.  ,\  Warren  girder  of  60  ft.  span,  composed  of  six  equilateral  triangles, 
<  arrics  upon  its  lower  chord  a  weight  of  2  tons  at  the  first  and  second  joints, 
!.■>  tons  at  the  centre  joint,  and  7i  tons  at  the  fourth  and  fifth  joints.    Find 
the  stresses  in  all  the  members. 


94 


THEORY  OF  STRUCTURES. 


Ans.  Stresses  in  tension  chord:  1st  bay-V  v  3;  2(1-^^3;  Sd-'^VVs" 
4th  =  V.'-V  3;  5th  -  V\  3;    6th  - 15  v'a^  ' 


Vv3;   2d- 


»5  * 
It     A 


Sd-WaT; 


130- 


•*L  V    M    N   n   p  r)    p  5 


Stresses  in  compr.  rhord:   1st  bay 

4th -Vv  3;  5th  =  Vv3. 
Diag^strosses  ,sj  and  2d  bays  -  Vy3-;  3d  and  4th  -  V  V3-;  5th  and 

Oth=  o\  3^ ah  and  Sth-VV3;  9th  and  10th -VVS-  Hth  and 

12th  =  '/v'3. 

I3t.  Determine  the  stresses  in  the  members  of  a  Fink  truss  of  24n  ff  .« 
and.  sixteen  panes;    depth  of  truss -SO  ft-    .,"/'";  ""^  o' 240  ft,  span 
64  tons.  ^  truss -dO  ft.,    umformly  distributed  load- 

Ans.  Stress  in   ^-4 -Vs-B.l/ -/).!/_ DO- FO - 
FQ^QH-HS;  in  C.4 -4V2-C0-G0-(?5;  in  EA 
e  c  'n^I  ^"^^      -Sv  5  =  £S;in,4A'-16v  17;  inBLi-DN-FP-HR 
%ro.^4°  "    '  ;"/o'5o-,ll^f  ^ '"  '^'''■'  -  A'.-32;  in  Xt-^ 
13a.  Determine  the  stresses  in  the  members  of  »  R«ii~       . 
long  and  12J  ft.  deep,  under  a  uniformlyXTri^uted  load  oJZt    ""*  'V' 
wuh  a  single  load  of  25  tons  con^-entrated  at  25  f t.1n.m  «„«  InT         '     '^'^' 
Ans.    Stress  in  ^B-^iiAv'2;    BL-TvJ;    ^Z)- Ws' Z)L-V  v/^r 

UK  -ete    Z)£-5G  tons;  compression  along  ^L-l93f  tons. 

f C  E  0  ^ 


Fio.  210. 
■»n.r...d  M "he  'e„r«     «,,  ™      .■"?,'■  ''"1'°  •  lo"!  «'  «)  ton,  (a)  con. 

Jo™,,.  .i«Hb„,T«Tdirur;ri" ',1'  i'^.'^rr-v  <-'  -■ 


Fi<i.  211 —ruse  rf. 


EXAMPLES. 


05 


134-  Draw  a  stress  diagram  for  the  A  truss,  Fig.  212,  (a)  due  to  a  dead 
load  of  600  lbs.  per  lineal  foot  of  truss,  (6)  due  to  a  live  load  of  1200  lbs.  per 
lineal  foot  of  truss  covering  the  first  two,  three,  jour,  and  six  panels,  respectively. 

135.  A  Baltimore  truss  300  ft.  long,  one  half  of  which  is  shown  in  Fig. 


Fig.  212. 


Fio.  213. 


!:'^;^*!r*'  *  '°*?°'  ^'^  "*•  "^  '""'^  panel-point  of  the  lower  chord  and 
40.000  lbs.  at  each  panel-point  of  the  top  chord.  Find  the  stresses  in  aU  the 
members. 

136.  The  lower  lateral  system  of  a  cantilever  arm,  Fig.  214,  carries  at  the 
outer  end  a  honzontal  wind  load  of  4/2,080  lbs.  from  the  suspended  span. 


B  panel*  ««7'7tf 


Flo.  214. 


Firul  the  stresses  in  all  members,  assuming  that  the  diagonal  stresses  are 
<  liially  divided  between  the  two  systems  of  diagonals. 

-,w/i3;  I^^  '"""t'l^'vpr  arm  shown  in   Fig.   215  carries  a  vertical   load  of 
.I«),(K)0  lbs.  at  Its  outer  extremity.     Find  the  stresses  in  all  members. 

138.  One  arm  of  a  swing  span  is  loaded  as  shown  in  Fig.  216  Find  all 
t  I-  sfrpsses  when  the  span  is  open,  assuming  that  of  each  panel  load  32  000 
II-.  IS  concentrated  at  the  top  chord  and  the  remainder  at  the  bottom  chord'. 

.,f  u^'ful^'»    "'1  !!?''  ^^TT.'''  ^^^  ^'■"^'  *"'«•  217,  for  an  upward  reaction 
"f  100,000  lbs.  at  the  end  of  the  span. 


PI*W^ 


96 


THEORY  OF  STRUCTURES. 


Fig.  215. 


7  pftnpls  &  k' 


Fio.  216. 


Fio.  217. 


i^:      ' 


iRl^n- 


EXAMPLES.  97 

140.  In  the  same  truss,  Fig.  218,  a  load  of  100,000  lbs.  is  applied  at  the 


Fio.  218. 

third  panel-point  from  the  left.    Assuming  that  the  reaction  at  the  left  sup- 
port  IS  42,000,  find  the  stresses  in  all  members. 

•    t'.*'*  A^!"*^'?'"  ^**'^*  ^"^  *'""^3  of  tfie  dimensions  and  loaded  as  shown 
m  fig.  219  has  the  joints  of  the 


lower  chord  lying  in  the  are  of  a 
parabola.  Draw  the  stress  dia- 
pam.  Verify  your  results  by 
using  the  method  of  sections  to 
determine  the  stresses  in  the 
members  intersected  by  the 
cutting-planes  MN,  M'N'. 


•«    -.    -,  'f 


Fig.  219. 


142.  Draw  the  stress  diagram  for  the  3-pin  arch.  Fig.  220,  the  left  truss 
being  loaded  with  1200  lbs.  per  lineal  foot. 

143.  Fig.  221  is  the  3-hinged  truss  for  a  bridge  hinged  at  C  and  at  the 


Fio.  220.  Pio.  221. 

two  abutments  A  and  B.  The  load  at  each  of  the  points  1  and  2  is  5  tons 
atiij  10  tons  are  concentrated  at  each  of  the  intermediate  ^anel-points.  The 
bracing  consists  of  isosceles  triangles  with  the  equal  bays  of  the  horizontal 
ihurds  as  bases,  and  the  joints  in  the  lower  chord  lie  in  the  arc  of  a  circle  of 
!Hi  ft,  radius.  The  depth  of  the  truss  at  the  abutments  is  18  ft.  Deter- 
nime  the  reactions  at  A,  B,  and  C,  and  draw  a  stress  diagram  for  the  truss. 

144-  Fig.  222  represents  one  of  the  two  symmetrical  trusses  for  the  central 
sjiaii  of  a  suspension  bridge.    The  joints  in  the  upper  chord  he  on  a  paraboUo 


98 


THEORY  OF  STRUCTURES. 


arc  and  consecutive  joints  are  connected  by  straight  links.  The  dead  load 
per  lineal  foot  of  truss  is  750  lbs.,  and  the  two  intermediate  trusses  are  pinned 
at  the  centre  and  at  the  piers.     Determine  the  stresses  in  the  members  met 


Fro.  222. 
by  the  sections  MN  and  M'N'.    Also  determine  the  stresses  in  the  same 
members  when  a  hve  load  of  1250  lbs.  per  lineal  foot  covers  (a)  the  first 
two  panels,  (6)  the  panels  from  A  to  C,  (c)  the  panels  from  C  to  B. 

145.  Each  shore  arm  in  the  suspension  bridge  of  the  preceding  example 
18  identical  with  one  half  of  the  central  span.  Determine  the  stresses  in  the 
corresponding  members  of  the  shore  arms  (a)  for  the  dead  load  of  750  lbs  per 
lineal  foot,  (h)  when  the  live  load  of  1250  lbs.  per  lineal  foot  covers  first  the 
shore  arm  and  second  one  half  of  the  central  span. 

U6.  A  braced  pier  is  of  the  dimensions  and  loaded  as  in  Fig  223  Deter 
mine  the  stresses  in  all  the  members  and  also  find  the  forces  in  the  vertical 
anchorage-bars. 

147.  Draw  a  stress  diagram  for  the  braced  pier  Fig.  224,  of  the  dimen- 

ettMl* 


0-»' 


JX}. 


Q^-» 


li 


B 

Fio    225 


Fig.  22.3.  pic.  224. 

Si?."/'  1'"'*';' »'"  "''T-     "^'f  u"^  '^'  ''^^"'*^"*  ^'^rti'^'^l  «"'!  horizontal 
forces  at  .1  an.l  B,  specifying  whether  they  are  in  tension  or  compression. 

148.  A  tmstlc  Fig.  22.5.  30  ft.  high  and  10  ft.  wide  is  loaded  as  shown. 
Draw  ih,.  s  ro...  ,l,,,.r.,n  and  find  ih,  character  and  magnitude  of  ,he  res^- 
ant  vertical  and  horizontal  forces  at  .1  \,hen  (a)P^2Q,  (b)  P^3Q  (r)  P-5Q 


EXAMPLES. 


99 


149.  The  figure  represents  a  pier,  square  in  plan,  supporting  the  ends 
of  two  deck-trusses,  each  200  ft.  long  and  30  ft.  deep.    The  height  of  the 

pier  is  .50   ft.  and  is  made  up  of  three  panels,  the      

upper  and   lower   being  each   17   ft.   deep.      Ten 

square   feet    of   bridge  surface  and  10  square  feet 

of  train  surface  per  lineal  foot  are  subjected  to 

a  \rind  pressure  of  40  lbs.  per  square  foot.    The 

centre    of    pressure  for  the  bridge  is  68  ft.,  and 

for  the  train  86  ft.,  above  the  pier's  base.      The 

wind  also  produces  a  horizontal  pressure  of  4000 

lbs.  at  each   of   the  intermediate  panel-points  on  Fio.  226. 

the  windward  side  of  the  pier.    Width  of  pier -17  ft.  at  top  and  33|  ft.  at 

bottom.    The  bridge  load -1600  lbs.  per  lineal  foot,  live  load -3000  lbs. 

per  lineal  foot.    Determine: 

(a)  The  overturning  moment.     (3180  ft.-tons.) 

(6)  The  horizontal  force  due  to  the  wind  at  the  top  of  the  pier.    (61.6  tons.) 
(f)  The  tension  in  the  vertical  anchorage-ties  at  S  and  T,    (Nil.) 
{(I)  The  vertical  and  horizontal  reactions  at  T.    (275  and  65.6  tons.) 
Draw  a  diagram  giving  the  wind  stresses  in  all  the  members,  and  indi- 
cate which  are  in  tension  and  which  in  compression. 

Ascertain  whether  the  wind  pressure  of  40  lbs.  per  square  foot  upon  a 
tram  of  empty  cars  weighing  900  lbs.  per  Uneal  foot  will  produce  a  tension 
anywhere  in  the  incUned  posts.  What  will  be  the  tension  in  the  anchoraue- 
tics?    (20.75  tons.)  ^ 

tso.  The  figure  represents  one  half  of  one  .  f  the  piers  of  the  Bouble 
•Viaduct.    The   spans   arf  sed   by   two  lattice  gir- 

ders 14'  9"  deep  and  ha\  ,  a  deck  platform.  The 
height  of  the  pier  is  183  9"  and  is  made  up  of 
eleven  panels  of  equal  depth.  Width  of  pier  at  top  - 
13'  li",  at  bottom  »  67'  7".  With  wind  pressure  at 
65.3  lbs.  per  square  foot,  the  total  pressure  on  the 
girder,  train,  and  pier  have  been  calculated  to  be  20, 
16.2,  and  20  tons,  acting  at  points  196.2,  210.3,  and 
92.85  ft.,  respectively,  above  the  base.  The  dead 
weight  upon  each  half  pier  is  222J  tons,  of  which  60 
tons  is  weight  of  half  span,  120  tons  the  weight  of  the 
half  pier,  and  42J  tons  the  weight  of  the  train.  As- 
suming that  the  wind  pressure  on  the  pier  is  a  hori- 
zontal force  of  2  tons  at  each  panel-point  on  the  wind- 
ward side,  and  that  the  weight  of  the  pier  may  be  con- 
sidered as  a  weight  of  6  tons  at  each  panel-point,  deter- 
mine: 

(o)  The  overturning  moment. 

(6)  The   total    horizontal  force  at  the  top  of   the 


pier 


Iio.  227. 
<tuL'  to  the  wind. 


100 


THEORY  OF  STRUCTURES. 


the  tinr^JsTut"  "  "'"'*  "'  '^^  ""'""'  «"'''»"™««-t-«  -^  S  and  T  due  to 
(rf)  The  vertical  and  horizontal  reactions  at  T 

that' iraliL'TStr'""'"  ^'"" """" '"  ^'^^  •"^'"^^  «^'  -^ 

Draw  a  stress  diagram  giving  the  stresses  in  all  the  members,  indicatine 
«Wh  are  -  te-- a.d  which  in  compression.    Width  of  pier  at  A  -S  ft.! 

as  shown'l^Tig.'yr"  '"''"  '"  "  '^'''  '""^"^  '"^  ''  *^^  *'°«'^°'» 


15a.  Determine  the  stresses  in  the  several  members  of  a  steel  trestle 
shown  by  F.g.  229  when  subjected  to  horizontal  ,vind  lo!ds  7f  43S 
24(»    3600  and  4800  lbs.  at  A,  C,  D.  and  E.  respectively!^ /if eS 

«  33J  ft.    The  width  of  the  trestle  at  the  top  is  8  ft.  and  at  the  bottom 


il 


j; 


CHAPTER  II. 


SHEARING  FORCES  AND  BENDING  MOMENTS. 

.Vote.-In  this  chapter  it  is  assumed  that  all  forces  act  in  one  and  the  same 
plane,  and  that  the  deformations  are  so  small  as  to  make  no  sensible  alteration 
either  in  the  forces  or  in  their  relative  positions. 

I.  Equilibrium  of  Beams. 

(^\SE  I.  AB  is  a  beam  resting  upon  two  supports  in  the  same 
horizontal  plane.    The  reactions  Ri  and  R2 
at  the  points   of  support  are  vertical  and 
the   resultant  P  of  the  remaining  external  ^^^      ^ 
forces  must  also  act  vertically  in  an  oppo-  fio.  231. 

site  direction  at  some  point  C.    Accordmg  to  the  principle  of  the  lever. 


lA 


T. 


m 


UC 


^2=fjg,    and    Ri+R2=P. 


Fig.  232. 


Case  II.  AB  is  a  beam  supported  or  fixed  at  one  end,  and  such 
a  beam  is  often  called  a  cantilever  or  semi-girder.    The  fixture  at  A 
tends   to   prevent  any   deviation   from   the 
,0  straight  in  that  portion  of  the  beam,  and 
the  less  the  deviation  the  more  perfect  is 
the  fixture. 

The  ends   may  be   fixed   by  means  of 
3B  two  props  (Fig.  232),  or  by  allowing  it  to 
rest  upon  one  prop  and  preventing  upward 
motion  by  a  ledge  (Fig.  233),  or  by  buUd- 
ing  it  into  a  wall  (Fig.  234). 
>B        In   any   ca.se   it   may   be  assumed   that 
the  effect   of   the   fixture,    whether  perfect 
or    imperfect,    is    to    develop    two   unequal 
torces,  Q  and  R,  acting  in  opposite  directions  at  points  M  and  AT. 

101 


1  1  *" 


Fio.  233. 


Ac 


4In- 


ip 


IT 


Fio.  234. 


102 


THEORY  OF  STRUCTURES. 


These  two  forces  are  equivalo.if  to  a  left-handed  couple  (Q  -0^ 
the  moment  of  which  is  Q-MX,  and  to  a  single  forco  R-Q^tN 
Hence  R-Q  must   =P.  ■«         »• 

Case  III.  AB  is  an  inclined  beam  supported  at  A  and  resting 
upon  a  smooth  vertical  surface  at  B. 

The  vortical  weight  P,  acting  at  the 
I)oint  C,  is  the  resultant  load  upon  AB. 
Lot  the  direction  of  P  meet  the  horizontal 
line  of  reaction  at  B  in  the  point  D. 

The   beam   is   kept    m   equilibrium   by 
the  weight  P,  the  reaction  Ri  tit  A,  and 
the  reaction  /?,  at  B.     Now  the  two  forces 
/(•_>  and  P  meet  at  D,  so  that  the  force  /?,  must  also  pass  through  D 
Honco  /?,  =P  soc  J/;^  and  R^  =P  tan  ADC. 
AVi.'e.— The  same  principles  hold  if  the  Ijoam  in  Cases  I  and  II 
IS  mclm.Ml,  and  also  in  Case  III  whatever  mav  be  the  directions 
of  the  forces  P  and  7?2. 

Casi;  IV.  In  yeiwrol.  lot  the  beam  AB  l)o  in  equilibrium  under 

the  action  of  any  number  of  forces  /',,  /\,,  />,, Q^^  q^^  q^ 

of  which  the  magnitudes  and  points  of  application  are  g"ivenl  "and 


?i  P. 


AC 


f 


y.   'p. 


M 

^ 


Q. 


I'Ki.  230. 


♦     * 

Q.  Q. 


which  act  at  right  angles  to  the  length  of  the  Imvuh.     Suppo-  the 
beam  to  bo  divided  into  two  s(.gments  by  an  imaginarv  |)lane  M\ 
Smco  the  whol.-  iM-am  is  in  o.|uilibrium,  oach  of  the  segments  mu.st 
also  1m'  in  (■(luilil)rium.     Consider  the  segment  .l.U.V. 

It  is  kept  in  <"(iuilibrium  by  the  forces  /',,  />.,,  /'j,  .  and  bv 
the  reaction  of  tli..  segment  /y.i/.V  upon  the  segment  AM\  at  th- 
plane  MX;    eall  tliis  rea.'tion  A',.     The  forces  /' 


/'2.  /-.-,. 


arc 


equivalent  to  a  smgle  resultant  R,  acting  at  a  ,M,int  <listant  r.  from 
.U.N.     Al,s„.    without    affecting    the   equilibrium,    two    forces    each 
oqunl  and  parallc-l  to  R,.  but  ..p,H.site  to  on,,  another  in  directi-.n 
ii«uy  IK.  applied  to  the  segm,.nt  AMX  ut  the  plane  MX,  and  the 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


103 


three  equal  forces  are  then  equivalent  to  a  single  force  Ri  at  MN, 
and  a  couple  (Ri,  -Ri)  of  which  the  moment  is  Rin. 

Thus  the  external  forces  upon  AMN  are  reducible  to  a  single 
force  Ri  at  MA',  and  a  couple  (/?,,-i?i).  These  must  be  balanced 
1)\  El,  and  therefore  Ei  is  equivalent  to  a  single  force  -Ri  at  MN 
and  a  couple  (-Ri,  Ri). 

In  the  same  manner  the  external  forces  upon  the  segment  BMN 
are  reducible  to  a  single  force  R2  at  MN,  and  a  couple  {R2,  -R2) 


I* 


Ac 


Ri 


12    *^ 


RrR. 

Fio.  237. 

of  which  the  moment  is  flara.    These  again  must  be  balanced  by  E2, 
i\w  reaction  of  the  segment  AMN  upon  the  segment  BMN. 

\()\v  El  and  E2  evidently  neutralize  each  otUor,  so  that  the 
force  /?,  and  the  couple  {Ri,  -Ri)  niust  neutralize  the  force  R2 
and  the  couple  (R2,  -R2).  Hence  the  force  i?i  and  the  coup'.o 
(If\,  -Ri)  are  respectively  equal  but  opposite  in  effect  to  the  force 
R2  and  the  couple  (/if 2,  -R2);   i.e., 

Ri'Rq,    Riri-R2r2,    and    n -rz. 

The  force  Ri  tends  to  make  the  segment  AMN  slide  over  the 
.'^■irnu-nt  BMN  at  the  plane  MN,  and  is  called  the  Shearing  Force 
with  res|X'ct  to  that  plane.  It  is  equal  to  the  algebraic  sum  of 
tlic  forces  on  the  left  of  MN, 


''P1+P2-P3+  ...  'I(P). 

So  R2~Qi-Q2-Q3+  .,  .  -I(Q)  is  the  algebraic  sum  of  the 
!  'K-is  on  the  rifjht  of  MN,  and  is  the  force  which  tends  to  make 
lli<-  M'jjinent  BMN  .slide  over  the  segment  AMN  at  the  plane  MN. 
/?.■  is  therefore  the  Sheariiuj  Force  with  respect  to  MN,  and  is  equal 
to  Ri  in  magnitude,  but  acts  in  an  opjxisite  direction. 

i\gain,  let  p,,  pj,  p3,...,qu  qj,  93, ... ,  be  respectively  the 


104 


THEORY  OF  STRUCTURES. 


distances  of  the  points  of  application  of  Pi,  P,   p,  n,    d 

(23,  .  .  .  from  MX.  i,  ^  2,  fa, . . . ,  Qi,  Q^, 

Then  ;?,r,=the  algebraic  sum  of  the  moments  about  MN  of 
all  the  forces  on  the  left  of  MX, 

=PiPi+P2P2-P3P3+  ...  =IiPp), 
is  the  moment  of  the  couple  (/?,,  -Ri). 

This  couple  tends  to  bend  the  beam  at  the  plane  MX  and  \U 
moment  is  called  the  Bending  Moment  with  respect  to  MN  of  all 
he  forces  on  the  left  of  MX. 

So  /?2r2=tlie  algebraic  sum  of  the  moments  about  MX  of  all 
the  forces  on  the  right  of  MX, 


-<?i?i-<?272- 


•^(Qq), 


IS  the  Bending  Moment,  with  respect  to  MX,  of  all  the  fo-  es  on 
the    ujht  of  M^,  and  is  equal  but  opposite  in  effect  to  AV. 

Zn      /T;T    r"  ""'  '"''  "^  •^^•^'  '"  ^^^  ^^h^--'  «"  that  to  d.>fine 
siderattr  "  ""'"'''"■-'  *"  '^"^''  ^^"^  '"^"'""^  ""^'^''-  ^«»- 

If  the  segment  CB  is  removed,  AC  may  be  kept  in  equilibrium 
by  applymg  to  the  face  of  the  secti...  at  C,  forces  of  precisely  th^ 
same  magmtude  and  character  as  those  which  aicte.l  when  \CB  was 
a  co„t.nuous  Ix^am.  These  forces  n.ay  Ik.  resolved  into  horizontal 
and  vertical  .o.nponents.     The  algebraic  sum  of  the  latter  i  e    the 

sun.  of  tlu.  vertical  fores  acting  c-xfrnallv  upon  AC,  ie     to  /?, 

a..,l  therefore  the  algehra.c  su.n  of  the  horizontal  com,K)nents  of  the 
orces  a,  „mst  1..  nil  ,,.„,,  „.es..  horizontal  con;,K,nents  n.ust 
Ik.  e,,u,valen,  to  a  coup!,,  wi.iel,  will  neutralize  the  couple  ,lue  to  the 
external  forces  actn.g  upon  .W,  of  which  the  n.o.nent  is  /?.r,  The 
mom..,.,  ,  ,]„.  horizontal  ....n.ponents  at  C  is  usuallv  cdle.l  the 
»>o»>vnt   of   rrs,s,„nrr,  |,„,    is  soUietimcs  also   known    as  (I,.  r/«.^»r 

Airain.  Fii?.  2.'<>{  r«'i>resf»nt'j   -i  hnriT/^t.*-!  i  *•    >      . 

--_    !    |.n»riiH    ,1  noriZOnt;ii    Ujai.i   of    depth    w^ing 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


105 


I 


Fio.  238. 


upon  supports  at  A  and  B.    The  vertical  reaction  at  4  is  72,  and 
ihc    resultant  load  upon  AD  (includ- 
iiif;  the  weight  of  AD)  \a  W  at  a  dis- 
tant ■  X  from  D. 

U't  a  rectangular  portion  pqrs  of 
tlic  Ix'am  be  removed  and  replaced  by 
tudiiicinbcrs,  pq  capable  of  bearing  a 
|.rc.<suie  only,  and  rs  capable  of  bearing  a  tension  only,  as,  e.g.,  a 
cliiun.  These  members  alone  are  not  sufficient  to  maintain  equilib- 
rium, and  it  can  be  easily  shown  experimentaUy  that  an  upward 
shearing  force,  S,  must  be  applied  at  r  and  that 

S-R-W. 

Also,  the  bending  moment  at  D  due  to  R  and  W 

-Ra-Wx, 

and  this  must  be  neutralized  by  the  moment  of  the  compression  C  in 
pq  «ith  respect  to  r  or  by  the  moment  of  the  tension  T  in  rs  with 
re.'ijK'ct  to  p. 

Therefore  Ra  -  Wx  -Ch-  Th. 

Hence  also  C  —  T. 

2.  Examples  of  Shearing  Force  and  Bending  Moment.— In  general 
the  terms  shearing  force  and  bending  moment  will  be  designated  by 
the  abbreviati.as  S.F.  and  B.M.,  the  shearing  force  and  bending 
mouieiit  at  any  point  distant  x  from  the  origin  by  5„  M^,  and  the 
shearing  forces  and  bending  moments  at  any  points  A,  B...  by 
So.  >'(, .  .  .  anil  Ma,  Mb  .  .  .  respectively. 

riie  beams,  unless  otherwise  specified,  are  horizontal. 
•v..  Ex.  a.  A  cantilever  0.4,  Fig.  239,  infixed 

•v  U  at  0  and  carriea  a  waght  P  at  the  free  end  A. 

^^ >-~^:i»,,^ -  5.  -p, 

')         !  *'^.  I    and  the  S.F.  18  tho  same 'at  evpry  point  be- 

'■'''-      ■'-  "'-  t^^W"   0  and   A.      Describe   the   rectangle 

I  "   pCFA,  taking  QC-P-AF.    The  line  (V 

is  the  S.F.  diagram   and  the  S.F.    at   any 

t„       .u  P**'"*  ^  '■  tl'e   vertical   distance    LN  be- 

t'^"  'I  that  point  and  the  line  CF. 

«hi,h  is  Pt  when  x-0,  l.e.,  at  0,  and  is  0  when  x-l,  i.e.,  at  A. 


4diN 


106 


THEORY  OF  STRUCTURES. 


Take  the  vertical  line  OG-PUnd  join  GA.     The  line  C4  is  thp  RM 
that  pomt  and  the  Ime  GA     The  B.M.  is  evidently  greatest  Tt  0,JZT 

the  max.  B.M.  -  PI. 
Also,  LU-M,-P(l-x)-&re&ALNF 

-the  total  S.F.  between  Z,  and  yl. 

Ex.  b.  A  cantilever  OA,  Fig.  240,  is  fixed 
at  U  and  carries  a  uniformly  distributed  load 
of  intensity  w.    Then 

Sx  -W{l-x), 

^•^^^^^-  ""i'^^  '«  «-'  when  x-0,  i.e,  at  0,  and  is  0 

j  — ^*   when  x~l,  i.e.,  at  A. 

Fio.  240  o^u  ^r ''*'  **"'  '''"■*'''*'  ""^  ^^  - «''  »"d  join  C^- 


? 


Y. 


--    'N 


Again, 


w 


Mr-~(l-x)*. 


the  equation  to  a  parabola  ifith  its  vertex  at  /. 

The  B..M.  is  'Ji'  when  x-0,  i.e..  at  0,  and  is  0  ..hen  x-l,  i.e.,  at  A. 
Take  the  vertical  line  0«-!l^'a„d  trace  the  parabola  AG.    The  .... 

greaJt  ;^  o!  so  that         '  "'''  ''''"'"'""  ""  -^^-    ^''^  B"^"  "  evidently 


curve 


the  max.  B.tf. 


a 


-Also, 


LU-Af,.^^(l-x)'-aTBaALN 

-total  S.F.  between  L  and  /4. 


Ex.  r.    4  cantilrver  OA,  Fig.  241 ,  is  fixed  at  0  and  carries  a  weiaht  P  at  A 
together  w,th  «  t^niformly  distributed  load  of  intensity  w.  ^  ' 

S^-P+u'<l~x), 
the  cqunfinn  to  a  straight  line. 

Th..  S.V.  is  /'+„./,  when  x-0.  i.e..  at  0,  and  is  P  when  x-/  1  e    at  ^ 

r'F ';  t'ho  s  F^r"     '"'"  .^^:'^^"'  -'»  ^i''^-/-  H".l  join  rV    •Thetc 


Again, 


.^f.-P(/-x)+^'(7-x)S 


SHEARING  FORCES  AND  BENDING  MOMENTS.  107 

which  may  be  written  in  the  form 

2w     2\w  I 

the  equation  to  a  parabola  with  its  vertex  at  H,  where  ^r  -  -  and  HT~~ 

w  2m,* 

The  B.M.  is  Pi  +  —  when  x  -0,  i.e.,  at  0,  and  is  0  when  x-l,  i.e.,  at  A. 
Take  the  vertical  line  OG-Pl  +  ~  and  trace  the  parabola  GAH.    The 


-4. 


Fio.  241. 


'urve  GA  is  the  B..M.  diagram  and  the  B.M.  at  any  point  L  a  the 
ilistuiife  LU  between  that  point  and  the  parabolic  arc  GA. 
The  B.M.  is  evidently  greatest  at  0,  so  that 


vertical 


Alio, 


the  max.  B.M. -Pi +  2 
LU'Mm-P{1-x)  +|-(/-jr)»-area  LNFA 
-total  S.F.  between  L  and  A. 


i'.x.  d.  The  beam  OA,  Fig.  242,  resting 
»l«'H  Mipportx  at  O  au,l  A,  rnrrics  a  weight  P  u/  j 

I'll,,;  ntrnted  at  a  point  li  dividing  the  beam  into  A    \ 

III-  segments  OB -a,  BA-b.    Then  *H^"1--r 

a+b-l. 

lot  /?,,  R,  Ik!  thf  reactions  at  O  and 
.1  KsjMMtively.  Taking  moments  about  A 
ill,,!  (), 


/?,- 


Tlun,  between  0  and  B, 


Ph 
I 


S.-R^-P- 


^^  THEORY  OF  STRUCTURES. 

and  between  B  and  A, 

80  that  on  passing  the  point  B,  the  S.F.  changes  sign. 

lake  the  vertical  lines  OC  -  R,  -  JiD  &nd  BF  ^  if  =  ajp.  ■  •     /^n      . 

the  equation  to  a  straight  line. 

The  B.M.  is  0  when  x=0,  i.e.,  at  0, 


R'a-Pj  when  X- a,  i.e.,  at  B. 


and  is 

So,  between  B  and  A, 

Mz-RiX-P{x-a), 
the  equation  to  a  straight  line. 


The  B.M.  is  R,a~P~  when  x~a,  i.e.,  at  B, 


and  is 


Rd-P(l-a)  -0,  when  x-f,  i.e.,  at  A. 


Take  the  vertical  line  BG.p"j  and  join  GO  and  G^.    The  lines  OO,  GA 

between  that  point  and  the  lines  0(7,  GA. 

The  B.M.  is  evidently  greatest  at  B,  so  that 


the  max.  B.M.  -  P 


r 


If  a~b--,  the  weight  is  at  the  centre  and 


PI 


them«z.  B.M.-  — 
Also,  between  0  and  B, 

LU-  .1/,  -  /?,x  -  area  OCNL 

-  total  S.F.  between  0  and  L. 
Between  B  ami  ,1, 

L6^-.»/,-/?,x-P(x-a)-/?.o  +  (ff.-P)(a,-a) 

-a/j/ctraif  sum  of  the  areas  OCDB  and  BLAT^ 
-total  S.F.  between  0  and  L. 

Ex.  r.  Thr  h-nm  OA,  Fig.  21.3,  resting  upon  ^upiMtrU  ai  0  and  A  came,  a 
unijomly  distributed  load  o}  intentity  w.  fi"      'i*  ^  ^^na  A,eamu  a 


SHEARINO  FORCES  AND  BENDING  MOMENTS. 


109 


Then 


which  is 


1(1  is 


wl 


S»-Ri—wx~wl^—x\, 


when  x-0,  i.e.,  at  O, 


0  when  x--,  i.e.,  at  the  middle  of  the  beam, 

wl 


-^when  x-Z,  i.e.,  at  A. 

The  S.F.  therefore  changes  sign  on  pws- 
imj  the  middle  point  B  of  the  beam  and  in- 
craisex  uniformly  from  its  mimimum  value  zero 

at  li  to  its  maximum  valixc-^  at  0  and  at  A. 

Take  the  vertical  lines  OC'-—-AF  and 

join  CF.     The  Hne  CF  is  the  S.F.  diagram,  Fm-  243. 

and  the  S.F.  at  any  point  L  is  the  vertiuai  distance  LN  between  that  point 

and  the  line  CF. 


Again, 
which  may  be  written  in  the  form 

'  8 


^.-«.*-f-|a^-x'). 


Mx 


ihy. 


the  equation  to  a  parabola  OGA  of  parameter  |  and  with  its  vertex  at  a  point 

defined  by  OB-i  and  BG-^\    The  parabola  OGA  is  the  B.M.  diagram, 

and  the  B.M.  at  any  point  L  is  the  vertical  distance  LU  between  that  point 
and  the  parabola. 

I'lie  H.M.  is  least  at  the  supports  and  gradually  increases  towards  the 
etntre  of  the  beam  where  it  is  greatest,  so  that  the 


max.  B.1I.' 


.\lso, 


•^•«-Y-l{H' 


'  8* 


-area  flOC  -  area  BLJV 

-area  OCNL 

-total  S.F.  between  0  and  L. 
r.x.  f.  The  beam  OA,  Fig.  244,  reslirj  upon  supports  at  0  and  A,carriea 
0  "ylormly  distributed  load  of  mtensily  w  together  unth  a  weight  P  concentrated 
"'  ili<  point  B  dividing  the  beam  into  the  segments  OB -a  and  BA  -6. 

Ri-  —  +  P-,      ft,-  — +  P-. 


Then 


/ 


r 


110 


THEORY  OF  STRUCTURES. 


Between  0  and  B 


the  equation  to  a  straight  line. 


The  S.F.  is 


Fio.  244. 


r,b     wl    , 


'HH 


whenx-a,  i.e,,atB. 


and  is 

Between  B  and  A 

Ss-Ri-wx-P--w(i-x)-P±, 
the  equation  to  a  straight  line. 

TheS.F.i8  «'(j-a)-Pywhenx-a,i.e.,atB, 

">•* '"  - 7 - ■Py  when  x-l,  i.e.,  at  A. 

Take  the  vertical  lines 

and  join  CD  and  F  .    The  broken  line  CDEF  is  the  S.F.  diaRram  and  the 

broko"n  YuIcDEF     "         ''"*'™'  '^'^'"'"^  ^''^  '^'*'''^'"  ^^*^  ^""*  ""'^  **«» 
Immediately  on  the  right  of  B 

the  S.F.  =^(^  -a)  +P^^-P_u,(l-a)  -pl 


SHEARING  FOllCES  AND  BENDING  MOMENTS. 
and  the  point  E  will  lie  above  or  below  B  according  as 


111 


w(i-a) 


>  or  </>-. 


In  the  latter  case  the  S.F.  changes  sign  at  B,  and  the  B.M.  is  found  to  have 
its  jjreatest  value  at  this  point. 

If  E  falls  above  B,  then  EF  cuts  OA  at  a  point  X  at  which  the  S.F.  is  nil 
and  the  B.M.  is  then  greatest  at  X.  '' 

Again,  between  0  and  B, 

which  may  be  written  in  the  form 

the  equation  to  a  parabola  OGH  of  parameter  — ,  and  having  its  vertex  at  a. 
point  defined  by 

Or-P^  +  I^    and    TH.Up^-A\ 

Tho  parabolic  arc  OG  is  the  B.M.  diagram  between.  0  and  B,  and  the  B  M  at 
any  point  L  is  the  vertical  distance  LU  between  that  point  and  the  arc  OG. 
Biiween  B  and  A 

Mz~R^x-—  -P(x-a), 

which  may  be  written  in  the  form 

2w\2     '^1/  2V2      ^l     'I' 

the  equation  to  a  parabola  AGK  of  parameter  |  and  having  its  vertex  at  a 
point  A"  defined  by 

The  paraMa  AGV  is  the  B.M.  diagram  between  B  and  A,  and  the  BM 
at  ;.ny  ,»,nt  L  is  the  vertical  dis.  n.e  LU  between  that  point  and  the  parabou! 
If  P  is  at  the  middle  point,  a~b-~-, 

and  the  B.M.  at  the  centre -P-  +  —. 

4      8 


112 


THEORY  OF  STRUCTURES. 


Again,  between  0  and  B, 


irx' 


LV-Mr^Ra-~- area  OCNL 

=  total  S.F.  between  0  and  B. 
Between  B  and  .4, 

L  L-  =  -l/x  =  Iia-~-  P(x  -  a) 

=  algebraic  sum  of  the  areas  OCDB  and  BENL 

=  totalS.F.  fromOto  A. 
An  examination  of  i:xs.  a  to  /  shows  that  the  S.F.  and  B.M. 
ordinatcs  in  Exs.  c  and  /  are  the  algebraic  .sums  of  the  corresponding 
ordinates  in  Ex.s.  a  and  h  and  Exs.  d  and  e.  Hence  it  follows  that 
the  effects  of  different  loads  upon  beams  may  be  determined  separately 
and  the  resultant  effects  are  then  found  by  superposing  the  corre- 
sponding results  thus  obtained. 

Ex.  1.  .1  cantilever  OA,  Fig.  24.5,  12  //.  long  carries  a  load  of  200  lbs.  at 
.1  and  a  uniformly  diMrihukd  load  of  600  lbs.  Draw  to  scale  the  S.F.  and  B.M. 
diagrams  and  determine  the  S.F.  and  B.M.  at  6  /.'.  from  0.  For  the  S.F.  dia- 
gram take  a  vertical  scale  of  measurement  so  that  1  in.  - 1600  lbs. 


The  S.F.  at 
and  at 


Fio.  245. 

0  =  600  +  200  =  800  lbs.  -  J  in. 

.4  =200  lbs.  =i  in. 

Take  the  vertical  lines  OC  =  i  in.,  AF  =1  in.,  and  join  CF.  At  6  ft.  from  0 

LN  =  /f  in.  by  measurement  -  500  lbs. 

=  S.F.  at  L. 

For  the  B.M.  diagram  take  a  vertical  scale  of  measurement  so  that  1  in. 
-6000  ft.-lbs. 

The  B.M.  at  0  -  600  X  6  +  200  X 12  -  6000  ft.-lbs.  - 1  in. 

and  at  ^4  =0. 

Take  the  vertical  line  OG  =  1  in.  and  join  GA.    At  6  ft.  from  0 

It/  -  J  in.  by  measurement  =  .3000  ft.-lbs.  =  B.M.  at  L. 


"^^•r* 


SHEARISG  FORCES  AND  BENDING  MOMENTS. 


113 


i:x.  2.  .1  beam  OA,  Fig.  246,  resting  upon  supporti  at  O  and  A,  30  ft.  apart, 
rnm,^  n  uniformly  di.itributed  load  of  6000  lbs.,  and  a  single  weight  of  600 
/'w.,  at  a  point  B  dividing  the  beam   into  segments  OB -10  ft.  and  BA  "20  ft. 


Fig.  246. 


Drrnr  to  scale  the  S.F.  and  B.M.  diagrams,  and  obtain  by  measurement  the  S.F. 
•iirl  B.M.  at  points  5  ft.  from  0  and  from  A. 


Therefore 


ft,  X30  =  600 X20 +  6000 X 15  =  102000 

/?.X30  =  600X10  +  6000X15  =  96000. 

/?i  =3400  lbs.  and  /?,  -3200  lbs. 


I- or  the  S.F.  take  a  vertical  scale  of  measurement  so  that  1  in.  -3000  lbs. 

The  S.F.  at  0  =  /?,  =3400  lbs. -Ij',  in.; 

immediately  on  left  of       B  =  /?,- 10x200  =  1400  lbs.  =^\  in.; 

iMimcdiately  on  right  of    B-ff. -10X200 -600 -800  lbs. -^  in. 

'''  ^-fli-30x200-600=-32001bs.--lAin. 

Take  the  vertical  lines 

or  -i,',  in.,  «D-iV  in.,  BE  =  ,^  m.^AF-l.?,  in.,  and  join CZ>  and  £F. 
i5.v  measurement      LN  at  5  ft.  from  0-|  in.  -2400  lbs. 

-S.F.  at/.. 
iW  mea.surement     UN'  at  25  ft.  from  O  -^  in.  =2200  lbs. 

,  .      ,  -S.F.  at  L' 

anil  IS  of  course  negative. 


1^'*  THEORY  OF  STRUCTURES. 

Again,  between  0  and  B, 

-1/x-3400j;-100x» 
o""  -»/x-28900--100(17-x)» 

a  parabola  with  a  vertex  at  a  point  17  ft.  measured  horiiontaUy  and  28  900 
ft.-lbs.  measured  vertically  from  0.  ' 

Also,  the  li.M.  at 

and  at 

Between  B  and  A, 


or 


0=0 

B- 24000  ft.-lbs. 

Mx  -  3400X  -  100x»  -  600(z  -10) 
-Ux-25600-  -100(14-x)», 


a  parabola  with  its  vertex  at  a  point  14  ft.  measured  horizontaUy  and  13  600 
ft.-lbs.  measured  vertically  from  0.  ' 


Also,  the  B.M.  at 
and  at 


5 -24000  ft.-lbs. 


Trace  the  parabolas  OGH  and  AGK,  taking  a  vertical  scale  of  meaaure- 
ment  so  that  1  m.  -24,000  ft.-lbs.    Then  «-««- 

BG-l  in.,  HT'1^%  in.,  KV-^j  in. 

By  measurement  LU  &t  5  ft.  from  0-ff  in. -14.500  ft.-lbe. 

-B.M.  atL. 
By  measurement  L'l"  at  25  ft.  from  0  -^  in.  - 13500  ft.-lbs. 

-B.M.  at  L. 

3-  Relation  Between  Shearing  Force  and  Bending  Moment.— Id 

the  examples  of  the  prececii.ig    -ide  it  l:ai>  been  shown  that  at  any 


— <«M»-*i 


rfl 


Ri 


Fig.  247. 


point  of  the  beams  under  consideration  the  B.M.  at  any  point  is  the 
total  S.F.  b(>tween  that  point  and  a  support.  The  following  is  a 
general  proof  of  this  statement. 

Let  the  beam  0.1,  Fig.  247,  .supported  at  0  and  4  be  acted  upon 


SHEARING  FORCES  AND  BENDING  MOMENTS.  115 

by  a  number  of  weights  concentrated  at  different  points  along  the 

beam ; 

U't  a  be  the  distance  between  two  such  consecutive  points  r  and 
r-^1,  and  take  Or  =  x. 

Let  q  be  the  weight  at  r  and  let  p  be  the  distance  between  r  and 
the  Ime  of  at  Lion  of  the  resultant  P  of  the  weight  between  0  and  r. 

Let  Ml,  Ri  be  the  B.M.  and  vertical  reaction  at  0. 

Then 


and 


Riix+a)-P(p+a)-qa+Mi"Mr+i 

RlX-Pp  +  Mi=Mr. 


Therefore 


(Ri-P-q)a  =  Sa='Mr+i-Mr 

-increment  of  B.M.  between  r  and  r+1 
'JM, 

S  being  the  S.F.  between  r  and  r  +  1. 

Now  S  is  nil  if  Mr+i  =M„  and  every  point  of  the  beam  between 
r  and  r  4- 1  is  subjected  to  the  same  constant  B.M.,  the  case  being  one 
of  simple  oending  without  shear,  as  in  a  car-axle. 

Again,  the  weights  on  the  beam  may  become  infinite  in  number 
and  i^o,  in  the  limit,  form  a  continuous  load.    Then 

dM  =  Sdx. 
From  the  equations 

^^      or 

-^  =  <s  for  concentrated  weights, 


and 


dM 


■^  =  <S  for  a  continuous  load, 


1'  '^  (It  once  evidmt  that  the  S.F.  at  any  point  is  measured  by  the  tangent 
oj  the  slope  of  the  B.M.  polygon  or  curre  at  that  point. 
Also,  designating  algebraic  sums  by  the  symbol  I, 

the         B.M.  =  i'(JM)=i'(5a) 

=  the  total  S.F.  for  the  concentrated  weights, 
and  for  a  continuous  load 

M^fSdx  =  the  total  S.F. 
iHtween  0  and  the  point  under  consideration. 


•■*M\ 


IK) 


TIIF.OHY  OF  .■^TRVcrUHEo 


Again,  J.U  and  dM  v-haii^o  sign  with  N  and  tlio  H.M.  is  a  nwx. 
(or  a  ))',}ii.  in  certain  special  cases)  at  the  point  where  the  change 
occurs. 

For  a  continuous  load  this  point  is  defincil  by  the  conditioa 


d]I 
di 


0. 


If  w  is  the  intensity  of  the  continuous  load  at  x  from  0, 

S^IU-fudi, 
dS    (PM 


and  therefore 


—  V 


dx     dx-  ' 


e(|uations  connecting  the  intensity  ir,  the  S.F.  and  the  B.M.  at  any 
point  in  the  l)eain. 
4.  Live  Loads. 

Kx.  ((.   -t   iiiiiih;  I'  trnvrh  from  A  to  B  over  a  horizontal  tjiraer  oj  length  I 
rc^linij  upon  ■supporl.-i  nl  A  and  li. 


V\u.  '.'4S, 
When  P  is  at  /•'  distant  i  from  ft,  the  mi<l(lli'  [hjhiI  of  die  girder,  the  re- 


ai'tioii  A'r  lit  A  --" 


(.;  -)• 


Ililunn  A  mill  F. 


S,-H, 


smd  the  iii'sitiir  .S.K.  !it  iiiiy  jHiitii  /•'  is  (jrenlml  nt  the  instant  after  P  passes  that 

point. 

ItitweiH  F  mid  H, 


S^^H.^P^J-K^L-r), 


SIlEARlSa  FORCES  AXD  BEN  DING  MOAfENTS. 


117 


mill  tlio  rtcgalivc  S.F.  at  any  [H)int  /•'  is  (jrfaloit  <it  the  instant  before  P  posset 

Ihr.i  j>„ilit. 

ALso,  the  positive  S.F.  is  /'  when  x  =  ~,  i.e.,  at  .1, 
and  is  0  when  x=  --,  i.e.,  at  B; 
the  negative  S.F.  is  {)  whori  r^-,  i.e.,  at  .4, 
and  is-P  uhcii  j- =  --,  j  o.,  at  B. 

■!;ikf  the  vrrtical  lines  .!/>  =  «£,'  =  P  and  join  Dli  and  .1J?. 

Then  l)H  is  tlic  /msilive  and  .lA"  the  »»(/«/ /ir  S.F.  diagram,  and  the  S.F. 
M  .my  |M)int  /•'  is  the  positive  vertieal  distance  /•Y,"  or  the  negative  vertical 
di-^i.iiice  FH  Ix'tween  DB  and  AB  or  Ix-twwn  AE  and  .4«  rcspeetively. 

Ajjain,  the  B.M.  at  any  point  A'  Ix'tween  ,4  and  F  distance  y  from  0  is 


My-li<[~-y), 


Hlii.il  is  the  Kreatpst  when  y  has  its  least  value,  i.e.,  when  y-x  and  /!  coin- 
iiiir>  with  F. 

S(,.  iHturai  F  and  B  the  H.M.  at  any  jwiint  /.  distant  y  from  F 


«lii.h  is  Rreatest  when  ;/  has  its  least  value,  i.e.,  when  i/-0  and  /.coincides 

With  /■'. 

Il-iiie.  Ihr  B.M.  lit  Hint  puiiil  is  gnat, si  „l  ll„   instant  l>  /Hisses  thai  point, 
""I  the  maximum  H.M.  diagram  is  tlicrefiire  given  by 


"■'<-')  ■':(':-'■)■ 


•' ••"'•">  ♦"  n  paralalia  AMi.  passing  through  the  points  .4  and  «  ami 

lin  ni;  its  vertex  al  .V  verti<ally  alx.ve  <>  the  ceiitr.'  of  the  girder  and  at  the 


■Im  Mice  O.V 


/'/ 


In  .Mldition  to  the  live  load,  let  .•.  single  weight  (,>  Im>  coniTiilrat.Ml  at  the 
"it  /    lisiant  a  from  ().     The  cornsponding  vertical  reaction  at  A 


-%-)■ 


118  THEORY  OF  STRUCTURES. 

At  X  from  0  between  A  and  T 


the  maximum  Sx-'jl^+xj  +7(5- +«)• 


At  X  from  0  between  T  and  0 


the  maximum  Sz-j(-+xj  +y/-+a)  -Q 


At  X  from  0  between  0  and  B 


the  maximum  'Si--|--x)  --(--o). 


The  S.F.  diagram  for  the  live  load  Is  D3,  and  that  for  the  concentrated 
load  the  broken  line  dgef. 

The  total  S.F.  at  any  point  is  the  algebraic  sum  of  the  vertical  distanrcs 
between  that  point  and  the  two  diagrams.    If  AdgT  and  TefB  are  inverted 


d      X 


Flo.  219. 

so  as  to  take  the  positions  Ad'tfT  and  Te'j'R,  the  S.F.  at  F-GY-  at  F'^O'P 
-  X'F'  -  d'X' ;   at  F"  - 0 ;   nt  /■'"'  -  f,'"/.'"'  - A'"'f'"'  -  _ a"'X"'. 
Again,  hrttirrn  A  and  T 


■ho  nmxiniun.  H.M.  at  j- from  ()-^'('^^t'\  +^^i +a)(--x); 
tween  T  and  O  '        ' 

he  maximum  H.M.  at  x  from  O  -  ^(f  -.r')  +  y  (,|  +  a)  (j- -x)  -Q(o-«) 


between  O  and  B 


the  ffiaxiffium  B.M.  at  *  from  0-y(['-x')  +-(.'  -<j)('  -x). 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


119 


In  Fig.  250,  ANB  is  the  maximum  B.M.  diagram  for  the  live  load  and  the 
B.M.  diagram  for  the  concentrated  load  consists  of  the  two  lines  AY',  BY'. 


Fia.  250. 

The  total  maximum  B.M.  at  any  point  is  the  algebraic  sum  of  the  vertical 
disiaiires  between  that  point  and  the  two  diagrams.  If  AY'B  is  inverted 
so  as  to  take  the  position  AX'B, 

the  total  maximum  B.M.  at   F'  -F'N'  +  F'Y'  -X'N'- 
"      "  "  "      "     F-FN+FY-XN;' 

"      "  "  "     "   F" -F"N" +'^"Y" -X"N" 

Kx.  b.  Let  a  continuous  load  weighing  w  per  unii  length  travel  over  th» 
gvdcr  from  0  towards  A. 


Pio.  2Sl. 


When  the  load  covers  a  length  OB-r,  the  reaction  st  A  =  ^. 

Thm  is  evidently  the  S.F.  at  all  ,H)int8  in  fnmt  of  the  advancing  load. 


^^  THEORY  OF  STRUCTURES. 

Jk'twecu  B  and  0,  at  any  point  distant  y  from  B, 

thoS.F.='|j'-„,y, 

which  is  prcatest  when  .v=0- 

Hence,  the  maximum  S.F.  at  ai  mt  due  to  the  live  load  is  in  front 
of  the  advancing  load,  when  the  lo  eov.rs  the  longer  segment  OB,  and  is 
given  by 

the  equation  to  a  parabola  OD  with  it8  vertex   at  O  and  intersecting  the 
vertical  at  .4  in  the  point  D  where  .ID-—. 

As  the  load  travels  off  the  girder,  the  S.F.  behind  the  load  is  found  in  pre- 
cisely the  same  manner,  the  S.F.  diagram  Ix-ing  the  parabola  AE  with  its 

id 


2" 


vertex  at  .4  and  intersecting  the  vertical  OE  in  the  point  E  where  OE 

Let  the  girder  alxo  earry  a  uniformly  distributed  dead  load  of  intensity  v/ 
Considermg  one  half  of  the  girder,  DH  is  the  S.F.  for  the  live  load,  and  KG, 


Fio.  2.-.2 

the  S.F.  for  the  dead  load,  may  l)o  traced  b,low  0.4.     The  total  maxijnum 
S.F.  at  any  point  B  is  then  \Y,  Mie  verticjil  distance  at  B  between  DH  and 

K(i. 


Also, 


and 


It  will  lie  sul)se(|uently  shown  that  the  design  o'  the  web  of  a  girder  prac- 
tic'iily  (ie|«'ti(ls  upon  the  S.F. 

Again,  the  U..M.  at  B  when  the  live  load  covers  OB  is 

"'■'''/>      \ 


SHEARI.WO  FORCES  AND  BEXDING  MOMENTS. 

liitircat  A  ami  li,  -at  y  from  .1, 

theB.M.='-g.'j,. 

wliiih  increases  until  y=l  —  x. 

If  the  load  advances  a  distance  ^  beyond  B, 

the  reaction  at  .1  =  — (z+z)', 
and  the  li.M.  at  /?=^(e+x)'(i-x)-^. 


121 


which  is  a  maximum  when 


0=-^{z+x){l-x)-wz, 

'"•  z-l-x, 

I.e.,  when  the  hve  load  covers  the  whole  girder. 

it  will  be  subsequently  shown  that  the  design  of  the  flanges  of  a  girder 
.l.|.-n,ls  essentially  upon  the  H.M..  and  the  maximum  flange  stresses  are 
tli.  nfore  to  be  found  when  the  live  load  is  uniformlv  distributed  over  the 
Iiilal  length  OA. 

Ill  the  results  deduced  above  it  is  assumed  that  the  length  of  the  live  load 
i^  N'lt  less  than  that  of  the  girder. 

Ia.  ;r  The  two  main  girders  of  a  single-track  bridge  of  80  //.  span  carry 
'I  '.  •!,!  loal  of  2.J(J0  lbs.  per  lineal  foot.  Determine  the  maximum  S.F  and  li  M 
wM(i  //.  from  a  support  when  a  live  load  of  3000  lbs.  per  lineal  foot  travels  over 
ll"   I'lidyi'. 

The  dead  load  on  each  girder 

2500 
-80X--  — 100000  lbs. 

riic  reaction  at  each  support  due  to  the  dead  load 

-"lOOOOlbs. 
Ihe  S.F.  due  to  the  dead  load  at  10  ft.  from  a  support 

-  mm  - 10  X " .  ~  -  37500  lbs. 

Thr  S.F.  due  to  the  live  load  at  10  ft.  from  the  support  is  greatest  when 
tl:'   live  load  covers  the  70  ft.  scgiiienf. 

The  corresiHMiding  reaction  at  the  unloaded  end 

70     .3000 

" SO  ^   •»"  ^  •''•'  ■" ■*''"'*^*  '^«- 

niprt-fon-  the  total  maximum  S.F.  al  10  ft.  from  a  support 

-  37oOO  +  4.W37  J  -  83437  J  lUi. 


122 


THEORY  OF  STRUCTURES. 


Jhc  B.M.  at  any  point  is  a  maximum  when  the  live  load  covers  the  whole 
The  total  load  on  the  girder  is  then 

-100000  +  80x^-220000, 
and  the  maximum  B.M.  at  10  ft.  from  a  support 

- 110000  X 10-""'^;^  ^^^-10.5 
-962o00  ft.-Ibs. 
5.  Moments  of  Forces  with  Respect  to  a  Given  Point  Q.— First, 

considor  a  singlo  force  Pj. 

De.scril)e  the  force  and  funicular 
polygons,  i.e.,  the  line  SiSa  and  the 
lines  AB,  BC. 

Through  the  point  Q  draw  a  line 
parallel   to   SiS^,   cutting   the   lines 
Fig.  253.  Fio.  254  '^^  ^""^  ^^  produced  in  x  and  y. 

n^'  ,  „  „  ^"""P  the  perpendiculars  BM  and 

U.\  upon  yx  and  SiSe  produced. 

Then  i:y._'^'"'^«_  Pi 

BM     ().\~0X 

^"^  I\BM^xy.OX. 

H..t  BM  is  equal  to  the  length  of  the  perpendicular  from  Q  to 
-  ""•  of  acfon  of  P„  an.l  the  pro.luct  xyOS  i.,  therefore,  equal 
>  t he  .no,n,.nt  of  /  ,  w.th  res,x>et  to  Q.     Honee,  if  a  .scale  i.s  so  chosen 

tiK.t  0.\  ^u„„y.  tl„s  .uon.ent  l)eco.nes  e.jual  to  xy:    i.e.,  it  is  the 

tnl,rni,l  rut  o/J  In,  the  two  s{,k.  n]  the 

jnniriuor    imh/f/ou    on    ,i    line   ,lr(,wn 

thronoh  the  ijiren  point  pnrollel  to  the 

ffinn  force. 

Xext,    let    tiierc    he    two    forces 

Di'scrihe  tho  force  and  funicular 
poly.'ons  S.S,S,;  and  ABd). 

I.<'t  the  first  and  last  si.les  (AB 
and  DC)  he  i)r()(luced  to  meet  in  G, 
and  i.'t  ;.  line  through  the  givm' 
point  (J  parallel  to  the  line  .%S,  inter.s,>ct  these  lines  in  .r  and  y 


Tlicn 


SHEARING  FORCES  AND  BENDING  MOMENTS.  123 

Draw  GM  perpendicular  to  xy  and  ON  perpendicular  to  Sz-Sg. 
xy  ^828^    resultant  of  P,  and  P2 


GM     ON 


ON 


anil  hence  (the  resultant  of  Pi  and  P2)  xGM  =  xyON. 

But  GM  is  equal  to  the  length  of  the  perpendicular  from  Qupon 
tli(-  resultant  of  P,  and  P2,  which  is  parallel  to  828^  ana  must 
iKTossanly  pass  through  G.  Hence,  if  a  scale  is  so  chosen  that 
0.\  =  unity,  xy  is  equal  to  the  moment  of  the  forces  with  respect 
t(.  (^;  i.e.,  it  is  the  intercept  cut  off  by  the  first  and  last  sides  of  the 
hmirular  polygon  on  o.  line  drawn  through  the  given  point  parallel 
to  the  resultant  force. 

A  third  force  P.,  may  be  compounded  with  Pi  and  P2,  and  the 
proof  may  be  extended  to  three,  four,  or  any  numljer  of' forces. 

The  result  is  precisely  the  same  if  the  forces  are  parallel. 

The  force  polygon  of  the  n  parallel  forces  Pi,  P2,  .  .  .  P„  becomes 
the  straight  line  8^8^82  ■ . .  8„.    Let  the  first  and  last  sides  of  the 


5    p. 


F:o.  257. 


Fio.  25S. 


fimi.-ular  polygon  m.«' t  in  G.  Drop  the  perpendiculars  GM,  ON 
upuii  xy  and  .SVS„,  xy  as  before,  l)eing  the  intercept  cut  off  on  a 
line  through  the  given  point  Q  parallel  to  8eS„.    Then 

xy  ■  ON  =  GM  ■  Sa8„.     Hence,  etc. 

Thus  the  moment  of  any  number  of  forces  in  one  and  the  same 
p'iinc  with  respect  to  a  given  point  may  be  represented  by  the  intercept 
' "/  "ff  by  the  firgt  and  last  siflcs  of  the  funicular  polygon  on  a  line 
drawn  through  the  given  point  parallel  to  the  resultant  of  the  given 

forces. 


124 


THEORY  OF  SriiUCTURES 

6.  Bending    Moments.     Stationary    Loads. -Let     a     horizontal 
l)oa.n  .1/i  support,.,!   at  .1  an,!  li  .-arry  a    number  of  weights  P, 
Pi,  Ps,  ...  at  the  points  .V,,  A';.,  .V3,  ...  e,  1, 


'R, 


"        N.  N..     N3      L  N.     N.    M 


^Rs        1 


Fio.  2.59. 

The  force  polygon  is  a  vertical  line    1234  . . .  n,  where  12=P, 
z.5  =  7  2,  etc.  ' 

Take  any  poI(.  0  an,l  describe  the  funicular  polygon  .1,.l2  I3 
Let  the  /,r.s7  an.l  ta.^f  si,l(.s  of  this  polygon  ho  pro.Iuced  to  meet 

Let  the  vertical  through  G  eut  .1/i  in  L  and  CZ)  in  A'-   LG  is 
the  Ime  of  acti,)n  of  tlie  resultant.  ' 

Draw  Oil  parallel  to  ('/;. 
From  the  similar  triangles  ()]H  and  GCK, 

on  CK- 


From  the  similar  triangles  OnH  and  GDK 


nU     GK 
OH  ~  DK- 


1//     />>A^     .ft/,     ft. 


Then'fore 
/?!,  fta  Ix'ing  the  reactions  at  A  and  D  respectively. 


'771 


BEMilNd  MOMESTS.  125 

But  l//  +  n//  =  ln  =  Pi+P2+  ...  =i?i+«2. 

Hoiipo  1//  =  /?,     and    nH^Rz. 

'lluis  //(c  h"ne  rfra?rn  //iro;//;/*  the  pole  parallel  to  the  clofting  line  CD 
'li  rides  the  line  of  loadf<  into  two  sefjments,  of  rchich  the  one  is  equal 
to  the  reaction  at  A  and  the  other  to  that  at  B. 

Let  it  now  1)0  required  to  find  the  hendincj  moment  ni  any  point 
.1/  of  tlie  beam,  i.e.,  the  moment  of  all  the  forces  on  one  side  of  M 
witli  resjK'ct  to  .1/. 

I.i  the  figure  these  forces  are  Ru  P,,  P.,  P^,  P^,  p^,  and  the 
(■'irresi)()iiding  force  polygon  is  //123456.  The  first  and  last  sides 
(it  tlie  funicular  polygon  of  the  forces  are  CD  parallel  to  0//, 
and  .l,5.ltt  parallel  to  06.  If  the  vertical  through  M  meet  these 
si.lcs  in  J  and  y,  then,  as  shown  in  Art.  5,  the  moment  of  the  forces 
A'l.  I\,  P-2.  Pa,  ^4,  Ps  with  respect  to  .1/,  i.e.,  the  bending  moment 
at  .1/,  ^OXxy,  OX  being  the  perpendicular  from  0  upon  1//  pro- 
duced. 

Hence,  if  a  scale  is  chosen  so  that  the  polar  distance  OX  is  unity, 
the  hending  moment  at  any  pn'nt  of  the  beam  is  the  intercept  on  the 
rrriiral  through  that  point  cut  off  by  the  closing  line  CD  and  the  oppo- 
>'ile  iiounding  line  of  the  funicular  polygon. 

7.  Moving  Loads.— IVams  are  often  subjectetl  to  the  action  of 
iiin\  ing  loads,  as,  e.g.,  in  the  case  of  the  main  girders  of  a  railway 
III  h lire,  and  it  becomes  a  matter  of  importance  to  determine  the 
ill  iHliiig  moments  for  ditTerent  positions  of  the  loads.  It  may  be 
a  .-miied  tliat  the  loads  are  concentrated  on  wheels  which  travel 
a'  iDss  the  bridge  at  invariable  distances  apart. 

At  any  given  moment,  let  the  figure  represent  a  beam  11  under 
ti."    loads    P,,    p.,,    Pa. 
l"l\«nn  CC'C"  .../),  the  closing  line  being  CD. 

let  the  loads  now  travel  from  right  to  left.  The  result  will  bo 
I'l'nscly  the  .same  if  the  loads  remain  stationary  and  if  the  sui>- 
I" Ms  II  are  made  to  travel  from  left  to  right. 

Thus,  if  tlie  loads  successively  move  through  the  distances  12 
-'•.  .{I,  .  .  .  to  the  left,  the  result  will  Ix-  the  same  if  the  loads  are 
i<'  i'l  >lalionary  and  if  tiie  supports  are  successively  moviv.  to  the 
TNhl  into   the   positions  22,  33,  44, . . .    The   new  funicular  poly- 


De.scril)e    the    corresponding   funicular 


126 


THEORY  OF  STRUCTURES. 


gons  are  evidently  C'C    . . .  D',  C'C"  . . .  D"   C"'C""        D"' 

the  new  closing  lines  l)eing  CD',  C"D",  C"'D"'  "  '        '       * 

The  hendinq  moment  at  any  point  M  is  measured  bv  xy  for  the 
first  distribution,  x'y'  for  the  second,  x"y"  for  the  third,  etc.,  the 

c 


Fio.  261. 

position  of  M  for  the  successive  distributions  beine  defined  br 
MM'  =  12,  M'M"  =  23,  M"M"'  =  34,  .  .  .  '^ 

Similarly,  if  the  loads  move  from  /e/<  /o  rt(7A/,  the  result  will  be 
tne  same  if  the  loads  are  kept  stationary  and  if  the  supports  are 
made  to  move  from  riqhl  to  left. 

It  is  evident  that  the  envelope  for  the  closing  line  CD  for  all 
distributions  of  the  loads  is  a  certain  curve,  called  the  envelope  of 
moments.  The  intercept  on  the  vertical  through  anv  point  of  the 
beam  cut  off  by  this  curve  and  the  opposite  boundary  of  the  funicu- 
lar polygon  IS  the  greatest  possible  bending  moment  at  that  point 
to  which  the  girder  can  be  subjected. 

Ex.  4.  Loads  of  12  and 
y  tons  are  conci titrated 
iipon  a  horizontal  beam  of 
12  //.  Kpan  at  dustanres  of 
3  and  0  //.  from  thv  right- 
hand  xiipiMiil.  Find  (a) 
the  B.M.  at  the  midtde 
point  of  the  Iteam,  and  also 
(/>)  the  maxiininn  B.M. 
,       ,,     ,      ,  Iir;.  2tiL/,.     print  need  at  the  same  point 

ulien  the  loads  travel  over  the  heam  at  the  fixed  distanees  of  6  ft.  apart. 


SHEARING  FORCES  AND  BENDING  MOMENTS.  127 

■Srri/fs  for  lengths,  i  in.  =  1  ft.;  for  forces,  A  in.  =  1  ton. 
Take  |W)lar  distance  =  f  in.  =  10  tons. 
<  Vivf  (I.  B.M.  =  xj/  X 10  =  3.15  X 10  tons  =  31  i  ton-ft. 
'  Vrvf  /..  B.M.  =j;'i/'X  10  =  3.6X10  tons  =  36  ton-ft. 

8.  Analytical   Method  of  Detennining  the  Mazimtmi  Shear  and 
Bending  Moment  at  any  Point  of  an  Arbitrarily  Loaded  Girder  AB.— 

At  any  given  moment  let  the  load  consist  of  a  number  of  weights 


^m^ 


-Q. 


w 

J2. 


one 


O    O 


-Q- 


7  -i-  w  ..  w    i  ;    .,.      , 


B 

^7- 


Fig.  263. 

u'l,  «'2, . . .  tr„,  concentrated  at  points  distant  oi,  02, . . .  a„,  respec- 
tively  from  B. 

The  corresponding  reaction  /?i  at  .4  is  given  by 

Ril  =  wiai+W2a2+  .  . .  +w„a„, 

I  boiiig  the  length  of  the  girder. 

Lot  ir„  =  u'i  +  w'2-(-  . . .  +w„,  the  sum  of  the  n  weights; 

Let  Wr  =  wi  +u'2+  ...  +UV,  the  sum  of  the  first  r  weights. 

The  shear  at  a  point  P  between  the  rth  and  the  (r-f-l)th  weights  is 

Si  =Ri-Wi-lV2-  .  .  .  Wr^Ri-Wr. 

9 

Lot  all  the  weights  now  move  towards  A  through  a  distance  x, 
and  let  p  of  the  weights  move  off  the  girder,  q  of  the  weights  be 
transforred  from  one  side  of  P  to  the  other,  and  s  new  weights, 
VIZ.,  v\^i,  w„+2,  •  ■  ■  w"n+.,  advance  upon  the  girder,  their  distances 
from  B  being  a„+i,  o„+2,  •  •  •  a„+„  respectively; 

I.ct  L  =  wi+W2+  ...  +Wp,  the  total  weight  leaving  the  girder; 

Let  7^  =  117+1 +w\ +2+ ...  +«v+„  the  total  weight  transferred 
from  one  side  of  P  to  the  other; 

I.i't  Rpl  =  wiai+xr2a2+  .  .  .  +trpflp; 

Let  R„l  =  Wr.^iar  +  i+U'r+^r+2+  ■  •  •   +MV+,a,.+,; 

let  RJ  =  w„+ia„+i+w„+^„+2+  . . .  +ii'„+,a„+,. 

i  liiis  Rp,  R^,  R,  are  the  reactions  at  A  due  respectively  to  the 
"■'■!-! it  which  leaves  the  girder,  the  weight  which  is  transferred,. 
aiiil  the  new  weight  which  advances  upon  the  girder. 


IL'8 


thkohy  of  STiticruiiKs. 


Tli(>  reaction  Ho  at  .1  witli  the  new  distribution  of  the  loads  i- 
given  by 

i?2i  =  "'p+iK+i+-r)+"p+2(flp+i.  +  x)+  ...  4-«v(a  +x) 
+  i(V+i(ar+i+J-)+  ...  +"'n(«n  +  J-)+«"„+ia„+i+ .  .. 
+  n-„+,a„+,  =  Ril-Rpl  +  x{W„-L}+R,l, 

and  hence 

(ff2-/?,)/  =  (/?.-/ifp)/+j(ir„-L). 

Also,  the  corresponiling  shear  at  P  is 

S.,  =  R.,-(trj,^.i+v'p^2+  .  .  .  +i(v  +  Mv+i+  .  .  .  +UV+,) 

^R'-nVr-i^+T). 

Hence  tho  shear  at  P  with  the  first  distribution  of  weights  is 
greattT  or  le.^^s  than  tlie  shear  at  .the  same  point  with  the  second 
distribution  accortling  as 

(Si  ^  S-j, 


or 
or 

or 


Rl-Wr>R2-Wr  +  L-T, 

T-L>R2-Ru 
T-L%R.-R^+j(]]\-L). 


(A) 


When  no  weights  leave  or  advance  upon  the  girder,  R„  Rp,  and  L 
are  severally  nil,  and  hence 


according  as 


x<    1    ' 


i.e.,  according  as  the  weight  transferred  divided  hy  the  distance  through 
which  it  is  Iransjcrrrd  is  (jreater  or  less  than  the  total  weight  on  the  girder 
divided  by  the  span. 

Again,  let  z  be  the  distance  of  7'  from  B,  and  let 

Rrl  =  Wta,  +  }r.,n2+   .  .  .    +!(VOr. 

The  bending  moment  at  P  with  the  first  distribution  of  weights  is 
Mi  =  R\(l-z)-rvi(ai-z)-W2(a',-i)—  ...  —w/^Ur-z) 
=  Ri{l-z)-RJ  +  z]\\. 


SHEARISG  FORCES  AND  BENDING  MOMENTS.  129 

riie  bending  moment  at  the  same  point  with  the  second  distri- 

Imtion  is 

.l/2  =  /?2(/-2)-»/V  +  l(ap  +  i    rX-z)-U'p  +  2(ap+2+X-z)-   ... 

-n'Aa,+x-z)-  .  .  .  -7(V+,(a,+,  +  j-2) 

-R2{l-z)-{Rrl-R,UR,l)-{x-z)i]\'r-L  +  T). 

Hence  the  bending  moment  at  P  with  the  fird  distribution  of 
mights  is  greater  or  less  than  the  bending  moment  at  the  same 
point  with  the  second  distribution  according  as 


Mi  %  M2, 


or 


Iil(J-^)-Rrl  +  zWr>R2(l-z)-{Rr-Rp  +  R,)l-ix-z)(Wr-L  +  T), 

or  zWr-{Rp-R,)l  +  {x-z)i]\\-L  +  T)  >  (R2-R,)(l-z) 


or 


'(f^-T+R.-Rr,)+liR,-R.)+x{W,-L  +  T)>'^il-z)m\-L).(B) 

\'nte.—U  no  weights  leave  or  advance  upon  the  girder  R„  R 
and  L  are  severally  nil,  and  '     *" 


according  as 


Ml  >  A/2, 


-zT  +  lR,+x{Wr  +  T)>Yl-z)W, 


If  also  the  point  P  coincide  with  the  rth  weight,  and  the  dis- 
tance of  transfer,  jri=  0^-0^+1,  then 

RJ^tPr+iOr+i,    7'  =  uv+i,    and    2  =  0,. 
Hence  Mi  ^  M2,  according  as 


or 


l-ar<    1    • 


i-e.,  according  as  the  sum  of  the  first  r  weights  divided  by  the  length 


130 


THEORY  OF  STRUCTURE-. 


oj  the  corresponding  seipnent  is  greater  or  less  than,  the  total  weight 
vpon  the  girder  dirided  by  the  t<pan. 

If  the  weights  are  cohcentratetl  at  the  panel-points  of  a  truss, 
the  last  relation  may  be  expressed  in  the  form 

first  (r)  weights  ->  total  weight 

r  {>anels        ^  total  number  of  panels* 

p.  Graphical  Determination  of  the  Maximum  B.M.  at  any  Point 
of  an  Arbitrarily  Loaded  Girder  AB.— When  all  the  weights  remain 
on  the  girder  the  livst  result  in  the  preceding  article  indicates  a 
simple  graphical  method  for  the  determination  of  the  max.  B.M.  at 
any  point. 

Erect  a  vertical  line  AC  at  one  of  the  supports  A,  and  scale  off 


lengths  .l/i,  /,/2,  .  .  .  l,-\lr,  .  .  .  ,ln-if  to  represent  the  w^veral 
weights  »',.  n'2,  .  .  .  ,  w.,  .  .  .  n\,  taken  in  order. 

.loin  fiC,  idid  from  am-  jM»int  I)  iK-tween  /,  and  i,_i  drawn  DO 
piiriiilcl  to  ('li  iii'.i'rscctiiig  Ali  in  O. 

Tlic  H.M.  ;it  O  is  a  max.  wiicii  tlic  weight  u\  is  concentrated  at 
that  point. 

II  till'  liii''  p.uallel  to  en  passes  tlirough  the  divisior.  jw)int 
iH'twecii  tw(t  adjac-nt  load  lengths  the  li.M.  at  the  |K)int  of  intersec- 
tion with  AIj  will  Ik'  a  max.  when  either  of  the  two  iouda  ruprusenif-d 


SHEARING  FORCES  AND  BENDING  MOMENTS.  131 

by  the  adjacent  load  lengths  is  concentrated  at  this  point.  For  exam- 
ple, the  B.M.  at  Oj  is  a  max.  when  either  uv_,  orw^  is  concentrated 
at  (h,  while  the  B.M.  at  O2  is  a  max.  when  either  u>,  or  «;,+  ,  is  con- 
cent rat«l  at  O2. 

.\gain,  the  S.F.  between  the  rth  and  r+Ith  loads  =  fti-TF„  and 
th  .s  is  zero  if  the  B.M.  is  a  maximum. 

Therefore  R^  =.  Wr. 

Ut  y  be  the  distance  of  the  C.  of  G.  of  the  n  weights  from  B. 
Thfn  Wn-y^R^.l^Wrl. 

Therefore  ?^  =.  HI  =  ^  _  ii^^      l-Or 


and 


I     Wn  "  AC  "  AB 


I 


Hence  the  rth  load  and  the  C.  of  G.  of  all  the  loads  are  equally  distant 
irt.iii  the  ends  (or  the  centre)  of  the  span. 

Is.I'kI"  hL""^',  V^u  "'  ^'  ^^•^'  ^'^^'  21.700,  22,900,  :8.550, 
IVtHK),  18,000,  and  18,000  lbs.  travel,  in  order,  over  a  truss  of  240  ft.  span  and 

l.<t  Ap,p, .      B  be  the  truss,  p.,  p,,  p, , . . .  being  the  panel-points.     Let 
h.  l.Kuis  travel  from  B  twards  .4,  ar     compare  the  shear  in  the  panel  p,p, 
»  .. ',    he  weight  of  3000  lbs.  has  «>aol..d  p.  with  the  shear  in  the  same  panel 
"ti.ii  the  weights  have  advanced  another  24  ft. 

«.-ixl85.50-1855Ibs.,    7?,-0,    £-1; 

H'.- 91300  lbs.,     L-0,     r-SOOOIba. 

Ill  II'  u  N,  ^  S„  according  as  (sec  A) 

3000-0^1855+ 1(91300-0)  ^10986; 

nn.!,'h,.rrfore,  s,<Su 

1^  t  the  weights  again  mlvancc  24  ft. 

ff.-r-.XlHOOO-lSOOlbs.,    l{„    n.    -^^-i- 
10  '  /     10' 

n,- 109300  lb...     L-O,     r  -  23000  ib« 


132 


THEORY  OF  STRUCTURES. 


Henro  i,  ^  S-,,  arrording  as  (sop  A,  .A-'.  S  ) 


23600-0^1800-0+1(109300-0),    or    23600^12730, 


and  thorcforc 


.S,>,S,. 


HotKT  the  shear  in  llio  panel  p.,.,  is  a  maximum  when  the  weight  of  3000 

lOH.    Is   lit    /'^. 

A^Min  let  the  3000  ll.s.  1h^  at  ,.:.  and  eompare  the  fv^ndinir  n.nment  at  p 
wth  the  U.ndmg  n.-nnent  at  the  same  point  when  the  weights  have  advance 
Iii8t  _'.  ft.  and  then  4S  ft.  towards  .1. 

First.   --120    ft.,  /.  =0.    r^L'i!(HX)    lbs.,    /?,/=  l,S000x24,  /?,-0     R  ,. 
22000v.)fi,  ^=..4  ft.,  ir.  =  f„S4(K)  ll,.s.,  »-,  =  l438.W  lbs.  '       ' 

Hence  -U,  <  .1/,,  according  as  (.sec  B,  Art.  8  ) 
1 20(0  -  22900  + 1 800  -  0)  +  22000  X  'Mi  - 1 8000  X  24 

+  24(r.K400-0  +  22«IOO)  >  .J^/ 240 -120) (145850-0), 
142.->(Woi^l7.-.O200, 
V,  >  .1/,. 


or 

and  therefore 


S,.n,ul   z-VM    ft.,    /.-.3(XK)    lb.s.,    7'-18.>W,    /?.-0,    «J-3O0OX216 

^^-i8.-i.wv.m.  x-24  ft.,  nv-.u.«K)  lbs.,  ir,- 163850  ibii.        •'^^^le, 

Hence  .l/i       .t/j,  according  as  (see  H) 
120  (3000  -  1,S.550  +0  -;«)00 --[^y)  +  240 (l 8.5.50 -.y^'  -o) 

>  24 


+  24(91.300-3000  +  18.5.50)  >  .:^-  (240- 120) (163850 -3000), 


or 

and  tliereforc 


240' 

21.5.5'2tK),'  19.{()200, 
.1/,  >.!/,. 


Hence  the  l)en(iing  moment  at  /.,  is  a  maximum  when  the  weight  of  3000 
llw.   IS  at   /..,   I.e.,   when  all   the   panel-point.s  are   loaded 

i:x.  <1.   Thnr  e,,u„l  whnh,  n.rh  h..ut„l  ,n,l,  „„.■  I„n,n,ll  at  Ihr  .„mr  c 

!'":;"" ;".''; "•''"■''"■  •'" •'' ■- /'•  ^/""'-  t'"  ••'' "'"' -•-' "•/"•./. nrr 4  n  ,,,,.1 
t- .;;:;:  f  "■*'''■'  -  /'■  "'"'^'-  ^''•""'  '*'•  '"''^"'"""  *'-^'-  (^''i/.  2«.5)  „„</  b  .u 

1 /*(<;.  .'IHO  iliaiiriims. 

\a'\  the  wheels  roll  from  A  to  W; 

!"'!  u   '  w  "M"',"";  '^•''-  '"''"'"'  *'"■  '"♦•  -■'•  ""•'  •'»<•  ^*-»'"«>K  resiK-etivelv 
-•t  .U,„  .1/*,  .1/.  Im.  the  H..M.  at  the  Is..  2,1,  un,|  M  wheels: 

iH-t  /i  lie  the  reaction  at  .1. 


''!*« . 


SHEARING  FORCES  AND  BENDING  MOMENTS. 

For  the  1st  wheel: 

I-t't  X  be  its  distatipe  from  .^. 

From  j;  =  1'  to  z  -4'  the  1st  wheel  only  is  on  the  girder. 


133 


Therefore 


R- 


,I2-£ 

12 


.l-JL 
12 


ari<l 


S.-H-1-f:,, 


which  ia     1     when     x~0    and  is    f    when     x-4'. 


Fio.  265. 
It>  this  case  ah  is  the  S.F.  diagram  from  A  to  4. 


Also 


.1  I  ..rabola  with  its  vertex   at   D,  C  horizontally  and  3  ft.-tons  vertically 

''"III  .1.  ' 

IheH.M.is    0     when     x-0    and  is     J  ft.-tons     when     x-4'. 
W  is  the  eorrcsixmdinK  H..M.  diiiKriini. 
Irom  x-1'  to  x-G'the  1st  and  M  whc-ls  are  on  the  girder.     Then 

^_12-j;  ^  U\-T    7     X 


12 


12       3     0' 


nil. 


S.-ft-l-l-£, 


"I"  Ills     J     when    x~4'    and  is     J|     when    x-6'. 
Ill  this  rase  be  is  the  S.F.  diajfram  fnim  1  to  0. 


Also. 


.U,-tf.r-lx4-'x-''-4 
3        0       ' 


iiinr 


134 


THEORY  OF  STRUCTURES. 


a  parabola  with  its  vertex  at  F,  T  horizontally  and  4J  ft.-tons  vertically 
from  A.  ' 

The  B.M.  is    }     when    x  =  4'     and  is    4    when    x-6\ 
CE  is  the  corresponding  B.M.  diagram. 


Fio.  266. 


From  X  -  6'  to  X  -  8'  t  he  t  hree  wheels  are  on  the  girder.    Then 


R- 


12-x  ,  16 


IH-x    23 


12 


6      4 


and 


So-R-2- 


11 


vhich  is     J     whrn    r  -  »i 


Inth 


and 


i.s  casp  al  is  the  S.F.  d 


-I     when    x-iy. 


liiigram  from  G  to  12. 


.\ls 


M.-Rx-lx4-lx&-' 


-.--10. 


a  p.inibolii  with  its   vcrtox 
from    .1. 


at  T,  7J'  horizontally  and  4jJ  ft..tons  vertically 


The  H.M.  j!^     1 


\vh 


uMi     x-6'    and  is    O     when    x-12'. 


Th.'  H.M.  ]•<    -l;     uhon 

E<;HH  in  the  forrcspondinjt  H.M.  diagram. 

f'..r  Me  M  irhrri: 

Let  X  »,<>  now  the  distance  of  this  whti-l  from  .4. 

From  x-0  to  x-2'  the  first  two  wheels  only  are  on  the  girder.     Then 


R~ 


S-x 
"    12    ■ 


i     6 


^ffil:: 


SHEARING  FORCES  AND  BENDING  MOMENTS. 


135 


and 


Si-R-L-' 
3     6" 


which  is    }    when    x-0    and  is    }     whea    x-2'. 
In  this  case  the  S.F.  diagram  is  a'b'  from  A  to  2. 


Also. 


3/.-«.-2.x-l. 


a  parabola  with  its  vertex  at  L,  5'  horizontally  and  4J  ft.-tons  vertically 
frnm  .1. 

The  B.M.  is    0    when    x-0    and  is    S     when    x-2'. 

AK  is  the  corresponding  B.M.  diagram. 

From  x-2'  to  X- 8'  the  three  wheels  are  on  the  girder.    Then 


and 


P     14-1      12-x^8-z     17     X 


which  is    I    when    x-2'    and  is     -|    when    x-y. 
Ill  this  case  the  S.F.  diagram  is  6'c'  from  2  to  8. 


Also. 


M^-Rx-l.2-^x---2, 

0         4 


8  p,ir;il)ola  with  its  vertex  at  Q,  5j'  horizontally  and  6f,  ft.-tons  vertically 

froni  .i. 

IhcH.M.  is  I   when  when   x-2',    is  «,♦    x-8',  and  is    V   when  x-4'. 

K\(}<i  is  the  corros|)onding  B.M.  diagram. 

Irom  X - 8'  to  X - 12'  the  2d  and  3d  wheels  only  are  on  the  girder.    Then 

p    12-J  .  14-x    13    X 


and 


5»-«-l-^-i 


whiihis     -I    when    x-8'     and  is     -|     when     x-iy. 
Ill  this  rase  the  S.F.  diagram  is  c'rf'  from  8'  to  12*. 


Also, 


.V.-ftx-lx2-^x-i-'-2, 


a  p  irahola  with  ita  vertex  at   64'  horijontallv  and  5^  ft.-tons  vertically 

frnlli  .1. 

I  hi-  H.M.  is  V   when  x-8',  it  0  when  x-12',  and  is  3   when  x-lV. 


136 


THEORY  OF  STRUCTURES. 


GSB  is  the  corresponding  B.M.  diagram. 

For  the  M  uhrcl. 

Let  x  be  now  the  distance  of  the  3d  wheel  from  A. 

Fromx-Oto  x  =  0  the  three  wheels  are  on  the  girder.    Him 


Ud 


12    "^    12    "^  12    "3  ~4 


which  is    I    when    x-0    and  is    |    when     x-6'. 
In  this  case  a"b"  is  the  S.F.  diagram  from  A  to  6. 


Abo, 


o        4 


from"^'''  ^^^  '**  ''^'**''  "^  ^'  *^'  *'°"^''"*''"y  «'"*  5t  ft-toM  vertically 

The  B.M.  is  0  when    *-0,    is  5  when    z-6',    and  is    V-   when    x-4' 

An  IS  the  corresponding  B.M.  diagram. 

From  X  -  6'  to  X  -  ly  the  2d  and  3d  wheels  only  are  on  the  girder.    Then 

/2_12z£  +  lp-f    Ll_i 
12  12    "  6     6 


ud 


Sc=R 


(J      0' 


which  ia    |     when     x-6'    nnd  is    J    when    x-10'. 
In  this  case  the  S.F.  diagram  is  b"c"  from  6'  to  IV. 


Also, 


.V.-ftx--x--, 


a  r^ar.alK,Ia  with  its  vertex  at  P.  r,y  horizontally  and  5,V  ft.-tons  verticafly 

The  B.M.  is    .-,     when    x-C    and  is     J     when    x-KY. 

I'll  is  the  <<>rn-;|)(iii(iliiir  W.W.  ■li:iur;iin. 

From  x-Ky  to  x-12'  the  3<1  wheel  only  is  on  the  girder.    Thea 

p_12-x  X 


and  s,'-R~.\-^- 

12' 

which  is    J     when    x-lO*    and  is    0    when    x-iy. 


SHEARIXG  FORCES  AND  BEX  DING  MOMENTS. 
In  this  case  c"b  is  iLc  S.F.  diagram  from  10  to  12. 


137 


Also, 


M^-Rx-x- 


12' 


'  ^''TK^n^r'- ** ''.  ''^'^''  **  ^'  ^'  »'°"^°nt«"y  '^nd  3  ft.-tons  vertically  from  A. 
TheB.M.  IS    J    when    x-Ky    and  is    0    when     x-12' 
DUB  IS  the  corresponding  B.M.  diagram 
The  maximum  B.M.  diagram  is  therefore  made  up  of  the  parabolic  arcs 


Fio.  267. 

AN,  NQG,  GHB,  and  the  absolute  maximum  B.M.  is  6^  ft.-tons  at  5#  ft 
from  .1.  *•  »  "" 

The  B.M.  may  also  be  obtained  in  the  manner  explained  in  Art.  7.    Take 
as  th..  scale  for  lengths  ,V  in.  -1  ft.,  and  as  the  scale 
for  f,.r(os  h  m.-l  ton.     Also  take  the  polar  distance 
'I,;  in.  =3  tons. 

Dr.iw  the  funicular  polygon  when  the  wheels  are 
<')in (lit rated  at  4,  6,  and  10.  Move  the  supports  right 
mill  l.lt,  1  ft.  at  a  time,  closing  the  funicular  polygon 
■111'  r  c.i.h  operation.  The  'envelope  of  moments,""  Fig. 
-'■r.  is  thus  obtained  and  the  maximum   B.M.  at   any 

i'"int    is   the   intercept   on   the   vertical   through  that 

P..,nt  .ut  off  by  this  envelope  and  the  opposite  boundary  of  the  funicular 
l-l  ......     Thm  m,cr«.pt  can  be  easily  scaled  and   the  B.M.  in  f.H.t-tTs  is 

•i    the  intercept  in  feet. 

10.  Hinged  Girders.- Any  point  of  a  girder  at  which  the  B.M  is 

"  '  .s  nillod  a  point  oi  contrary  flexure,  and  on  passing  .uch  a  point 

n.'  H..M   noco.s.sunly  changes  sign.    A  hinge  (or  pin)  may  thorefore 

1;'  -mroduced  at  th.s  ,K>int  an.i,  if  it  is  strong  enough  to  bear  the 

-  :i. .  the  eciuihbnutn  of  the  girder  is  not  affected.    Consider  a  hori- 

-mal  >r,rder  re«tn.g  upon  four  supports  at  A,  B,  C,  and  D  and  hinged 

at  u.r  pomts  h  and  F  m  the  side  spans 

Let  AK-a,  EB-b,  BC-c,  CF~e,  I)F~d- 


Fio.  268. 


138 


THEORY  OF  STRUCTURES. 


Lot  Wi,  W2,  W3,  M\,  TFs  be  the  loads  upon  AE,  EB,  BC,  DF, 
FC,  respectively,  and  let  x,,  xa,  X3,  X4,  X5  be  the  several  distances  of 
the  corresponding  centres  of  gravity  from  the  points  E,  B,  C,  F,  C. 

K' 


Fio.  269. 

Since  the  B.M.  is  nil  at  ^  and  ^  and  also  at  D  and  F,  the  two 
portions  AE  and  DF  are  in  precisely  the  same  condition  as  two 
independent  girders  carrying  the  same  loads  and  resting  upon  sup- 
ports  at  the  ends. 

The  portion  EF  may  also  he  treated  as  an  independent  girder 
supported  at  B  and  C  and  carrying  in  addition  to  the  weights  W2, 

a  weight  Wih-^j  at  the  cantilever  end  E 
"     "       "'^(l-l')  "     ' F, 


and 


these  two  weight,'^  l^inp  equal  to  the  reactions  at  E  and  F  respectively 
on  the  a.ssuniption  that  AE  and  DF  arc  indeix-ndeiit  girders. 

Lot  Ri,  /?2,  /?3,  /?,  be  the  reactions  at  .4,  B,  C,  D,  respectively. 
TJion  Ria-Wixi 


and 


Rid-WiZt. 


Ri  an.!  /?4  are  therefore  always  jwsitive  and  there  is  no  tendency  on 
the  part  of  the  girder  to  rise  from  off  the  supports  at  A  and  D,  and 
tons(.(|uently  no  anchorage  is  needed  at  those  jwints. 
Take  moments  about  C  and  B.    Then 

-(Tri-«i)(6  +  c)-UVxa  +  c)4«2O-ir3X3  +  ir5Xs  +  (ir4-ff4)e-0 


SHEARISG  FORCES  AND  BENDING  MOMENTS. 


139 


and 


+  ir5(a:5+c)  +  (Pr4-A<)(c+e)  -=o, 

two  equations  giving  R,  and  R^,  since  R,  and  R,  have  been  already 

determined.  •' 

The  pier  moments  Pi  at  B  and  P^  at  C  are 


and 


^/<c;r  ra/«es  devending  solely  upon  the  loads  on  the  spans  containing 
the  hinges.  ^ 

The  bending  moment  at  any  point  in  BC  distant  x  from  B 

'=R2X-iWi-Ri)(b  +  x)-W2(X2+X)-M 

=  Pi  +  x(fli  +  ^2  -  Wi  -  W2)  -  M, 

M  t.oing  the  bending  moment  due  to  the  load  upon  the  length  x 

I  he  shearmg-force  and  bending-moment  diagrams  for  the  whole 
glider  can  now  be  easily  drawn. 

For  any  given  loads  upon  the  side  spans,  let  AEH  and  DFL  be 
til.'  hendmg-moment  curves  for  the  portions  AB,  CD-  BH  and  CL 
nrresentmg  the  pier  moments  at  B  and  C  respectivelv  The  bend- 
ing n.onients  for  the  least  and  greatest  loads  upon  BC  will  be  repre- 
sHit.Ml  by  two  curves,  HKL,  HK'L,  and  the  distances  TT'  VV 
t  Mot,Kh  ^vhich  the  points  of  contrary  flexure  must  move,  irldicate 
tlH..;  portions  of  the  girder  which  are  to  be  designed  to  n.sist 
'"iiiliug  actions  of  opposite  signs. 

Affiiii,  let  the  two  hinges  Im;  in  the  intermediate  span 

Ix't  AB^a,  BE^b,  EF^c,  FC^e,  CD^d; 

[.(•t  Wu  W2,  W3,  ]\\,  ]\\  be  the  loads  upon  AB,  BE  EF  CD 
(  .  respectively  and  let  x„  x„  x,,  x,.  x,  be  the  several' distances 
''Ml,r(.orre.si)onding  centres  of  gravity  from  the  points^  B  F  C  C 

hr  ovi.lontly  may  be  treated  as  an  independent  girder  supi'mr'ted 
ai  I  lie  two  ends  Hn<l  carrying  a  lo:u{  W3. 

.■\''  ;»»'»^^'"'^y  be   treated  lus  independent  girders   carrying 
tH"  loa.ls  U.,  U.  and  W„  W„  respectively,  and   also  loaded  at 


^^0  THEORY  OF  STRUCTURES. 

the  cantilever  ends  E   and  F   with  weights  equal  to  the  reactions 


at  E  and  F  ilue  to  the  load  W3  upon  girder  EF,  which  is  assumed 
to  be  independent.    Thus 


theloadat^  =  ]r3-: 


(<      It    (I 


F=Tr 


■('-?)■ 


The  pier  moments  Pi  at  B  and  P2  at  C  are 


Pi  =  Tr2X2  +  Tr3J3- 


and 


P2  =  Tr5j5+Tr 


.(>-?> 


//letr  ro/M6.s  depending  sokhj  upon  the  loads  on  the  span  containing 


the  h 


infics 


Let  lii,  R,,  /i*:,,  7^4  he  the  reactions  at  .1,  B,  C,  I), 
and  take  inonicnts  about  the  points  B,  A,  D,  C.    The 


respectively, 


liiii  -  M'lX,  +  ( H'aJ-a  + 1' Va^)  =0  =  Pio  -  M'lXi  +P, 
-/?2fl4-]r,(a-j,)  +  ir2(a  +  J2)+Tr3J3— =0: 


ftad 


-Rid-]V 


(e  +  (/)  -  TrsC/s + c/)  -  TJVd  - 1,)  =  0  ; 


Trja-o  +  WiXt  =  0=-R^d  +  Wtn  -  Pj 


EXAMPLES. 


141 


/?2  and  7?3  are  always  positive, 

Ri  is  positive  or  negative  according  as  WiXi^l'i,   and 

Thus  there  will  be  a  downward  pressure  or  an  upward  pull  at 
oacli  end  according  as  the  moment  of  the  load  upon  the  adjoining 
spun  is  greater  or  less  than  the  corresponding  pier  moment.  The 
(•nils  must  therefore  be  anchored  down  or  they  will  rise  off  their 

supiKirts. 

riio  shearing-force  and  bending-moment  diagrams  for  the  whole 
girder  can  now  be  easily  drawn. 

Lot  HEFL  be  the  bending-moment  curve  for  any  given  load 
upon  the  span  BC,  BH  and  CL  being  the  pier  moments  at  B  and 
C  r('si)octively. 

The  bending-moment  curves  for  the  least  and  greatest  loads 
on  the  siile  spans  may  be  represented  by  curves  ATH,  AT'H  and 
DVL  DV'L,  and  the  distances  TT,  V\"  through  which  the  points 
01  tontrary  flexure  move  indicate  those  portions  of  the  girder  which 
arc  to  be  designed  to  resist  bending  actions  of  opposite  signs. 

Reverse  strains  may,  however,  be  entirely  avoided  by  making 
the  length  of  EF  sufficiently  great  as  compared  with  the  lengths 
of  tlie  side  spans.    . 

The  preceding  examples  serve  to  illustrate  the  mechanical  prin- 
ciples governing  the  stresses  in  cantilever  bridges. 


EXAMPLES. 


I.  A  bfiam  20  ft.  long  and  weighing  20  lbs.  per  lineal  foot  is  placed  upon 
support  dividing  it  into  segments  of  IG  and  4  ft.,  and  is  kept  horizontal 

a  .  owtiward  force  P  at  the  middle  point  of  the  smaller  segment.     Find 

value  of  /'  and  the  reaction  at  the  support. 

Sli'.w  that  the  required  force  P  will  be  doubled  if  a  single  weight  of  150  lbs. 
^  i-|M'ni|cd  from  the  end  of  the  'anger  segment.     Draw  shearing-force  and 


si-tnoMient   diagrams   in    both   cases. 


Am.  IJOOlbs.;    1<)(H)  lbs. 


rca 


^.  A  ;nu„  and  eight  boys  carry  a  sfi.k  of  timk-r,  the  man  at' the  end  and 
ri-lit  boys  at  a  common  point.  Find  the  iwsition  of  this  point,  if  the  man 
'    arry  twice  as  much  as  each  boy. 

Am.  Distance  between  supix)rt8  =  J  length  of  l)eam. 
'.  A  timljor  beam  is  .supported  at  the  end  and  at  one  other  point  ■    the 
Mon  at  the  latter  is  double  that  at  the  end.     Find  its  position. 

Ans.  Distance  between  supports -i  length  of  beam. 


Imm 


142 


THEORY  OF  STRUCTURES. 


i^JL  ^T\   f'  f ^^  "'^  ^^^"^  "^  ^  ^''^  C  80  as  to  act  as  one 

t^^Lr^r^  "'  1^::^  ^'  ^^-^-  ^^-  ^^-^  ^'•'  ^^-^^  ^t.;  each  of 
the  bolts  will  bear  a  bending  moment  of  100  Ib.-ft.  Find  the  greatest  weight 
which  can  be  concentrated  on  the  portion  BC. 

Abo  find  the  greatest  load  which  can  be  uniformly  distributed  from  4  to  D 
Draw  the  corresponding  S.F.  and  B.M.  diagrams  in  each  case. 

,    A  K-        ^n    ^o«.  ^««.  IVf  lbs.;  251b8. 

5.  A  beam  ^B  of  30'  span  rests  upon  a  support  at  A,  is  fixed  in  a  wall 
at  B  and  « Jimged  at  a  pomt  C  dividing ^B  into  the  two  segments   'C-20' 

In  each  case  give  the  maximum  S.F.  and  maximum  B.M.  on  AC  and  on 
^-  ^'»»-  («)  10  tons,  50  ft.-tons;  20  tons,  175  ft.-tons 

....         ...  ('')    2  tons,  20  ft.-tons;  6  tons,  40  ft.-tons. 

6.  A  horizontal  girder  of  length  21  rest?  upon  supports  and  carries  N  weights 
each  equal  to  W.  If  2a  is  the  distance  ^etweeri^Snsecutive  weightrprace 
the  weights  so  as  to  throw  a  maximum  bending  moment  on  the  girder.  Find 
this  moment. 


Ans.  if  N  is  even,  maximum  B.M.  - 
is  l-a(N-i)  from  support 


WN 

81 


(4l*+a'-2NaD  when  1st  weight 


If  iV  is  odd,  maximum  B.M.—:4^-^(Af._l)   ^hg^   1st 


WNl     Wa 

~2         4"(^'-l)   when   1st   weight 
isi-o(iV-l)  from  support. 
7.  Two  weights  P  and  Q  «P)  are  carried  by  a  horizontal  girder  of  length  t 
resting  upon  supports  at  the  ends,  the  distance  between  the  weights  being  a 

fnTfi  'i'.J^^'S'l*'  «°  '^.t°  t»>™*  ^  maximum  bending  moment  on  the  girder 
and  find  the  value  of  this  moment. 

An..  MaximumB.M.-Mzg£L'  when  P  is  ^iz^  from  support.  PT  being 


P+Q. 


iWl 


2W 


8.  Three  loads  of  P,  Q,  and  R  tons,  spaced  6'  and  4'  apart,  are  carried 
by  a  girder  of  20'  span.     If  the  B.M.  at  the  middle  point  of  the  girder  fs^Se 
same  when  either  P  or  Q  is  concentrated  there,  show  that  PO:fl-5-3   > 
Also  find  m  terms  of  P  the  maximum  B.M.  at  5',  10',  and  15'  from  iheend" 

9.  Three  loads  of  J,  4,  and  J  tons,  spaced  4'  apart,  are  to  be  carried  bv  a 
prder  of  18' span.  Place  the  loads  so  as  to  throw  a  maximum  B.M  on  fhe 
beam  at  4',  6'.  9',  12',  and  15'  from  the  end,  and  find  the  values  of  the  ^veral 
bending  moments.  «-veraj 

.4n..  19S  ft-tons;  26,Ht.-tons;  30i  ft.-tons;   25i  ft.-tons;    17i  ft.-tons. 
10    I-ig. -M  is  a  step-ladder  in  which  .4B  =  9'.. 4C'- 10',. 4J?  =8'  and  BC- 
6 .     A  man  weighing  160  lbs.  stands  at  a  point  3'  from  .4.     Find  the  tension 

!;;  fmo  th  ^■''-  '^'*  ^•"-  ^^""^  '"'  '"'^  '^«-  "^"'"•-^  '^^  «  "" 

°™°°    '  Aw,.  17  lbs. 


EXAMPLES. 


II.  A  man  of  weight  IT  ascends  a  ladder  of  length  /  which  rests  against  a 
smooth  wall  and  the  ground  and  is  inclined  to  the  vertical 
at  an  angle  a.  The  ladder  has  n  rounds.  Find  the  bend- 
itiK  moment  at  the  rth  round  from  the  foot  when  the 
man  is  on  the  pth  round  from  the  foot.  (Neglect  weight 
of  ladder.) 


Atu. 


**^P'7r-rTTi  sin  a. 


'(n+l)» 
I  J.  A  regular  pnsm  of  weight  W  and  length  a  is  laid 
upon  a  beam  of  length  2l(>a).  If  the  prism  is  so  stiff  as  to 
bear  at  its  ends  only,  show  that  the  bending  action  on  the 
beam  is  less  than  if  the  bearing  were  continuous  from  end 
to  end  of  prism. 

Ana.  1st.  Max.  B.M. 


2d. 


13.  A  railway  girder  50  ft.  in  the  clear  and  6  ft.  deep  carries  a  uniformly 
distributed  load  of  50  tons.  Find  the  maximum  shearing  stress  at  20  ft.  from 
one  end  when  a  train  weighing  1 J  tons  per  lineal  foot  crosses  the  girder. 

Also  find  the  minimum  theoretic  thickness  of  the  web  at  a  supp  >;  i,  4  tons 
being  the  safe  shearing  inch-stress  of  the  metal.         Ans.  l6j  tons;  .195  in. 

14.  A  beam  is  supported  at  one  end  and  at  a  second  point  dividing  its 
length  mto  segments  m  and  n.  Find  the  two  reactions.  Also  find  the  ratio 
of  TO  to  n  which  will  make  the  maximum  positive  moment  equal  to  the  maxi- 
mum negative  moment.  .        vi  w  — 

2^('»'-»»').2^('n+n)',  m:n:  A+V 2:1. 

15.  One  of  the  supports  of  a  horizontol  uniformly  loaded  beam  is  at  the 
end.  Find  the  position  of  the  other  support  so  that  the  straining  of  the  beam 
may  be  a  minimum.  ^       ^.  .  l«nirfh 

Ana.  Distance  from  end  support-    ^^ 

V2 

16.  A  steel  plate  girder  of  80  ft.  span  carries  a  uniformly  distributed 
load  of  80  tons  and  also  loads  ot  4,  6,  and  8  tons  concentrated  at  points  on 
the  Rirder  10,  40,  and  60  ft.,  respectively,  from  one  end.  Draw  the  S.F.  and 
li.M.  diagrams.  State  the  B.M.  at  each  of  the  concentrated  loads  and  find 
the  position  and  amount  of  the  maximum  B.M. 

Ana.  435;   1020;   790  ft.-tons;  at  the  centre,  1020  ft.-tons. 

17.  A  beam  of  80  ft.  span  carries  weights  of  4,  6,  6,  6,  6,  and  5  tons  at 
points  10,  20,  30,  40,  50,  and  60  ft.,  respectively,  from  one  end.  Determine 
the  supporting  forces  at  the  two  ends  and  draw  the  S.F.  and  B.M.  diagrams. 
Als„  state  the  B.M.  at  the  centre.  Ans.  ISJ,  14}  tons;   430  ft.-tons. 

18.  A  beam  of  80  ft.  span  carries  a  uniformly  distributed  load  of  80  tons 
a  fid  also  a  concentrated  load.  Find  the  amount  and  position  of  the  latter, 
th.  supporting  forces  at  the  ends  being  55  and  45  tons.  Also  find  the  B.M. 
and  S.F.  at  the  concentrated  load  and  at  the  centre. 

■  \ns  20  tons  at  20  ft.  from  support;  900  ft.-tons;  35  and  15  tons:  1000 
ft -tons;  —5  tons. 


144 


TIIEORY  OF  STRUCTURES. 


Stona 


i").  The  total  load  on  ilic  axle  of  a  tnirk  is  fi  tons.     Thn  wheels  are  6  ft, 
ap.'i'-'  he  two  a\l('-l)o\*i's  .")  ft.  apart.     Draw  the 

<■  .  iidin^  nioinent  oti  the  a.\le  ami  state  whut 

it  IS  at  the  centre.  Ans,  1}  ft.-tons. 

20.  In  the  triL^s  (Fip.  27'-)  the  points  of  trisection 
P  and  /•.'  (if  the  liori/ontal  member  BC  arc  con- 
y;  nect('(l  wiili  the  middle  points  F  and  G  of  the 
rafttis  liy  the  iiii'miM'Vs  1)1'  and  L(l.  Weights  of  2, 
'.i.  and  I  tuns  are  concent  rated  at  the  points  F,.\, 
and  a,  ros)>cctivcly.  Find  thi"  siip(H'rtin(r  forces  at  H  and  C  and  draw  the 
curves  of  S.F.  and  H.M.  for  the  mcml«r  /<('.      Tlie  span  is  (M)  ft.  and  the  rise 


30  ft. 


Ai 


4  tons;  .")  tons. 


Ji.  The  post  OH,  1  t  ft.  in  Icnpth,  is  pivoted  at  O  and  is  acted  upon  by  a 
horizo'ital  force  /'  at  (1.  wliirc  Ad  OO  =  tj.  The  member  -IF  is  pivotr<i 
at  .1  a;id  is  loaded  with  .">  tons  at  F.     The  joints  of 


the  memtMTs  IU\  II).  ni:,  and  FF  lii 


th< 


of  a  ciri  le  witli  ii.s  centre  at  O.  HI'  subtendint; 
an  atiftle  *'Af  at  O.  AF  is  connected  with  lU'hI'.F 
by  means  .if  parallel  tics  inclined  at  'M°  to  the 
vertical  ami  sp.aced  Jit  cipial  distances  apart  alon^r 
-IF.  iK'tcrmine  the  sfrcs-scs  in  these  lies  and  draw 
S.F.  and  M.M.  diapra-iis  for  the  post  OH  and  the 
nicMrixT  AF. 

22.  \.  JM'am  JO  ft.  long  is  loaded  nt  four 
|K>int.s  eijuidistant  from  each  other  and  the  (>nds 
with  e<|ual  weights  of  ,\  tons.  Find  the  U'tiding 
monienv  at  each  of  these  |Hiint.H  and  draw  the  curve 
of  shearing  foric 

J3.  .V  uniform  lieam  '_'0\  :{  ft.  in  length  rest.!! 
wth  one  end  on  the  gntund  and  flic  nilicr  against  a  smooth  vertical  wall: 
the  !i,\i.n  i.s  i.i  liticdat  til)  totlie  vertical  and  has  a  joint  in  the  middle  which 
ean  Inar  a  Uiiding  inument  of  ;{(»,(K)0  Ib.-fi.  Find  the  greatest  lojid  which 
iiiiiy  lie  umformly  clisiributed  over   the   U'am.     .\lso  lind  how   far  the  fiwit  of 


Fio.  273. 


the    lieam   sle.nid    be    moved   t( 


ds   th. 


in    order  that   an  udditiunul 


2000  lbs.  mav 


oncentrated  at  the  loinl. 


Draw  curves  of  shearing  fori'e  and  U'liding  moment  in  each  case. 

Alls.  ,HO(K)lhs.;  distance -10  ft. 

H.  \  girder  AH,  Fig.  271,  .'«)  ft.  long,  carri.«s  a  brick  wall  Id  ft.  hi.rfh  by  1  ft. 
thick  at,d  weighing  120  lbs.  |)er  ciibii-  f(H)i.  A  i|<Mirw,iy  H  -Ml  ft,  wide  is  rut 
in  the  wall  with  its  centre  line  10  ft.  from  the  end  supiMirt.  Draw  S.F.  atid 
iVM.  diagr.'ims,  stating  the  scales  used. 

»5-  A  girder  >*\  ft.  long,  Fig.  27"i,  carries  a  brick  wall  1  ft.  thick  and  wpiKh* 
Jng  rjO  lbs,  |n  r  i  iir.i.  f.ii.t.  Th.  r.  iiiiiiiiihg  dimensiiHis  of  the  wall  are  »iiuwU 
in  the  sketch.     Draw  S.F.  and  M.M.  diagrams,  staling  the  scales  used. 


ffli 


EXAMPLES. 


145 


j6    V  Warn  rarrios  a  load  of  2  tons  per  foot  run  over  thf  central  half  of 
„s  Lnpth,  and  a  load  of  1  ton  per  foot  run  over  the  remaining  portions.     It 


A 

►  -IJ- 

] 

.1      ,     ... 

83 

fc-l 

:»' < 

r..  274. 

P 

Fi 

>—«i>r-»|«" --»)'- T**^ 


=1 


P'lo.  275 


h,-  f.  1h-  supported  bv  two  props  so  that  the  preatest  stress  induced  in  the 
l„,,m  is  the  least  |wssihle.  Assuming  the  beam  of  uniform  section,  find  the 
pr.iit.  r  ix>sition  of  the  props. 

27  In  a  Ijeam  MiCDE  the  length  (AE')  of  24  feet  is  divided  into  four 
r,m  ,1  panels  of  0  ft.  each  bv  the  points  /?,  C,  D.  Draw  the  diagram  of  moments 
fur  the  following  conditions  of  loading,  writing  their  values  at  each  panel- 
point  •  (I)  Heain  supported  at  .4  and  K,  loaded  at  D  with  a  weight  of  10  tons; 
,  l..iin  supported  at  B  and  D,  load.><l  with  10  tonf  at  C,  and  with  a  weight 
„i  .'  tn,H  at  each  end  .4  an.l  E;  (in)  beam  encastn?  trom  .1  to  B,  loaded  with 
;i  \\Ai\n  of  2  tons  at  each  of  the  points  C,  D,  and  E. 

i8  \  iH'am  ARCD,  2S  ft.  in  length,  overhangs  its  two  supports  at  B  and  C 
•„v  111,'.  l..ii.'ths  l/i-S  ft.  and  (7)-4  ft.  Th:>  length  of  BC  is  10  ft.  and  O 
,.  ,1-  niiddlc  iviint.  \  load  of  1(K)  lbs.  is  susiK-nd.-d  from  .4,  and  loads  of 
ntid  JOO  ll)s.  are  uniformly  distributed  over  BO  and  CD  respectively. 


S(M( 


i  ;:„!  111,.  iH.sition  and  amount  of  the  maximum  positive  B.M.  on  the  beam 
uii.i  ,lraw  the  S.K.  and  B.M.  diagrams. 
.Ills,  ll.".''  ft.-tons  at  t'»l  ft.  from  B. 

S.t.  </t  tons:  100 at  .4  and  B,  +WJat  B,  -175  at  Rand  C;  +200 at  C, 

Oat  D.  ^  r. 

BM.  in  It. -Urns:  0  at  .1,  -SOO  at  R,  + 1000  at  0,  -  100  at  C,  0  at  D. 
10  V  girder  .1  />  of  1 S  f I.  span  is  to  carry  thn<e  weights  of  2,  4,  and  3  tons, 
t  ,k,  t,  1.1  order  un.l  8l>a<rd  4  ft.  apart.  The  iK>inls  B  and  <'  divide  the  sptm 
i„t..  the  thnv  Hegmetils  .l«-  t  ft..  /«'-S  ft.,  and  r/>-(l  ft.  Show  that  the 
!■.  \1  ill  aiiv  rx'int  in  -1»,  B(',ot('1)  is  a  ma\imurn  when  the  weight  of  2.  4, 
„r  :!  i.m^  wi.icht,  n-siH'clivelv,  is  i-oiKTiitrate*!  at  that  l>oint.  Alsii  find  the 
UN-  p  ,mf«  at  which  the  H.M.  is  a  maximum  for  two  distributions  of  the  load. 
i».i,nuiiie  the  corres|H>ndiiig  Utidiiig  moments. 

.\ns.  V.))  ft.-tons  at  4  ft.  from  .1;  2.".J  ft.-tons  at  IJ  ft.  from  .t. 
,»o.  A  uniformly  louded  U-ain  n>sls  u|>on  two  sup|K)rls.      Plaw  the  sup- 
I«.ris  v,>  that  the  straining  of  the  U-am  may  l)0  a  minimum. 

Ant.  iMstanee  of  each  supi^irt  from  orntre  -/M  -      -j, 

31.  Two  liars  AC,  CB  in  the  same  horinontal  line  are  jointed  at  C  and 

,Mi rf.d  upon  two  pro|)s.  the  one  at  .4,  the  other  at  some  point  in  CB  ilis- 

Ut.t  I  fn.m  V.    The  joint  C  will  wifely  licar  n  Ib.-fl.;  the  bars  are  each  I 


146 


THEORY  OF  STRUCTURES 


ft.  in    lengih    and   w  lbs.  in  weight, 
lie. 


Find  the  limits  within  which  x  must 


Ans.  l~— — -. 
Sul-f  -.'« 

3J.  A  nniform  load  PQ  moves  .alonp  a  horizontal  beam  rosting  ujKin  sup- 
por'  at  it.s  ends  .1  and  /{.  Prove  that  the  l)endiiiK  moment  at  a  given  pi»M  O 
is  a  maximum  when   I'ij  oeeupies  such  a  |«>sifion  that  OP:0Q::<)A.OH. 

Draw  etirvTS  of  maximum  shearing  foree  and  bending  moment  for  all 
points  of  the  beam. 

33.  A  beam  is  supported  at  the  ends  and  loaded  with  two  weights  mil' 
and  nW  at  points  distant  a,  h,  respertively,  from  the  ronserutive  supports. 

Show  that  the  bending  art  ion  is  greatest  at  mH' or  nil' according  as  -  ^  -. 


1+- 


m+nl 


and 


34.  Find  the  maximum  B.M.  on  a  horizontal  Iwam  of  length  I  supported 
at  the  two  ends  and  carrying  a  load  which  varies  in  i.itensity  from  w  at  one 
end  to  M'  +  pl  at  the  other. 

Ans.  Maximum  B.M. -|^(2i?-^),  where  x(px  +  2u) -2R,  ft  being  the 

reaction  at  the  end. 

35  ,\  wh(H»l  supporting  10  tons  rolls  over  a  beam  of  20  ft.  »pap.  Place 
the  wheel  in  such  a  position  as  to  give  the  maximum  bending  moment,  and 
find  its  value.  .^'i'-  .\t  the  centre;  50ton-ft. 

36.  Two  wheels  a  ft.  apiirt  support,  the  one  mW  tons,  the  other  nil'  tons. 
m  U'ing  >n,  and  roll  over  a  beam  of  /  ft.  .«pan.  Show  that  the  be^^ing 
moment  is  an  almitliUc  maximum  at  the  centre  or  at  a  point  whose  distance 

from  the  nearest  supfxtrt  is     -  —  - acctirding  as  '>  "( 1 

find  its  value  in  each  cas«'.     il<'2a.) 

4        Till'/.       ,,     m+n,,.i,       nn     )'.„„«, 

Ans.      ,     ton-ft.;  -~  — II   .  / v    ton-ft. 

4  4/         (       m  +  »i  ) 

37.  Two  equal  wheeU  4  ft.  apart  and  loaded,  the  one  with  4  and  the  other 
with  t'l  ton^,  roll  in  thi'<  ord'-r  ni  -i  i-oimtant  rate  from  left  to  right  across  a 
girder  of  0  ft.  span.  Fiml  the  maxitmiiii  IVM.  at  '>  ft.  and  at  8  ft.  from  the 
left  sup|Nirt,  and  also  finij  the  iHisitimi  .nid  :iinount  of  the  absolute  maximum 
B.M.     Draw  the  maximum  .S.K.  and  H.M.  diagrams. 

.l/».^.  1.!^  II. -I' .I-  ••'.  Ii.-i"ti*:  .■bs.ilute  maxitiuitii  U.M  ^  l.">',!  ft.-tons  at 
3.7  ft.  from  left  r.iii>|»ti. 

38.  Two  wheels,  each  siipfxirting  7  ton*  ,iiid  spaced  4  ft.  apart,  roll  over 
a  girder  of  7J  ft.  spin.  Find  the  rn.ixiiniiin  MM.  ;it  the  centre  and  the  «b«v 
lute  maximum  M.M.  for  ih.-  wli  iie  span.  .Mso  slinw  that  there  are  two  dis- 
tributions of  tlx-  I'lads  which  will  uive  the  same  niaMiniim  H.M.  at  two  {)ointS. 
Find  the  (Misitions  of  thes<'  jMiinis  and  llu'  eorres|M)n(iing  B.M. 

A-ir.  !.;;  ft.  iiiis.  14;'.,  ft.-tuiisat  1  ft.  fr"?!!  the  centre;  13,^  ft.^.ma  at 
3  ins.  from  the  ifiitre. 


EXAMPLES. 


147 


3Q.  Two  wheels  supporting,  the  one  11  tons,  the  other  7  tons,  and  spaced 
fi  fi.  aiuirt,  roll  over  a  fjirdor  of  l-'i  ft.  span.  Draw  the  B.M.  and  S.F.  dia- 
p:m<.  and  find  the  absolute  maximum  H.M.  for  the  whole  span.  Also  show 
that  liuru  are  two  points  at  each  of  which  two  distributions  of  the  loads  give 
the  SMinc  maximum  B.M.s.  Find  these  jMiints  and  the  eorresponding  H.M.s. 
l/,v.  ;{7.21  ft.-tons  at  IJ  ft.  from  the  eontre,  34.32  ft.-tons  at  3  ins.  from 
tlic  (.ii^'e,  and  2('>.(1(>  ft.-to:H  at  9.212  ft.  from  the  centre. 

40.  In  the  preceding  example  .show  that  ^he  maximum  negatiw  sh.-ar 
at  I.',  ft.  from  a  support  when  the  7-toii  wlu-el  oidy  is  on  the  beam,  is  the 
SHiiii  !is  the  maximum  negative  shear  at  the  same  point  when  both  of  the 
wiiii  U  arc  on  the  Ix-am,  and  find  its  value.  Als<i  show  that  the  maximum 
ui./rilnr  shear  at  OJ  ft.  from  a  support  is  the  same  when  only  the  11-ton  wheel 
is  nil  the  beam  as  when  the  two  wheels  are  on  the  beam,  and  find  its  value. 

Ans.  IHJ   tons;  ',V»*  to'XS- 

41.  Solve  Ex.  39  when  the  girder  carries  an  additional  load  of  J  ton  per 
lined  fiK)t. 

.\ns.  49.12a5  ft.-tons  at  .959  ft.  from  the  centre;  46.5075  ft.-tons  at  3  ins. 
and  MM'}  ft.-tons  at  3|V  ft.  from  the  ce'.u  \ 

41.  A  rolh-d  joist  weighing  150  lbs.  per  .  •■  il  foot  and  20  ft.  long  carries  a 
unif.irtiily  distributed  load  of  (lOOO  lbs.,  and  two  wheels  5  ft.  apart,  the  one 
btHrmtr  .VWO  lbs.  and  the  other  3000  lbs.,  roll  over  the  joist.  Find  the  maxi- 
niiiih  siicars  at  the  supports,  at  the  centre,  and  at  5  ft.  from  each  end. 

AtiK.  10,250  lbs.;  9750  lbs.;  3250  lbs.;  6750  lbs.;  6250  Iba. 
43.  A  rolle<l  joist  17  ft.  long  is  supporteil  at  one  end  and  at  a  point  13  ft. 
disi.iiit  from  that  end.     Two  wagon-whw'ls  5  ft.  apart  and  each  carrying  a 
l<m.|  of  littX)  U>8.  [Jass  over  the  joist.     Find  the  maximum  positive  and  nega- 
tive moments  due  to  these  weights,  and  also  the  corresponding  reactions. 
Ans.  Maximum  pnnitivf  B.M.  -5512i  Ib.-ft.; 

reactions  -  15.')0  and  1050  lbs. 
Maximum  negaliif  B.M.  -5200  Ib.-ft.; 

reactions  -  1700  lbs.  and  -400  ll)s. 
or  -'2900  lbs.  and  -;UM)  ll«. 
Thi'  maximum  B.M.  diagram  for  each  half  of  the  V.i-(i.  span  is  given  by 
.»/t- 100(21 -2x)x, 
xl»  m:  tnciwun-d  from  a  8up|K)rt. 

44    I'wo  wheels  loaded,  the  one  with  ft  and  the  other  with  4  tons,  and 

I  ft,  apart,  roll  from  left  to  right  across  a  girder  of  IS  ft.  span.     Draw 

~  I',  and  B.M.  diagrams.      Determine  the  position  and  amount  of  the 

iuti'  maximum  B.M.,  ami  als<»  the  maximum  B.M.  at  ft  ft.  from  the  left 

p'Tt.     Ill  the  latter  cii.oe  show  that   the  maximum  B.M.  is  the  same  for 

1  .!ilTcrciit  distributions  of  the  loads. 

.{lis.  23/j  ft. -tons  at  Oj  ft.  from  left  support;  1S|  ft.-tons. 
4";  rhn><>  whe<ls,  each  loadi-d  with  a  wi'ight  H'  and  s|)aced  5  ft.  a|.rfrt, 
n\ir  .1  Inniiu  of  IH  ft.  span.  I'lace  thi'  wlii-i-ls  in  such  a  |<oMition  as  to 
I  III'  maximum  iN-nduig  moment,  and  find  its  value..  .Mso  place  (a)  the 
.vii-;-  -.1.  itiiii  i'l.M.  at  any  piint  Ix'tw/M  ti  ihr  two  hindmi^t  whrrls  may  iw- 
(uii'-iii.t,  and  find  its  value.    .\lso  (h)  determine  all  the  pnaitiona  of  the 


til' 

iil~. 

Iw,. 


Its 


THEORY  OF  STHLCTCKES. 


wheels  which  will  givo  the  same  Ix-tuhn)?  moment  at  6  and  12  ft.  from  one 
end,  ami  fitul  its  value. 

Am.  8Jir  when  middle  weight  is  at  eentre  of  Ix^am. 
(n)  1st  whiH'l  at  1  ft.  from  support;    H.M.  =71V'. 
(/))   When  distance  Ix'tweeii  end  wheel  and  support  is  ^  2  ft.  and 
.-  (ift.M.M.-7ir. 

46.  Thr(>e  wheels  loaded  with  S,  <),  and  10  tons  and  spaced.')  ft.  apart,  a  rp 
pl.'iced  uiMin  .1  Ix'am  of  l.">  ft  span,  the  S-ton  wheel  In-ing  .3  ft.  from  the  left 
ahutnietil.  Determine  graphically  the  H.M.  at  (i  ft.  from  the  left  abutment. 
Al.-io  find  the  gre.itest  li.M.  at  the  same  point  when  the  weights  travel  over 
the  iH-am,  and  the  <tb.'<olutc  wdtimum  lx>nding  moment  to  which  the  beam 
is  suhjeited. 

Ans.  17?  ton-ft.;  '>.'j;  ton-ft.;  absolute  maximum  B.M.  tilJi  ton-ft.  at 
2d  wheel   when   1st   is  JJi   ft.  from  su(i|)ort. 

47.  Three  loails  of  .'),  :i,  and  4  tons,  spaced  t  ft.  apart,  travel  in  order  over 
a  girdiT  of  12  ft.  span.  Draw  niaximimi  S.F.  and  H.M.  di.igrama.  Show 
that  ilu'  U.V.  is  an  .•ii)solute  maximum  and  ((pial  to  ISj  ton-ft.  at  5J  ft.  from 
on,'  end  when  the  .Vtnn  loa<l  is  concentrated  at  that  |M>int.  .\l.so  show  that 
the  absolute  maximum  H.M.s  at  <»and  S  ft.  from  one  end  are  IS  and  14§  ton-ft. 
res|MMtivcly,  and  that  they  are  each  produced  by  two  different  distributions 
of  the   loads. 

48.  Tiirec  loads  of  3,  ."»,  and  4  tons,  s()aced  1  ft.  af)art,  travel  in  order  from 
lej  (I  right  over  a  ginlcr  of  12  ft.  span.  Find  the  |)<>sition  and  amount  of 
the  abt'olule  m.iximum  H.M.  .\lso  show  that  at  4  ft.  from  the  left  su,  ,K)rt 
tlicre  are  two  distributions  of  the  loads  which  will  give  the  same  maximum 
H.M.     Find  the  ar.ount  of  this  iKiuling  moment  at  the  |xiinl  in  question. 

Aug.  22,'i  ft. -tons  at  •>,'.;  ft.  from  left  .s»j>|>ort;  17J  ft.-tonn. 

49.  Thnc  loads  of  3,  4,  and  ."•  tons,  spaced  4  ft.  apart,  travel  in  order  from 
left  to  right  o\cr  a  ginlcr  of  12  ft.  sp.in.  FincI  the  |Hi>:itio?i  and  amount  nf 
the  ;il)soluic  Mci\iniuiii  II. M.  Show  ili.ii  I  ft.  from  the  Icfi  sup|Hirt  then' are 
two  distri'iutioiH  of  the  louls  for  wh;.  li  ihc  H.M.  has  the  same  maximum  value, 
and  find  its  amount.  .\lso  show  tli.it  at  some  jMiint  U'twi-en  (i  ft.  aid  7  ft. 
from  the  left  sup|Mirt  the  maximum  H.M.  is  the  same  for  two  distribution.s  of 
the  lo.'iii.     Find  the  |Hisiti<in  of  this  (Hiint  and  the  corres|)onding  H.M. 

Aii.i.  2U  ft. -tons  at  .">j  ft.  from  left  sup|xirt ;  17J  ft. -tons;  l!>.4ft  ft.-t<iii< 
at  li.472  ft.  from  h'fi  sup|M>rt. 

50.  Four  wlieels.  each  carrying.')  tons,  travel  over  a  girder  r»f  24  ft.  clear  span 
at  ci|Ual  di-itaiici'^  I  ft.  jiiiart.  l)cterinine  graphically  the  maximum  H  M. 
at  H  ft.  from  a  sup|»>rt,  and  also  the  ab^olute  maxiimim  H.M.  on  the  girder. 

Alls.  »J»  ton-ft.;  S()J  ton-ft. 
ji.   Four  wheels  each  loaded  wiiii  a  weight   il'  and  spaciHi  't  ft.  M|>url  rol: 
over  a  Ix'am  of  1  s  ft.  span.     Place  the  whc<'l.s  in  such  a  fiosition  as  to  (tive  the 
tiiaxinnim  Ih'ikIIiii;  motncni,  and  lind  its  value. 

.bi.f,  Ohi-  whi-ij  o.T  the  iM-.am  and  iniildle  wheel  of  remaining  three  at  the 
eontn';  maxiinuiti  H..M. -Kilt  .  If  all  wheels  are  on  b(*am.  maximum  B.M. 
-HfilF. 


/ 


EXAAfPLES. 


149 


52.  All  the  wheels  in  the  preceding  example  being  on  the  beam,  the  B.M. 
t  thJ  centre  for  a  certain  range  of  travel  is  constant  and  equal  to  that  for  a 

parii.ular  distribution  of  the  wheels  when  only  three  are  on  the  beam.     Find 
•,ii   i;iiiiie,  the  H.M.,  and  the  position  of  the  three  wheels. 

.\S.  While  the  end  wheel  travels  3  ft.  from  the  support;  SW;  the  first 
,li  ,1  .")  ft.  from  the  sup|H)rt.  ,    •,     u 

53.  If  the  load  on  each  of  the  wheels  in  Ex.  51  is  5  tons,  and  if  the 
Uiin  also  carries  a  uniformly  distributed  load  of  20  tons,  and  two  loads  of 
'  aiul  3  tons  concentrated  at  !K)ints  distant  5  and  0  Ft.,  resi)ectively,  from 
on,,  end,  find  the  maximum  shearing  force  (both  positive  and  negative)  and 
ill,,  maximum  iK'iiding  moment  for  the  whole  span;  also  find  the  loci  for  the 
ii.:i\inium  shearing  force  and  lx;nding  moment  at  each  jwint. 

Ins.  Denoting  the  distance  from  support  by  x,  the  maximum  positive 
shtarmg-force  diagram  is  given  by  equations  l&.Si  =  443-40x  from  j=0  to 
3  -  tJS  -  35x  from  x  =  3  to  ."),  ■=  392  - 3.->x  from  x  - 5  to  8,  - 352  - 30x  from  x  - 8 
t,' !t.  2'J8-30x  fromx-9to  13,-233-25x  from  x  =  13  to  18;  and  the  moxi- 
mnm  najiUivf  shearing-force  diagram  by  equations  1H.S, -233-2.'>x  from  x-0 
t.).-.,  172-30X  from  x  =  5  to  9,-118-30x  from  j-9  to  10,-6H-35x  from 
Z'  10  to  13,  -fv8-3.'>x  from  x  -  13  to  15,  -  -7  -40x  from  x  - 15  to  18. 

Maximum  positive  shear  -  V."  tons;  maximum  negative  shear -V,'  tons; 
mnitmum  IxmUmi  moment  curve  is  given  \>y  .!/»  -"iVx  - '»'  x'  from  x  -0  to  x -3; 
M,  '  i  'X  -  f  {x=  from  x  - 3  to  x  -  .■. ;  M,  =  '«•  x -  V-r'  -  »•'>  from  x  =-  5  to  x  - 8; 
M,    x\\*  -  JiSx'  +  12  from  x  -8  to  x  -9;  abs.  max.  U.M.  -  142  ton-ft. 

54.  Four  wheels  loaded  with  4.  4,  8,  and  8  tons  an>  placed  u|wn  a  girder 
of  Jl  ft.  span  at  distances  of  3  in.,  6i  ft.,  SJ  ft.,  and  9  ft.  fmm  the  left  8UpiH)rt. 
F;iiil  l.y  scale  measurement  the  iK-nding  moment  at  the  centre  of  the  girder. 
If  ; 111  wheels  travel  over  the  girder  at  the  given  distamrs  apart,  find  th; 
luiMiiiuin  U.M.  to  which  the  girder  is  subjected. 

An».  Max.  U.M.  -  122.22M  ton-ft. 

55.  Four  wheels  loaded  wiih  5.  2,  4,  nn<l  3  tons  travel  in  order  from  left 
t.i  ii'.'lit  over  a  girder  of  12  ft.  span  at  di.st.inces  4.  3,  and  2  ft.  apart.  Draw 
■\i  max.  S.F.  and  H.M.  diagnnns.  C.ive  the  maximum  fiositive  and  negative 
«!,.  ,:iMK  forces  at  0,  3,  7,  9,  and  12  ft.  from  the  left  8upiH)rt.  Find  the  pt«ition 
ill:  I  iinounl  of  the  nhs.  max.  H.M.  Show  th.-it  there  lire  two  distrihuticms 
..f .),.  wlii'<'ls  for  whi.h  the  H.M. at  5  fi.  from  the  lelt  siip!>ort  i.s  a  maximum, 
..si  iiiid  its  value;   also  show  that  there  is  a  lM>int  l)etween  2  and  3  ft.  ami 

11.  luiweeii  Sand  9  ft.  from  the  left  sup|Mtrt,  at  which  the  maximum  B.M.  is 
III    -  i!iie  for  two  distributions  of  the  wImh-Is.     Find  the  i>.)sitions  and  amounts 
f  'isi'  niaximiim  iM-ndiiig  moments. 

1  ..  ^Hi  and  -5J  tons;  +.■>!  iiiiil  -21  tons;  +3,»,  and  -  ,',  tons;  +5^ 
111!  2J  tons;  +8J  and  -  4J  tons;  21  ft. -tons  at  the  centre;  20J  ft.-t«iiis; 
!l      ft. -Ions  lit  2!  f'    and  l7.(H'i|  ft. -Ions  at  8.292  ft.  from  the  left  sui>i«irt. 

■;(..  .\  rolled  j.-i  weighing  t.V)  lbs.  |H'r  lineal  font  and  M  ft.  long  carries 
;h.  1  .iir  wh.-<.|s  of  a  lucomolive  at  3.  8.  13,  and  IS  ft.  from  one  end.  Find 
h  :;t\injum  Irf^rling  mntnf'Ut  jtsid  !'•-=•  m.-s\!!nu!!!  shear*,  both  p<>«itive  and 
luk'nive,  the  load  on  each  whe«'l  Uing  lO.lMM)  lbs. 

I  IS.  Max.  B.M.  -  I'23,(i0t)  Ib.-ft. ;  max.  shears  -  J.3..i(K) lU.  nid  25,.')00  lb*. 


150 


THEORY  OF  STRUCTURES. 


57.  Solve  the  preceding  example  when  a  live  load  of  21  tons  per  lineal 
foot  is  substituted  for  the  four  concentrated  weights  on  the  wheels. 

Ans.  Max.  B.M. -2,202,rjOO  ton-ft. 

58.  The  loads  on  the  whci-ls  of  a  locomotive  and  tender  |)a.s8ing  over  a 
Ix'am  of  (10  ft.  span  are  H.1S0,  14, ISO.  21,2(i(),  21,2(10.  21,2(>0,  21,2(50,  10,900, 
Kl.'.NH),  1(1,(H)0.  ltl,',HK)  lbs.,  counting  in  order  from  the  front,  the  intervals 
N'ing  5,  5i,  ."),  'y,  '),  S],  .">,  4,  .">  ft.  Place  the  wheels  in  such  a  [wsition  as 
to  give  the  maximum  iK-nding  moment,  and  find  its  value.  Also  find  the  maxi- 
mum IxMiding  inonieiits  for  spans  of  :H).  20.  and  1(1  ft. 

Ans.  For  (k)  ft.  span,  max.  K.M.  is  at  .")th  wheel  and  - 1  ,.559,925.4  Ib.-ft. 
when   1st   wheel  is  7.S.")  ft.  from  supjiort. 
For  'M)  ft.  span,  max.  H.M.  at  .'>th  wh(>el  when  2d  wheel  is  .596  ft. 

from  sup|>ort  and    =43(i,7<>l.4  Ib.-ft. 
For  20  ft.  span,  max.  H.M.  at  centre  when  3<i  wheel  is  2i  ft. 
from  sup|K»rt  and    =212,(100  Ib.-ft.  =  mix.  U.M.  at  same  point 
when  4th  wheel  is  .">  ft.  from  supix)rt. 
For  It)  ft.  span.  max.  H.M.  is  at  .5th  wheel  and   -132,875  Ib.-ft. 
when  4th   wheel  is  5  ft.  from  supjiort. 
5Q.  If  the  00  ft.  l)eam  in  the  preceding  example  also  carries  a  uniformly 
distributed  load  of  (10,(K)0  lbs.,  find  the  eur\e8  of  ma.ximum  shearing  force 
and  Ix'iiding  moment  at  cai  h  |H)int. 

60.  .\  span  of  /  ft.  is  cros.sed  by  a  beam  in  two  half-lengths,  supported 
at  the  centre  by  a  pier  whose  width  may  Ix'  neglected.  The  successive  weights 
on  the  whiH'ls  of  a  locomotive  and  tender  pa.ssing  over  the  Ix-am  are  14,(KX), 
22,0tX»,  22,000,  22.0(KI,  22,000,  14,(K)0,  14,000,  14,(M)0,  14,000  lljs.,  the  inter\-al8 
Ix-ing  71,  4J,  4i,  41,  101,  •"».  •"'.  •">  ft-  I'lace  the  whtx-ls  in  such  a  |H)sition  as 
to  throw  the  gri'alesl  (M>ssibie  weight  uinm  the  centre  pier,  an<l  find  the  mag- 
nitude of  this  weight  for  s|wiw  of  1 1 )  ."lO  ft.;  (2)  2.5  ft.;  (3)  JO  ft.;   (4)  IS  ft. 

An.i.  (I)  ,S.5,;{'20  ll>s.;    (2)  .5«1,S.S()  llw.;    (3)  4.S.4(K)  llie.;    (4)  44,000  lbs. 

61.  lioads  of  ;JJ,  (1,  (1,  t(,  and  0  tons  follow  each  other  in  order  over  a  ten- 
panel  truss  at  distances  of  S.  .5}.  41,  and  41  ft.  apart.  Apply  the  results  of 
.Art.  S  to  determine  the  (xisition  of  the  lr)ads  which  will  give  the  maximum 
diagonal  and  llaiige  stre.sst's  in  the  third  and  fourth  panels. 

6j.  a  truss  <if  2»0  ft.  span  and  ten  panels  has  loads  of  12J,  10,  12,  11,  9, 
9,  9,  9,  and  9  tons  concentrated  at  the  panel-|ioints.  Finil  by  scale  meiuturr- 
mriit  the  U-nding  moments  at  the  four  panel-|»oints  which  an'  the  most  heavily 
loaded,  and  determine  by  Art .  H  whether  these  are  the  gn-atest  lx<nding  momenta 
to  which  the  truss  is  subjected  as  the  weights  travel  over  the  truss  at  the 
panel  distances  jipart.  ,\„.i.  114(1,  1992,  2.59M.  2yi('  ton-ft. 

63.  l/iads  of  71,  12,  12,  12,  12  Ions  an>  nmcentrated  u|¥>n  a  horiiontal 
Ix'am  of  2.5  ft.  span  at  distances  of  IS,  los,  \M,  21(1,  and  272  in.,  n»S|»eetively. 
from  the  left  supiMirf.  Find  graphically  the  U-nding  moment  at  the  et-ntrc, 
of  the  spun.  If  ihe  loads  travel  over  the  truss  at  the  given  distanecs  apart. 
iMid  the  niaxnniim  M..M.  at  the  same  section.  Ann.  J.M9  ft.-tons. 

64.  \  s|Htn  of  /  ft.  is  crossed  by  two  i-antilevers  fixed  at  the  ends  and  hinged 
at   the   'cntre.     Draw   diagrams   of  shearing   force   and   bending   moment 


ill  N 


EXAMPLES. 


151 


,1,  far  a  single  weight  W  at  the  hinge,  (2)  for  a  uniformly  distributed  load 

of  iiitoiisity  If.  ,  ,      J-  t 

Am.  Taking  hingp  for  origin,  the  shearing-force  and  bending-moment 

diagrams  are  given  by 

(1)  S.'—;    Mz'-^- 


(2) 


Sx-wx;    Mz-  - 


65  A  beam  for  a  span  of  100  ft.  is  fixed  at  the  ends.  Hinges  are  mtro- 
,1u,.m1  nt  points  30  ft.  from  each  end.  Draw  curves  of  shearing  force  and 
iKiMii.g  moment  (1)  when  a  weight  of  5  tons  is  concentrated  on  each  hmge; 
(.'  wh.n  a  uniformly  distributed  load  of  i  ton  per  lineal  foot  covers  (a)  the 
ttMitrr  length,  (6)  the  two  side  lengths,  (c)  the  whole  span. 

.1«».  (1)        Side  span:  max.  B.M.  -  -  1.50  ft.-tons;  max.  8.F.  -6  tons 


Centre  span: 
(2)  (o)  Side  span: 
Centre  span: 
(6)  Side  span: 

Centre  span: 
(c)  Side  span: 

Centre  span: 


-  + 


"      _  — 


0 
75 
25 

0 

75 


-  +  25 


min. 

max. 
min. 
max. 
min. 


-0 

-2J 

-2i 

-3i 

-0 

-0 

-41 

-  I 

-2i 

-0 


66  A  beam  A  BCD  is  supported  at  four  poinU  A,  B,  C,  and  D,  and  the 
int.  rmodiate  span  BC  is  hinged  at  the  two  points  E  and  F.  The  load  upon 
til,  iHiim  consists  of  15  tons  uniformly  distributed  over  AB,  10  tons  umform  y 
,1  .iiil.utod  over  BE,  5  tons  uniformly  distributed  over  FC,  30  tons  ""'fwrmly 
.li«tril)ut»(l  over  CD,  and  a  single  weight  of  5  tons  at  the  middle  pomt  of  EF. 
.1/;  1.-,  ft.;  BE-hh.;  EF-15  ft.;  FC-10  ft.;  CD -25  ft.  Draw  curves 
of  l;.M.  and  8.F.,  and  find  the  points  of  inflexion. 
.Ir.s.  A  to  B.  S.F.  in  tons:    +5  at  .4,  -10  at  B. 

B  to  C.     :   +  12J  at  B,  +2i  at  E,  i2J  at  H,  -2i  at  F. 

CtoD.     "     "    "    •■   +17at  C,  -13at  D. 

A  to  B.  Max.  n.M.  in  ft.-tons:  + 12 J  at  5  ft.  from  .1,  -37 J  at  B. 

BtnC.      "       ' :  +18iat  W,   -50at  r. 

Cu>D      "       "     "       "      :  +70i'i  at  lOJ  ft.  froniD. 
Points  of  inflexion  are  10  ft.  from  A  in  AB  and  21  §  ft.  from  /)  in  CD. 
f,7   Solve  example  «9  when  the  gir«ler  carries  an  additional  load  of  1.30  tons 
„,„i,„>uly  .rwtribute«l  over  IM)  ft.  from  .1,  and  also  a  concentrated  load  of  20 
t  II.  Mt  ihc  middle  |K>int  of  (7i. 
Ans.  S.F.  in  ton.:  at  .4  -  -87i,  at  fl-  ilB7i.  at  W-  +87J,  at  C-  -02^ 
and  +91t,  at  20-ton  load  -51  i,  at  />  -  -8|. 
B  M  in  ft.-tons:  at  fl  -  -5100,  at  43|  ft.  from  H-+ 1914A,  at  centre 
of  BC  -  1.S7.5,  at  C  -  -3300,  at  20-ton  bad  -  -450, 
atRi  ft.  from  D-+38A- 
PU.  of  contrary  flexure:  in  AB  none;  in  CD  at  17i  ft.  from  D 
Required  length  of  CD -71.854  ft. 


if'; 


If 


152 


THEORY  OF  STRUCTURES. 


68.  A  girder  AC,  SO  ft.  lonp,  carries  a  uniformly  distributed  load  of  80 
tons  and  is  supported  at  the  three  p<jiiits  .1,  li,  and  C.  A  hinge  is  introduced 
at  //,  and  the  distances  are  shown  in  Fig.  276.    Draw  to  scale  the  S.F.  and 


H    B 


T 


c 


II  B 


D    K 


4»— so — -111 


~1(4— W^— |-l»'ij.    w'  4^ 


Fia.  276. 


Fio.  277. 


B.M.  diagrams.  Compare  the  results  with  those  which  would  be  obtained 
if  the  p')rtion  BC  is  hinged  at  K,  Fig.  277,  and  a  fourth  support  is  introduced 
at  D,  the  liistaiu'cs  being  as  shown. 

Alls.  Fig.  1.   Inactions  =  10  at  .4,  -4SatB,  - 2'.' at  C,  in  tons. 
Total  B.M.  area-  +00().")i. 
Pt.  of  contrafiexure  in  BC  at  44  ft.  from  C. 
Fig.  2.   Reactions  - 10  at  .1  and  C  -30  at  ^  and  D,  in  tons. 
Total  U.M.  arca--i;«3J. 

Pt.  of  ( ontratlcxuro  in  Bli  none,  in  ZX:"  at  20  ft.  from  C. 
69.  A  span  of  .'UK)  ft.  is  cnisscd  l)y  a  girder  ABCD  carrying  a  uniformly 
distributed  load  of  .JOO  tons  and  resting  upon  suplX)rts  at  .1,  B,  C,  and  D.  The 
length  of  .1«=40  ft.,  of  rtr  =  ISO  ft.,  and  of(/>=.SOft.  The  portion  BC  is 
divid.  d  by  hinges  at  //  and  K  into  two  equal  cantilevers  BH  and  C'A",  and  a 
si.sjH-iuled  span  HK  of  1(K)  ft.  Draw  Id  sntlf  the  curves  of  S.F.  and  B.M.,  and 
determine  the  points  of  contrary  flexure  in  the  side  spans.  What  must  be  the 
length  of  CI)  to  make  the  reaction  at  />  e(|ual  to  nil/ 

A11.S.  Mux.  S.F.  in  tons:  at  .1  =  -M,  at  B  -  ±90,  at  i/  =  50,at  K  -  -50,  at 
r=  -90  and  +7.'),  at  />  -  -."). 

Miu.  B.M.  in  I'l.-tons:  at  .1  =0.  at  B 2cS00,  at  centre  of  BC- 

+  I2.")(),  at  ('  -2S00,  at  />=0. 
/•/>■.  (>/  ciinlrnnj  flcxun :  in  AB  none;  in  CD  10  ft.  from  D, 
I{e(iuired  length  of  C'Z>- 74.834  ft. 


CHAPTER  III. 
MOMENTUM.    ENERGY.    BALANCING. 

1.  Velocity— Acceleration.— The  idea  of  velocity  involves  both 
sp,.(..l  and  direction.  The  velocity  of  a  body  is  its  rate  of  change  of 
jH>.ition,  and,  if  constant,  it  is  measured  by  the  number  of  units  of 
kiifitli  described  in  a  unit  of  time. 

llie  units  of  length  ordinarily  used  are  a  foot,  a  metre,  or  a  cen- 
timetre, and  the  unit  of  time  is  a  second.  Thus,  if  s  feet  are  described 
in  /  seconds,  the  constant  velocity  v  is  given  by 


vr-    or    s-r<. 

Tho  velocity  may  not  be  constant,  but  may  be  always  changing 

(luiiiig  an  inter^'al  of  time,  however  short,  and  then  'j-  is  the  average 

nWity,  the  actual  velocity  at  any  point  l)eing  the  average  velocity 

nwr  a  very  small  distance  including  that  point. 

riic  average  velocity,  for  example,  of  a  train  travelling  at  the 

4  'i  X  '')280 
rat.,  of   45   miles  i)er  hour  is   -^^^^   -66  ft.  i)er  second,  but  its 

artml  velocity  at  any  point  may  l)e  very  different  from  this  amount. 
At  sei»  the  sjioed  is  always  measured  in  knot)^,  a  knot  being  6080  ft. 

60S0 
p<.r  hour,  or  an  average  velocity  of  very  nearly  r^^^QQ^'l-SQ  ^t-  per 

fst'cmid. 

W  hen  a  l)ody  has  passed  over  s  ft.  in  t  sees.,  the  average  velocity 

,    Js         .  ., 
in  Ining  an  adiUtional  distance  of  Js  ft.  in  Jt  sees,  w  jj-,  ami  it  Jf 

i>  .iiininished  indefinitelv,  the  average  velocity  becomes  the  actual 

l.-)3 


154 


THEORY  OF  STRUCTURES. 


velocity  at  the  instant  t  and  is  indicated  by  the  relation 

ds 


r  = 


dt 


The  acceleration  of  a  body  is  its  rate  of  change  of  velocity,  and 
if  it  is  constant,  it  is  measured  by  the  change  of  velocity  in  a  unit 
of  time. 

Thus,  if  the  velocity  changes  from  V  f/s  to  r  f/s  in  <  sees.,  the 
acceleration  a  is  given  by 


r 


a  = 


or 


v  =  V+at. 


In  Montreal,  e.g.,  the  velocity  of  a  body  falling  freely  from  rest  is 


32.1705  f/s     or      0,S().7.3  cm/s     after  1  sec. 
64.353     "       "     1961.46     "  "     2  " 

96.5295    "       "     2942.19     "  "     3  " 


There  is,  therefore,  an  increase  in  the  velocity  of  32.1765  f/s 
or  980.73  crii/s  tvcry  second.  The  acceleration  m  this  case  is 
the  same,  and  is  called  a  uniform  acceleration. 

Again,  let  U  be  the  average  velocity.    Then 


and  therefore 


and 


r''^ 


v+y 


='V+\at, 


The  acceleration  may  not  be  constant  but  may  be  always  chang- 
ing in  any  interval  of  time,  however  small  that  interval  may  be.  and 
in  tliis  case  (r-V)/t  is  the  arernije  acceleration.  Let  the  velocity  change 
from  V  after  t  sees,  to  r  +  Jc  after  (t  +  Jt)  sees.,  so  that  Jr  is  the  change 

of  velocity  in  an  additional  jx-riod  of  Jt  sees.    Then  —  is  the  average 

acctlfration  jht  seeimd.  and  this  is  true,  however  small  Jt  may  be, 
so  that  in  tiic  limit  it  l)ecomes  the  actual  acceleration  at  the  instant 
t  and  is  indicated  by  the  relation 

dv    <Ps 


':!• 


ENERGY. 


155 


The  weights  of  various  materials  are  given  in  the  tables  at  the 
m\  of  the  chapter,  nnd  the  specific  gravity  (sp.  g.)  of  a  substance 
is  its  weight  as  compared  with  the  weight  of  an  equal  bulk  of  water. 

riic  woight,  for  example,  of  a  cubic  foot  of  water  is  approximately 
IIXH I  ounces,  and  the  weight  of  a  cubic  foot  of  cast  iron  is  7200  ounces; 
its  sp.  g.  is  therefore  7200/1000  =  7-2. 

The  unit  of  force  is  generally  assumed  to  be  the  attraction  of  the 
Partli  on  a  iwund  weight  and  is  called  the  force  of  a  pound.  Thus  if 
a  is  tlu>  acceleration  of  a  weight  of  W  lbs.  under  the  action  of  a  force 
of  r  lbs., 

g      W 

Now  the  value  of  g  is  different  in  different  parts  of  the  world,  being, 

for  cxii^inle, 


.2.1765    f/s 
32.19078   " 
32.182 
32.152       " 

32.088       ' ' 


or 


980  73  cm/s  in  Montreal, 

981.17      "  "  Greenwich, 

980.9       "  "  London, 

980  "  "  Baltimore, 

981  "  "  Paris, 
978.04      ' '  at  the  Equator. 


Tn  order  to  avoid  the  use  of  the  variable  g  a  n-w  unit  of  force 
cailrd  the  poundal  and  one  gth  of  that  jast  defined  has  been  intro- 

.liK,  .1.    Thus  if  Q'^Pg,  &  force  of  Q  iwundals  is  a  force  of  -(-■?) 

]iomi'ls. 

2.  Work.— Work  must  be  done  to  overcome  a  resistance.  Thus 
l.n,lii  s,  or  systems  of  liodies,  which  have  their  parts  suitably  arranged 
til  I  \(Trnine  resistances  are  capable  of  doing  work  and  are  said  to 
IKN^css  energy.  This  energy  is  termed  kinetic  or  poteiUial  according 
a^  it  is  due  to  motion  or  to  position.  A  pilenlriver  falling  from  a  height 
upon  til.'  head  of  a  pile  drives  the  pile  into  the  soil,  doing  work  in 
vii  Ml.,  of  its  motion.  Examples  of  potential  energy-,  or  enffrgy  at  rest, 
arr  alfoidod  by  a  bent  .<<pring,  which  does  work  when  allowed  to  resume 
it-  M.uural  form;  a  raised  U'dght,  whieli  pan  do  -"ork  by  falling  to  a 
low.  r  level;  gunpowder  and  dynamite,  which  do  work  by  exploding; 
a  I.ryden  jar  charged  with  electricity,  which  does  work  by  being 


156 


THEORY  OF  STRUCTURES. 


discharged;  cod,  storage  batteries,  a  head  of  water,  etc.  It  is  also 
evident  that  this  potential  energj-  must  be  converted  into  kinetic 
energy  before  work  can  be  done.  A  familiar  example  of  this  trans- 
formation may  be  seen  in  the  action  of  a  common  pendulum.  At  the 
end  of  the  swing  it  is  at  rest  for'a  moment  and  all  its  enorg\-  is  poten- 
tial, men,  under  the  action  of  gravity,  it  has  reached  the  lowest 
point,  it  can  do  no  more  work  in  virtue  of  its  position.  It  has  acquired, 
however,  a  ccTtain  v(>locity,  and  in  virtue  of  this  velocity  it  does  work 
which  enables  it  to  rise  on  the  other  side  of  the  swing.  At  mter- 
modiate  points  its  energy  is  partly  kinetic  and  partly  potential. 

A  measure  of  energ>-,  or  of  the  capacity  for  doing  work,  is  the 
work  done. 

The  energy  is  exactly  equivalent  to  the  actual  work  done  in  the 
following  case.s : 

(«)  If  the  effort  exerted  and  the  resistance  have  a  common  point 
of  application. 

(/')  If  the  points  of  application  are  different  but  are  rigidly  con- 
nected. 

(r)  If  the  energy  is  transmitted  from  member  to  member,  pro- 
vided the  nu>ml)er.s  do  not  change  form  under  stress,  and  that  no 
energy  is  absorbeil  by  frictional  resistance  or  restraint  at  the  connec- 
tions. 

Generally  speaking,  work  is  of  two  kinds,  viz.,  intenud  work,  or 
work  done  agamst  the  mutual  forces  exerted  between  the  molecules 
of  a  body  or  system  of  bodies,  and  external  work,  or  work  done  by  or 
against  the  external  forces  to  which  the  bodv  or  bodies  are  subjected. 
In  cases  (n),  (h),  (c),  above,  the  internal  work  is  necessarUy  nil. 

A.S  a  matter  of  fact,  every  body  yields  to  some  extent  under 
stress,  and  work  umst  be  done  to  jiroduce  the  deformation.  Fric- 
tional resistances  tend  to  opjwse  the  relative  motions  of  members 
an.l  nuist  also  al..sorb  energy.  If,  however,  the  work  of  deformation 
and  the  work  absorbed  by  frictional  resistance  are  included  in  the 
term  work  done,  the  relation  still  holds  that 

Energy  =  work  done. 

A  measure  of  work  done  i.s  the  j^oducf  of  the  resistance  by  the 
distanre  Ihrowjh  which  it  is  overcome.  When  a  man  raises  a  weight  of 
one  i)ound  one  foot  against  the  action  of  gravity  he  does  a  certain 


^l! 


MOMENTUM.    ESERGY.    BALASCING. 


157 


amount  of  work.  To  raise  it  two  feet  he  must  do  twice  as  much  work, 
aiul  ton  times  as  much  to  raise  it  ten  feet.  The  amount  of  work 
iMii  thorcforp  be  proportional  to  the  number  of  feet  through  which 
the  wvipht  is  raised.  Again,  to  rai.se  two  pounds  one  foot  requires 
twice  as  much  work  a.'!  to  raise  one  pound  through  the  same  distance; 
while  live  times  as  nmch  work  would  be  required  to  raise  five  pounds, 
aii.l  ten  times  as  much  to  raise  ten  pounds.  Thus  the  amount  of 
wosk  must  also  be  proportional  to  the  weight  raised.  Hence  a 
nuasure  of  the  work  done  is  the  product  of  the  number  of  pounds  by 
the  number  of  feet  through  which  they  are  raised,  the  resulting  num- 
1m  r  lieing  designated  jool-poundK.  Any  other  units,  e.g.,  a  pound  and 
an  inch,  a  ton  and  an  inch,  a  kilogramme  and  a  metre,  etc.,  may  be 
chosen,  and  the  work  done  represented  in  inch-pounds,  inch-tons, 
kilnp-am-metrcs,  etc.  This  standard  of  measurement  is  applicable 
to  all  classes  of  machinery,  since  every  machine  might  be  worked 
liv  means  of  a  pulley  driven  by  a  falling  weight. 

3.  Oblique  Resistance.— Ix't  a  body  move  against  a  resistance 

/;  inclined  at  an  angle  0  to  the  direction  of  motion  (Fig.  278).    No 

wdtk   is   done   against   the   normal   compo- 
nent R  sin  0,  as  there  is  no  movement  of      ^- 

the  point  of  application  at  right  angles  to   ^ 

the  direction   of   motion.    This  coinixjnent 

i-,  thiTcfore,  merely  a  pressure.    The  work 

.lone    against    the    tangential     component 

/;  cos  0  between  two  consecutive  points  M 

aii.l  .V  of  the  path  of  the  body  is  R  cosdMN. 


Fio.  278. 


ill  nee  the  total  work  done  between  any  two  points  A  and  B  of  the 


iatli 


.  J  (R  COS  0  •  MA')  -  jfft  cos  Ods, 


.-■  being  the  length  of  AB. 

It  AB  is  a  straight  line  (Fig.  279),  and  if  R  is  constant  in  direc- 
tion and  magnitude, 

the  total  work  -R  cosdAB^'R-  AC, 

AC  l)eing  the  projection  of  the  displacement  upon  the  line  of  action 
,t  the  resistance.    Let  the  path  be  the  arc  of  a  circle  (Fig.  280) 


158 


THEORY  OF  STKiCTURkS. 


subtending  an  uiifjh"  a  at  the  centre.     If  R  and  d  remain  constant 
till'  WDpli  done  from  .1  to  B. 

-R  cos  tfXarc  AB-^R  cos  OOA  a^^ROM  cosO  a^  Rpa-Ma, 

p  being  the  jxTiH-iuhcuhir  from  O  uinm  the  direction  of  R,  ai:  I 
M  =  Rp  Ix'ing  the  nioincnt  of  resistance  *)  rotation. 

If  there  are  more  resistances  tiiaii  one,  they  may  be  treated 


Y       *■ 


Fio.  279. 


separately  anil  their  several  efTects  superposet!.  In  such  case, 
.1/  will  Ih'  the  total  moment  of  resistance  and  will  be  equal  to  the' 
algebraic  sum  of  the  separate  moments. 

The  normal  eoiniKineut  R  sin  0  produces  a  pressure. 
4.  Graphical  Method.  — h>t  a  Ixxly  describe  a  |)ath  AS  (Fig.  281) 
against    a    variable    resistance    of    such    a 
_B  chatucter  that   its  mjignitude  in  the  direc- 
tion of  motion  may  l)e  represented  at  any 
rK)int  M  by  an  ordinate  MX  to  the  cune  CD, 
IaA    the    path    AB    W   8ub<livided    into   a 
rmmlK-r  of  parts,  .<aeh   |«irt   MP  being  so 
small  that  the  resistance  from  M  to  /'  may 
Ik'    consideretl    uniform.     The    mean    value 
MX  +  HJ 


of  this  resist aiiei 
MX  '  f'(J 


,  and  the  work 


done   in   .ivcreoming   it -"'.,'  'U/' > the    a:va     MXiJl'    in   the 

limit,     ll.iicc  tlif  tot.d  Work  done  fioiii  .1  |<>  B     the  area  l>ounded 
by  llif  (Miiv.s  .1/^  CI),  and  the  ordiiialc-;    \('.  Bl) 

5.  Energy.     Impulse.   Momentum.     Th.-  in,rk  done  in  overcoming 
a  re.sistaiiei'  ol  U  lbs.  Iiir<iugh  i\  dis'ciu-e  of  s  ft.  is  Rs  ft.  Uw. 


ENERGY. 


159 


The  pmi?er  required  to  overcome  a  resistance  of  R  lbs.  at  a  velocity 
of  r  f./s.  is  Rv  in  ft.-lbs.  per  second. 

.    horne-pov.'er,  usually  written  H.P.,  is  an  arbitrary  unit  which 

\       IrHned  by  Watt  an  the  power  which  accomplished  33,000  ft.-lbs. 

n.        mute  or  550  ft.-lbs.  per  second.    Thus  the  H.P.  required  to 

Rv 
over,  fiiie  the  resistance  of  R  lbs.  at  a  velocity  of  v  f./s.  is  -^. 

One  foot-pound -1.35^  joules. 
One  British  thermal  unit  - 1058  joules. 
One  watt  ^  one  joule  per  second,  and  is  the  work  done 
by  a  current  of  1  ampere  at  1  volt. 

In  electricity,  however,  power  Is  usually  reckoned  in  kHowatU, 
each  of  1000  watts. 

One  H.P. -550 XI. 356 -746  watts,  so  that 
One  kilowatt -1000 +  746 -1.34  H.P. 

.\n  average  of  14  watts  per  candle-power  ip  required  for  an 
incandescent  electric  lamp,  and  a  lamp  of  2000  CiJidle-power  must 
tlitrtfore  absorb  8  kilowatts,  or  about  11  H.P. 

Hy  the  preceding  article,  when  the  body  starts  from  rest,  i.e., 
whfu  r-0, 

i-a  Pg 


and 


t>-a/-|rf(. 


Tlirrffore 


9  i 


:m\ 


Pt- 


w 


g 


rims  the  terms  Ps  and 
till'  t.rriw  Pt  and  — r. 


H'  I"' 

r  are   convertible   and   so   also  are 


Kilt  Ps,  the  protluct  of  a  force  of  P  pounds  acting  jt  its  mm 


160 


THEORY  OF  STRUCTURES. 


direction  through  a  ilistaiicc  of  ,s  ft.,  is  the  uork  done  in  ft.-Ibs.     There- 
fore       T7  '•''  »'»*»  luojisun'd  in  ft.-Il)s.  ami  is  calhnl  the  kinetic  energy 

or  ston-d-uj)  work  of  a  ImmIv  <»f  U'  \W.  moving  with  a  velocity  of  v  f/s. 

Afraiii,  /*/,  the  ])ro(hict  of  a  force  of  /'-lbs.  acting  for  <  secomis 

is  its  i//i/)(//.>';  and  is  expres^'d  in  scrond-poiinds.     Hence,  too,  the 

jircMhict  -   r,  whicli  is  the  nicchanical  (Hjuivalent  of  Pi,  is  expressed 

in  s<vond-pounds  and  is  callcil  liie  motnentinn  or  quantity  of  motion 
of  a  lM)dy  of   ir  ll>s.  moving  witli  a  vch>eity  of   j-  f/s. 

Tiic  fornmhi'  obtained  may  now  Ik'  tai)ulated  as  follows: 
AVheu  the  body  starts  from  rest,  i.e.  when  V-0, 

As  =  77  —  ("^  in  fof)t-pounds, 
Qn  ■=  JU'i-^  in  foot-|M)unilal8, 

Pt  =—r  m  second-iKumda, 
.'/ 

(Jt"  Wr  in  s»cond-iM»undala. 
When  the  IkkIv  starts  witii  a  vi-locity  of  l'  f./s.' 


It" 


\r--\V-     ai.d     at^v-V. 


Therefore 


/',•<•=  ,       I —   -      I  -  m  tixii-iM Hinds, 
-    'I         -    .'/ 

y.H™  JMc-- \U'r-  in  f<Mit  poundals, 

I't^      V 1    in  seconil-ixtunda, 

.'/         f/ 


Qt -•  ll'i'-irr  in  s«'cond-|M)nndul». 
The  nlation 
18  the   anaivtiral   statement   of   Newton's  Second   Taw  of  MotiiW, 


y 


MOMESTVM.    ESEIiOY.    BALANCISG. 


161 


wliif}i  li!is  l)Pon  exproRscd  by  Clork  Maxwell  in  the  following  form: 

•  Tills  rliuru/e  of  momentum  is  nuinerieally  equal  to  tlie  impulse  which 

j.roiliicfs  it,  and  is  in  the  xamc  direction." 

riiis  n'sult  is  also  true  for  two  or  more  IxMlies  or  systems  of  bodies 

X  \(  i:illy  acted  upon  by  extraneous  forces,  and  the  etjuation  may 

!«■  written 

Imr-lFt. 

llnico,  the  total  change  of  momentum  in  any  assigned  direction  is 
cirml  In  the  aUjelmiic  sum  of  the  impulses  in  the  .same  direction. 
Tlirrct'ore.  als«),  if  there  are  no  (>xtraneou.s  forces,  the  total  momentum 
ill  itiii/  (is.4(ined  di recti oi  is  constant,  which  is  the  •)rinciplc  of  the 
'•()ii-<rvati(in  of  linear  momentum. 
riic  relation 

-'  V  f/       f/    / 

i~  In.  analytical  expression  of  the  statement  that  Ps,  the  work  done, 
h  r|ii;il  In  the  rhant/e  of  kinetic  energy  in  a  giv«'n  interval. 

Il  the  iMMly  is  a  material  particle  of  a  connecteil  system,  a  similar 
;r-iili  holds  for  ev(>ry  other  jtartide  of  the  system,  and  denoting 
'  <jiliriiir  b.im  by  the  symlK»l  I, 

2\    g  g     I 

-n  ih;il  the  sum  of  the  work  done  by  the  s«'veriU  forcos  is  equal  to 
til.  tntiil  change  of  kini'tic  energy-.  This  is  a  |mrticular  casi-  of 
lii.  iinmijilc  oj  comerration  oj  eneriry  which  asserts  that  energy  is 
■I  i/i   liiirHhtc. 

I  his  principle,  like  Newton's  laws  of  motion,  a<hnits  of  no  g«>n 
II'  jiiM>f,  but  every  ex|»erim«'nt  verifies  its  truth. 

\  part  of  the  work  -I'/N  may  l>e  ex|K-tiiied  in  doing  what  is  calle<l 

>ijiiiln ar  useful  irork,  .la,  e.g..  in  overci>ming  an  external  resi.sl- 
:i and  lb}  wuMe/i  irork,  as,  e.g.,  in  overcomir>g  frictional  resistance. 

1 1. noting  l>y  T,  the  total  elTi'Ctive  work  and  by  T^  the  toUd 
aviiilalile  or  m(Aice  work, 

1  /    W  W     \ 

-th'  total  change  of  kinetic  en«i^. 


L 


tarn 


162 


TIIEOUY  OF  STRUCTURES. 


If  it  requires  an  exp«'n»liture  of  T„  (t.Ahs.  of  work  to  drive  a  machine 
giving  T,  ft.-ibs.  o{  useful  u-ork,  the  efficiency  of  the  machine  is  defined 

to  bo  the  ratio  of  the  useful  to  the  total  work  or  -^. 

'  m 

In  the  ciuse  of  a  inachini'  working  at  a  normal  speed,  the  velocities 
of  the  different  part*  are  iK'riodic.  In'ing  the  same  at  the  beginning 
and  end  of  any  jH>riod  or  number  of  periods.  For  any  such  interval 
f-V*,  and  therefore 


so  that  there  is  an  equivalence  between  the  motive  and  effective 
work. 

It  retjuires  an  exjienditure  of  Wh  ft.-lbs.  of  work  to  raise  a  weight 
of  ir  j founds  from  rest  to  rest  through  a  vertical  distance  of  h  feet. 
The  weight  will  then  |X)S8ess  an  ecjuivalent  anK)unt  of  potential 
enertjy.  and  if  it  is  allowed  to  fall  freely  through  the  vertical  dis- 
tance of  h  feet  it  acquires  a  velocity  of  v  f/s.  given  by 


anil  therefore 


2  y 


This  shows  the  equivalence  l)etween  the  work  done  in  raising  the 
weiglit,  the  ymtentinl  enertjy  of  the  w<  ight  at  its  highest  point,  an(* 
the  work  given  out  in  the  fonn  of  hnetic  enerify  in  falling  freely 
Sup|K(s<'  that  the  weight  inosing  with  a  velocity  of  v  f/s  strikes 
a  second  IkmIv,  and  that  tlu"  fMiint  of  application  moves  in  the  din-c 
tioii  of  the  blow,  through  a  .small  distance  j  against  a  mean  resi^i- 
ttuce  It'.    Then 

R'x  "  work  n'quin>d  t(»  overcome  IV, 
-kinrtic  energy  of  W, 

Within  a  certain  limit,  called  the  limit  of  ela.iticily,  the  actual 
retiislaiirr  is  din-ctly  pro|)<»ru<)nal   lo  the  tli.-'taner  througli  whidi 


TRIASGLE  AND  PARALLELOGRAM  OF  VELOCITIES  163 

till'  [Kiint  of  application  inovcs,  and  thorcfore  varies  uniformly  from 
n,!  Ill  a  niaxinmni  resistance  R.    Then 

7 

aii'l  therefore 

Rx-2\Vih+x). 

llcnee,  it   W  is  suddenly  npjAied  from  re.'<t, 

an.i  ft  =  2ir, 

Ml  ,liat  the  effect  of  the  sudden  application  is  to  (ieveloi)e  a  resistance 
n|ii;il  to  tuice  the  weijjht. 

6.  Triangle  and  Parallelognun  of  Velocity  and  Acceleration.— So 
iiir  the  motion  of  a  body  on  a  straight  line  lias  only  Ixmmi  consideretl. 
lit  a  steamer  in  (  sees,  move  from  0  to  A  with  a  velocity  of  r\  f/s., 
ail' I  in  the  same  time  let  a  bo<ly  move  across  the  deck  in  a  direction 
I'MaWo]  to  O/i  with  a  velocity  of  r^  f/s,  Fig.  282.  In  /  sees,  the 
ImmIv  will  be  at  a  |)oint  T  defined  by  0.4  -  vil  and  .4r-r2/,  aC  bring 

paritllel  to  OB.    The  ratio  'm  "^.""^  '«  constant,  anil  therefi  -e  C 
'  OA     vit    v\ 

oc 

•ii'M  descril»e  th<*  straight  line  OC.    Also  the  ratio  ^^   is  constant, 

aiil  ^ince  OA  is  descrilied  at  a  constant  rate,  so  also  is  OC,  and 
iirnlore  the  n'sultant  of  ii  and  r^  is  a  constant  velocity.  Taking 
•  ii.  Ik-  (me  s<'cond,  then  O.l-vi.  AC -' r-^.  and  0(\  the  actual  di»- 
I'lai.  iiii'iit  of  tin*  lM)dy  in  a  unit  of  time,  is  the  resultant  velwity  in 
iliKi  rmn  ami  magiiitutle.  It  is  the  diagonal,  f/rflirn  from  the  .ttarttng- 
/»'    '  (f.  of  the  |>:irallel<igram  0.\CH. 

Ajiain,  if  O.l  and  AC  are  taken  to  repres<'nt  the  incn'awe  (or 
ijfn.r'ii,   ,,f  (•(  ,ii|Miiieiit    vel(KMti('s.  OC  must    n'present  the  increase 
I  :    iriiirth)  of  the  n*suifant  veliM-ity. 

Ihii.x  the  |>;>rall<'lograms  of  veliH-ity  and  acceleration  are  eotab 
li-ii'  I  and   veloc'ilics  and  aJt'i'lcralions   may   Im-   resolved  and  com 
I"  I'l'ii'l   in  accordance  with   pre<'i.s«'ly  the  miuw  rules  that  govern 
l!i'    mangle  «)f  force?*. 


164 


THEORY  OF  STRVCTLRES 


lielnlire  Velocity. — Tho  velocity  of  one  body  relatively  to  another 
is  till'  velocity  with  which  the  first  hotly  would  ajrpear  to  move 
if  the  ohserMT  were  moving  with  the  second  Innly.  The  relative 
motion  is  of  course  uiiafTected  if  the  same  velocity  is  applied 
in  the  same  direction  to  each  body.  Ix't  a  body  a'  0  move  in  the 
direction  O.l  with  a  velwity  of  c,  f/s.,  and  let  a  IkkIv  at  D  move 


li«..  -^s-i. 


Vui.  2S3. 


Hi- 


in  the  direction  l)K  with  a  velocity  of  r^  f/s,  Fig.  283.  Apply  to  0 
and  to  I)  a  v<  locity  of  r^  f/s.  in  a  direction  opjxisitt'  to  the  motion 
of  I).  Then  I)  is  brought  to  rest,  while  O  has  two  simultaneous 
velocities,  the  one.  r,,  in  the  dinM-tion  of  OA,  and  the  other,  ro,  in 
the  direction  of  (Hi,  parallel  to  l)K. 

lakiiiir  '^.l  r,  and  OH  ij.  the  diagonal  0('  is  the  resultant 
of  tiirx'  two  velocities  and  rejjreseiits  in  direction  and  magnitude 
tile  relative  velocity,  i.«'..  the  vehM'ity  of  0  as  seen  from  D. 

7.  Equation  of  Motion  and  Energy.   T-et  X\,  y\.  Z\   be  the  co- 

onlinales  of  t!ie  ('.  of  (i.  of  a  moving  Inxly  of  m/l,s8  M  with  respect 
to  three  re<M,iiigular  axes  at  any  given  instant. 

b  (  Xj.  (/..,  22  be  the  co-ordinates  of  tlie  .same  |K»mt  after  a  unit 
of  time. 

let  J).  ?/i.  e,  Ik"  the  co-ordinates  of  any  particle  of  mass  m  at 
till'  jji.i'ii  instant. 

!■< '  J".'.  //.'•  ^.'  Ix'  the  co-ordinutes  of  the  same  |«rticle  after  a 
unit  of  time.      Then 

Mil     I'imx,),     .Vi/i --iw^i),     Mzx-Knui); 

Mj-j    l^mjTj).     Miij- litfiyj),    A/«a-2'(m«g); 

therefore     M'Xj    xi   -  2'm(x-i  -Xi),    Miyt-yi)  -  i'm(yj-|fi), 


EQUATIONS  OF  MOTIOS  AND  ENERGY. 


185 


or  Mu-Imu,    Mvlmv,    Mw-Imw, 

7i.  ~r.  w  being  the  component  velocities  of  the  C.  of  G.  at  the  given 
iii<l:iiit  with  respect  to  the  three  axes,  and  u,  v,  w  the  component 
vildcities  of  the  particle  m  at  the  same  instant. 
From  these  last  equations, 

Mu"  -  Imuu,    A/v'  -  Imvv,    Mvi?  -  Imww. 
Tlicrefore  A/(u* +  v' +  u.-*)  -  Im{uu->rw-irww), 

which  may  be  written  in  the  form 

l/,„-'+7  +  w*)+i'm  |(u-u)»  +  (»-r)»  +  («>-w)»| -JmCua+va+u)*), 
or  Mt/'  +  JfrtV-i-mr*, 

r  Ining  the  resultant  velocity  of  the  C.  of  G.,  v  that  of  the  particle, 
unit  r  that  of  the  particle  relatively  to  the  C.  of  G. 
The  last  equation  may  be  written 

~2~' 


■+ — :r— ■ 


2 

Thus  the  energy  of  the  total  mass  collected  at  the  centre  of 
crux  it y,  together  with  the  energy  relatively  to  the  centre  of  gravity, 
i«  i(|uiil  to  the  total  energy  of  motion. 

If  the  botly  revolves  around  an  axis  through  its  C.  of  G.  with 
:in  angular  velocity  to,  the  second  term  of  the  last  equation  becomes 


Ijmr'.^-^-W-l/, 


r  Ixiiip  the  distance  of  the  particle  m  from  the  axis  and  /  the  moment 
' !  iniTtia  of  the  body  with  respect  to  the  axis. 

Aguiii,  let  X,  Y,  Z  be  the  forw-s  parallel  to  the  axes  of  x,  y,  z, 
r'-|«ctiv(>ly,  acting  upon  a  particle  of  mass  m.    Then 

rf-'x 


IX, 

■  jrr, 


wr 


1G6 

and 
Therefore 


THEORY  OF  STRUCTURES. 


Im 


dt^ 


IZ. 


'"[dt  df^  ^dt  df^  ^  dt  dt^f  ^  Vdt  ^^dt^^dt) 

Intogruting  with  res|x'rt  to  /, 

^- '"(d^' ^%^%  )  -f^^^dx^ydy^Zdz)  +H, 
H  being  a  constant  of  integration, 

or  iJwf«2=-/i(A'rfj  +  rrfy+Zd2)+H. 

Hence,  if  t'o  is  the  initial  velocity, 

J(2"mi>2 - -iwro^)  -  fliXdx  +  Ydy  +  Zdz) 

"  the  work  done  on  the  system. 

8.  Angular   Velocity.    Centrifugal  Force. — Angular  velocity  may 

1m'  ilcliiied  iis  the  miiiilxT  of  radians  \x^r  second,  a  radian  l)eing  the 

au'rle  subtended  at  the  centre  of  a  circle  by  an  arc  equal  in  length  to 

1S()° 
the  radius.    This  angle  is  -57.2958  degrees,  and  if  a  wheel 

2t.V 
makes  .V  n'volutions  jx'r  minute,  its  angular  velocity  is   ^   radians 

|KT  second. 

,\  ImmIv  ntiislrnitied  to  move  in  a  plane  curw  exerts  upon  the 
l)ody  which   constrains  it  a  force  called  ceutrijurjnl  forte,  which  is 
e(jual  and  op|K»site  to  the  de^'iaiiiuj  (or  retUri]*- 
etah  force  exerted  by  the  constraining  bo<ly  ujwn 
the  revolving  ImmIv. 

Ix't  a  particle  of  maao  m  move  from  a  point 
/*  t<r  a  consecutive  [wiint  Q  (Fig.  284.)  of  its 
path  during  an  interval  of  tiiiif  t  under  the 
n"tion  of  -i  normid  deviating  force. 

b't  the  normals  at  /'  and  Q  meet  in  O;  I'lJ 
may  Ix'  considered  as  the  imlelinitely  small  arc  of  a  circle  with  ita 
centre  at  O. 


CENTRIFUGAL  FORCE. 


187 


If  there  were  no  constraining  force,  the  body  would  move  along 
th.  luiiRpnt  at  P  to  a  point  T  such  that  PT-vt,  v  being  the  linear 

vc  citv  at  P. 

Indr-r  the  deviating  force  the  body  is  pulled  towards  O  through 
i,  aistaiipe  P.V-i/<^  /  being  the  normal  acceleration  and  QN  being 
♦Irawii  i)erpendicular  to  OF. 

\ho,  in  the  limit,        PQ-PT-QN-vt. 

Hut  QN'-PNX20P. 

TlwnTore  v'l^-hft^2R, 


R  Uing  the  radius  OP,  and  hence 

t. 
R 


f'^'<^R. 


oj IxinK  the  angular  velocity. 

Ilonce  the  deviating  force  of  the  mass  m 


-  wi/  -»»„--  >n  u/'R, 

and  is  equal  and  opposite  to  the  centrifugal  force. 

\cuin,  if  a  solid  InMly  of  mass  M  revolve  with  an  angular  velocity 
.,  lU.ui  m  axis  pa.>«iiig  through  its  C.  of  CI.,  the  total  centrifugal 
tniv.  ^iH  U"  nil,  provided  tlie  axis  of  rotation  is  an  axis  of  synmietry, 
or     'ftti'  of  th»-  principal  axes  of  inertia  at  the  C.  of  G. 

II  th«'  axis  of  rotation  is  parallel  to  one  of  these  axes,  but  at  a 
li-taiiw  U  from  the  C.  of  G., 

-    ir    - 

the  centrtfi^  force  -  ^mruP  -  (i?£mr  -  tt?MR  -  —'^R> 

r  i .  ii!K  the  tlistance  of  a  particle  of  mass  m  from  the  axis  and  W 
111.'  weight  of  th<'  b«Hiy.    Thus  the  centrifugal  force  is  the  same  as  if 
!  ••  wliule  mass  were  concentrated  at  the  C.  of  G. 


1G8 


THEORY  OF  STUUCTURES. 


If  the  axis  of  rotation  is  inclined  at  an  angle  0  to  the  principal 
axis,  the  IkmIv  will  be  constantly  sub- 
jeeted  to  tiie  action  of  a  couple  of  moment 
2/s'taii'',  E  Ix'ing  the  actual  energj-  of 
the  IkxIv. 

Consider,  for  example,  the  case  of  a 
ring  of  railius  r  rotating  with  angular 
velocity  uj  ai)out  its  centre  0.  Ijei  p  be 
the  weight  of  the  ring  i)er  unit  of  length 
of  jHTiphery.  Consic ler  any  half-ring  A FB. 
The    centrifugal    force    of    any    element 


Fio.  285. 


<J 

The  comiwnent  of  this  force  parallel  to  AB  is  balanced  by  an 
e(|ual  and  ojJiKisite  force  at  ("',  the  angle  ("'OB  being  -the  angle 
CO  A.    Thus  the  total  centrifugal  force  parallel  to  .405  is  nil. 

The  comjwnent  of  the  force  at  C,  perixindicular  to  AB, 

= ' —  urV  sin  COD  -  ?— -w2r  cos  C'CE 
'J  9 


pCC      CE       uPr 


Hence,  the  total  centrifugal  force  jxTpendicular  to  AB 

'J  9 

If  T  is  the  force  develoix-d  in  the  material  at  each  of  the  points  A 
and  li, 

U 
sino«>  the  direction  of  T  is  evidently  p<'rpendicular  to  AB, 

and  thtvefor."  T --uPr^^-i*.  ' 

9  9   ' 

r  l)eing  the  circumferential  velocity. 


EXAMPLES. 


169 


Lot  /  be  the  intensity  of  stress  at  A  and  B,  and  w  the  specific 
«( ijiht  of  the  material. 

Assuming  that  T  is  distributed  uniformly  over  the  sectional 
;iri;is  at  A  and  B, 

'     9 

Thus,  the  stress  is  independent  of  the  radius  for  a  given  value 
of  r.  and  the  result  is  applicable  to  ever\'  point  of  a  flexible  element, 
wliiitcver  may  be  the  fonn  of  the  surfaces  over  which  it  is  stretched. 

0.  Ex.  1.  A  train  of  W  i6».  grm»  weight  staHs  from  a  ttation  at  A  and  runa 
,„  li  let^l  to  a  station  at  D,  I  ft.  away.  If  the  average  speed  is  not  to  exceed  v 
j  ..  I'tinl  the  time  ()€tween  the  <u'o  stations. 

1 1  the  time  is  not  limited,  fiiut  the  least  time  in  which  the  run  from  A  to  D 
(tin  /«  made  and  the  maximum  n/teed  attained. 

Let  /'  Ib8.  be  the  average  pull  exerted  by  the  engine; 
li   "     "     "         "        road  resistance; 
H   ''     "     "        "        brake  n^sistance. 

r.ider  the  action  of  the  force  P~li  the  speed  of  the  train  (cradually  in- 
,n.is,.s  from  nil  at  .1  to  v  f/s  at  B,  and  the  train  then  possessos  a  kinetic 

,  of  '  JL„i  ft..lb8.    The  train  runs  at  the  uniform  speed  of  r  f/s  from 

2  (J 

i;  til  C,  when  steam  is  shut    off,  the      h 1 ^ 

Inik.s  applied,  and  the  train  is  gradually  a| ? ? io 

hnmuht  to  rest  at  D,  its  kinetic  energy     ^...^.4, j_p.jf »i*-«-^ 

li.^iiij;    l)een    absorbed    by  the     force 

/;  *  l{  acting  through  the  distance  C  D-q.  ^ "*•  28o. 

Ilcnce,  taking  AB-p, 

the  time  between  A  and  D 

-  time  from  AtoB+  time  from  B  to  C  +  time  from  C  to  D  ^ 


.Mso, 


(P-R)p-l-—v'-(B+R)q. 


(1) 
.(2) 


Substituting  in  (1)  the  values  of  p  and  q  from  (2),  the  time  between  A  and  D 

I     I  Wv         P  +  B 
"»'*'2    g   (P-R)(B  +  n)' 

Secondly,  if  the  speed  is  not  limited,  it  gradually  increases  from  nU&t  A 
.   niaximum  at  E.  where  AE-m.    Steam  is  then  shut  off,  the  brakes 


MICROCOPY    RESOLUTION    TEST   CHART 

ANSI  and  ISO  TEST  CHART  No    2l 


1.0  :i- 


I.I 


WW* 

1^ 


m 

2.0 
1.8 


125  III  1.4    i  I 


^ 


^     .>1PPLIED  IIVHGE 


170 


THEORY  OF  STRUCTURES. 


i 


applied,   and    the    train   is   gradually  brought    to   rest    in    the    distance 
ED  =  l-m, 

Taking  Vmax.  f/s  as  the  max.  speed  attained, 


and  therefore 


(P-/?)m---(iWx.)'-(B  +  ft)a-m), 


t'max 


r-2lg  (P-ff)(B +  /?),* 


P  +  B 
Also,  the  least  time  between  .1  and  D 

-time  from  .4  to  i5  +  time  from  EtoD 

2m       2(1 -m)        21 
Vmax.  t'max.       t'max. 


(3) 


/  21 W         P  +  B        1  ♦ 

"1    g    (P-R)(B  +  R)1  ^^^ 


(4) 


//  the  trains  runs  up  an  incline  of  1  in  m,  the  pull  P  of  the  engine  has  to  do 
the  additional  work  of  lifting  the  weight  W  lbs.  of  the  train  through  the  vertical 

distance  -  every  second.     The  times  and  maximum  speed  can  therefore  be  found 

from  the  expressions   2),  (3),  and  (4)   by  substituting  ft  +  -  for  ft. 

m 
Asain.  if  E  jhs.  is  the  weight  on  the  locomotive  drivers,  and  if  the  adhesion 
18  /I  times  the  wtight,  then 

B-/<n*, 

and  the  max.  pull  P^/tE. 

The  stecjH'st  incline  up  which  an  engine  can  climb  is  -— ■ 

f'E' 

In  (iracticc  n  usually  varies  from  J  to  J. 

The  II. P.  rofiuircd  to  run  a  train  of  If  ton.s  at  a  speed  of  .1/  miles  per  hou; 
upt.r  down  an  incline  having  a  sloix<  of  n°.  and  against  an  average  road  resist- 
ance on  the  level  at  this  siK-od  of  ft  lbs.  jxt  ton,  is 

(ft  cos  a  i  2(K)0  si  n  r, )  - -^-^ 
375 

which  for  a  light  incline  of  1  in  m  Ijocomes 

/„     .'(KK)  \  ».!/ 

\  m    I  -i-r^  •  npl"""Mnmtely. 


m 


Ex.  2.  The  rim  of  a  fty-wherl  weighing;  .-»0()0  lbs.  has  a  vrlncity  nf  40  //i; 
what  IS  its  kinrhr  rnrrgyf  What  is  the  toss  of  kinetic  energy  due  to  a  reduc- 
tion of  A  per  cent  in  the  rim  velocity  f    How  much  i*  the  rim  velocity  reductd 


EXAMPLES. 


171 


by  a  ,liminution  of  45,000  ft.4bs.  in  the  kinetic  energy  t  I }  .08  is  the  coefficient 
„/  uj/c  friction,  and  if  the  diameter  of  the  fly-wheel  is  12  timet  that  of  the  axle, 
how  many  H.P.  will  be  required  to  turn  the  wheel  f 

The  kinetic  energy-^  —- 125,000  ft  .-lbs. 

The  rim  velocity  reduced  4  per  cent  becomes  (40-1.6) -38.4  t/r  ,  and 

5000  (38.4)' 
the  corresponding  kinetic  energy  -  — ^ — . 

Therefore  the  loss  of  kinetic  energy 


32  V  '-'  2    / 


If  V  f/8.  is  the  rim  velocity  when   the   kinetic  energy  is  diminished  by 
45,000  ft.-lbs., 

5000/40^ _  A  _45QQQ     and  therefore    v-32f/8., 
32  V  2       2/ 

so  that  the  reduction  of  velocity  is  8  f /s. 

The  circumferential  velocity  of  the  axle-JJ-V  ^/•• 

Ttic  weight  on  the  axle -5000  lbs. 

The  frictional  resistance  at  the  axle  surface  -.08X5000 

-400  lbs. 
Therefore  the  H.P.  required  to  turn  the  wheel 


400  xy 
550 


-2H. 


Ex.  3.  A  weight  of  Wt  tons  falls  h  ft.  and  by  n  siuxetsive  blows  drives  an 
imhi'tir  pile  weighing  W,  tons  a  feel  into  the  ground.  Find  the  mean  effective 
rtsKUvirr  of  the  ground.  If  the  grouml  reniitance  is  directly  proportional  to  the 
,l,liii:  „l  iH-ndration,  how  far  will  the  pUc  sink  under  th<  rth  hlowt  If  the  head 
oj  II:'  pile  is  crushed  for  a  length  of  z  ft.,  x  being  very  small  as  compared  with 

ti,'  .1'  pth  -  (./  i>enrtration,  find  (1)  the   mean  thrust,  during  the  blow,  between 

IL  .n  „,ht  and  the  pile;  (2)  the  time  of  penetration;  (3)  tiie  time  during  which 
ll„    iJuir  nrl.t. 

\^  the  pile  is  inelastic,  the  weight  docs  not  rebound,  but  the  pile  and  weight, 
iiMin.  .liiitely  the  blow  is  struck,  move  along  together  with  a  common  velocity 
oi  ■!  f  s.,  given  by 

u- — V, 

9  9 


Sltll'C 


the  momentum  does  not  change. 


^"2  THEORY  OF  SrHUCTVRES. 

Thus  the  available  energy  of  the  pile  and  weight 


1  ir,  +  ir 


't/'- 


n\' 


This  energy  is  sufficient  to  carry  the  pile  and  weight  through  a  distance  of 
-ft.,  against  Rt,  the  mean  effective  ground  resistonce  in  tons,  and  therefore 


R, 


or 


R,- 


n','      nh 


ir,+ir,  a' 


w.  that  the  pile  and  weight  will  slowly  sink  under  a  superposed  weight  of 

To  allow  for  gravity,  the  value  of  R,  should  be  increased  by  TT.  +  ir. 

I^t  r  be  th,.  moan  thrust  between  the  weight  and  pile  for  the  very  small 

interval  of  t  .seconds,  during  which  the  head  of  the  pile  is  crushed  through 

wnth  Re,  ^^  hich  may  be  disregarded  without  causing  any  sensible  error.    Hence 

Tx-the  work  expended  in  the  crushing 


and 


'  "  11^   ,  w  -  tons. 


Also   r  is  necessarily  very  great  as  compared  with  R,  which  may  be  di^ 
regarded  without  causing  any  appreciable  error. 

Therefore  7'<- impulse 

-change  of  momentum 

9  '     n\  +  Wtg 

«    <J 


and 


,      X      V       X        . 

'  "-^  '-  y  -  ^  Xtime  of  fall  of  weight. 


EXAMPLES. 


173 


Alter 


the  blow  the  pile  sinks  —  ft.  with  an  average  velocity  of  —  f/s,  and 


ihetuneofpenetratum-—-^-^--  -^—  ^^^ 
Wi  +  W,      a 


Wx      4nVh 
if  the  ground  resistance  is  proportional  to  the  depth  of  penetration,  let 
xr -penetration  produced  by  r  blows; 
,      »         "  "       "   r-1  blows. 

Then  x.-Xr.  -        "  "       "the  rth  blow  against  an  average 

•  i  t  Xr+Xr-i    R» 

resistance  of 


Hence 


nr 


8.), 


T«+Xr-,  fi«y(jj_T^    )  -work  done  by  each  blow 
2         a 


ft. 

a' 

•J 

n 

K 

-<- 

-i"n' 

• 

-» 

• 

x'- 

3 

• 

X,-0--. 


t  herefore 


xj-r-,    or    x,--7=Vr, 


tiiat  the  penetration  under  the  rth  blow 


-x.-x      - 


/. 


V  « 


If  Rm»M..  is  the  maximum  ground  rrsiatance, 


174 


THEORY  OF  STRUCTURES. 


The  time  of  penetrating  the  depth  x-ar-r-.yt). 


'4\/A' 


If  X  is  not  small  as  pompared  with  the  depth  of  penetration  of  a  blow 
let  y  and  2  be  the  distances,  in  feet,  through  which  the  pile  moves  durina  the 
action  of  the  blow  and  ajtcr  the  blow.     Then 

ra;  +  /?f^  =  work  done  in  the  crushing  and  in  overcoming  the 
ground  resistance  in  t  seconds 

-        9 
T/- change  of  momentum  of  Wi 

Tt  —  Ru  -change  of  momentum  of  pile 
IF, 


so  that 


R*z  -  work  done  after  the  blow 

-^  ~^'u'-W,h-Tx-R,y, 

Tx  +  R,(y+z)-W,h, 
Tx  +  RM~n\h. 


Ex.  4.  Let  a  body  of  might  W,  moving  in  a  given  direction  with  a  velocUu 
v^stnke  a  ho,ly  of  weight  11%  moving  in  the  same  direction  with  a  vdocituZ 
Afcr  unpad  !H  the  boilies  continm  to  move  in  the  same  direction  w^th  a  common 
velocity  u.     t  ind  the  energy  lost  in  impact.  '-vmmon 


or 


~g  "'  ■'■"7''-  """lonipntum  before  impact 
-momentum  after  impact 

Energ\-  before  impart         - '— '  —  +  "''  — 


after 


"V      g      )2' 


i 


EXAMPLES.  175 

1  m' 

Energy  lost  by  impact  -jiWrn'  +  W,v,')  —^  (Wi  +  Wt) 

W,W,iv,-v2)' 

It  t  .  of  the  bodies  is  subjected  to  any  constraint,  energy  must  be  ex- 
|.i  luled  to  overcome  such  constraint,  and  the  loss  of  energy  by  impact  will  be 

F.x.  5.  Let  a  hammer  weighing  Wi  lbs.,  moving  ivith  a  velocity  of  v  ft.  per 
,if>nd,  strike  a  naU  weighing  Wt  lbs.  and  drive  it  x  ft.  into  a  piece  of  timber,  of 
wnij'ii  IF,,  against  a  mean  resistance  of  R  lbs. 

First,  assume  the  timber  to  be  fixed  in  position. 

Let  ui  be  the  common  velocity  acquired  by  the  hammer  and  nail. 

(PFj  +  Wt)^  -energy  expanded  in  overcoming  R. 

'Rx (1> 

^^        X.           t             *          W,+Wt  ... 

IJut  — »- change  of  momentum- ui (2> 

9  9 

Therefore  ¥7?^^"^ ^^> 

&nd  the  time  of  the  penetration-— - -a~ sec (4> 

iiecond,  let  the  timber  be  free  to  move,  and  let  u»  be  the  common  velocity 
aKiuired  by  the  hammer,  nail,  and  timber. 

{W,+Wt)^  -energy  expended  in  overcoming  R  plus  the  energy  expended 
2g 

in  producing  the  velocity  ut 

'Rx  +  (W,+W,  +  W,)^ (6> 

Wl      W,+W.        W,+W,  +  W» 

But  — V- ui- Ut. (6) 

9  9  9 

Hpnce,  substituting  these  values  of  u,  and  u,  in  eq.  (5), 

W,'W,  _i^ 

(W,  +  \\\)(\V,+W,  +  W,)2g'^''' ^^ 

^ilso,  the  lime  of  the  penetration 

x_        ^,W, _v_ 

~'iu,~  W,+W,-^W,gR'^'' ^^^ 


a|» 


176 


THEORY  OF  STRUCTURES. 


and  the  distance  through  which  the  timber  moves 

■ 

i>'"(ir,  +  ir,-i-ir,)'2</«" (®) 

Ex.  6.  An  nccumulntiir,  loaded  to  a  prcsKiirr  of  750  lbs.  per  square  inch, 
has  a  ram  oj  L'l  xuk.  diamdcr  with  a  stroke  uf  24  //.  How  much  H.P.  can  he 
obtained  for  a  period  of  oO  seconds? 

UP  •     1     '    --',.„ x'7'>0X 24 

lI.I>.rec,u,rod=-^(21).^^^^ 

-226.8. 

Ex.  7.  Ten  thousatul  'jQ-ieatt  incandescent  and  two  hundred  and  fifty  4.50- 
ivatl  arc  lamps  are  to  be  supplied  with  power  from  a  waterfall,  20  miles  away, 
having  an  effective  head  of  40  ft.  The  efficiency  of  the  converting  apparatus  is 
92  per  cent,  of  the  turbine  8.1  per  cent,  and  the  losses  are  .">  per  cenl  between  the 
lamp.s  and  converters  at  the  receiving  end  of  the  transmission,  10  per  cent  on  the 
line,  and  10  per  cent  in  the  generators  and  transformers  between  the  line  and  the 
turbine-shaft.     Find  the  necessary  flow  of  water  per  hour. 

The  total  watts  required  =  (.'jO.1000O  +  4.")0.25O)Vir" -V/'W-W-W- 
Hence,  if  Q  is  the  water-supply  in  cubic  feet  per  second, 

6no  i5  =„ p^.fii-'.-'^oo      (100)' 

"'    '  5.J0        ■    ■        740      92-8.5-95-90» 
and  (?=.3(K)17.")c.f/sec., 

and  the  supply  in  cubic  feet  p<'r  hour 

=oof)OQ  =  i,o.8o,r);}o. 

Ex.  8.  .1  l-«c.  Itulld  mnvinij  u-ith  a  vihwity  of  ,8f)0  ft.  per  second  strikes  a 
tarijft  and  is  slopjud  dead  in  the  space  of  ;\,inch  (7 -32).     Then 

and  /?',  the  mean  resistance  overcome  by  the  bullet,  =50,000  lbs.    The  time 
in  which  the  bullet  i.s  brought  to  rest 


momentum  _  i'f,-Ji'800_      1 
force  50,000    ~  32,000 


sec. 


Kn.  0.  \  mhimr  iif  water  of  Icnqth  I  fnt  mores  along  a  pipe  mth  a  velocity 
of  v  I  .s  ((//(/  is<ii,irlhi  and  uniformh/  xhul  off  by  the  closing  of  a  valve.  Find 
the  increase  in  the  pri  x.iiire  per  xifiiore  inch  near  the  valve. 

If  /)  is  the  iiiin':isc(l  picssuif  prr  s  jumic  font,  and  t  sees,  the  interval  in 
which  the  closing  is  etTcctcd, 

/)< -momentum  of  the  lluid  mass 
62i, 


EXAMPLES 


177 


ami  therefore 

the  pressure  per  square  inch  -  .^^  =7x1  T' 

I'.x.  10.  A  weight  of  W  pouiidn  of  water  passes  through  a  turbine  wheel 
jrmn  ii  l<>  /.  It  fiitcrK  at  a  with  a  vdocily  of  v,  f/s  in  a  (tircction  inclined  at 
nil  nnijU-  of  !)0°-/'  to  the  rwtiui  Oa=Ti.  //  leaves  at  f  with  a  velocity  of  v,  f/s 
jh  It  ilir^dion  fh  inelincd  ai  an  angle  3  to  the  radius  Of^ri.  If  co  is  the  angular 
vil'irllii  of  the  wheel,  find  the  work  done  every  second  by  the  water  on  the  wheel. 

I-ct  /)i,  />j  bo  the  perpendiculars  from  the  axis  0  upon  the  directions  of 
I:,  I-,  respectively.    Then 

i'i/)i  —  vjp?)  =  change  of  angular  momentum  per  sec.  between  a  ivnd  / 


-   !l 
!/ 

=  ,!/,  the  moment  of  impulse. 
Therefore  the  work  done  in  foot-pounds  per  second 
W 

=  3/(0  -  —  (U  (Vipi  —  Vipi) . 

9 

Let  i''„,  be  the  component  of  v,  along 
the  tiiiigent  at  a;  , 

1.(1  i/'  be  the  component  of  vt  along     r^ 

the  tangent  at  /. 

Tlien     /),  ti,  "=  p,v^  seer  -  r,  "^ 
and  Pa^j  ■" PjiV  cosec  3  =  r^v'^ , 

so  that  the  work  in  foot-pounds 

-Mw 'o{r,v^-  rjvj 

y 


'T  '*'"'"    " 

whire  I/,  =riw  and   Wj-rjw,  are  the  ji^riphoral  lineal  velocities  at  a  and  / 
rcsi.iMlively. 

Thr  components   t',',  and  v'^  are    usually  called    the  whirling    velocities. 

I  A.  11.  .\n  ice-yacht  traveh  in  the  direction  of  its  keel  with  a  velocity  of 
r  '  iiidrr  the  action  of  a  wind  blowing  with  a  velocity  of  w  f/s  in  a  direction 
mnl:':i.,i  nn  angle  ?  with  the  keel.     The  sail  is  set  at  an  angle  a  with  the  keel, 


178 


THEORY  OF  STRUCTURES. 


and  it  j's  assumed  that  there  is  no  resistance  to  motion  along  the  keel.  Find  the 
maximum  speed  of  the  yacht. 

In  Fig.  288,  take  OA  =  v  and  BO^w.  Then  AB  is  the  apparent  or  relative 
velocity  of  the  wind. 

The  yacht  will  be  moving  at  full  speed  when  AB  is  parallel  to  the  sail  OD, 


Fig.  288. 

and  the  components  of  the  velocities  of  the  wind  {OB)  and  the  sail  {AO)  at 
right  angles  to  the  sail  are  then  equal. 

Therefore  i'  sin  a  =  U)  sin  (/?  -  o). 

For  given  values  of  w  and  a,  r  is  a  ma.ximum  when  P  - 90°,  and  then 

r  sin  a  *  U!  cos  a, 

or  v-w  cot  a. 

The  velocity  of  the  boat  to  windward 

.  cos  y?  sin  ^  —  a 


-  V  cos  f?  =  W 


sin  a 


which,  for  a  given  set  of  sail,  is  greatest  when 

2,3-0 -<K)°   or    ^  =  45°+!. 
Thus  the  maximum  speed  to  windward 

— — (cosec  a  —1). 

T.x.  1 2.  The  charge  of  powder  for  a  27-ton  breech-loader  with  a  9-ton  carriage 
is  .'?()0  Ihs.:  the  weight  of  the  projectile  is  .'300  lbs.,  its  diameter  ts  10  in.,  and  its  radius 
of  tinratiiin  .3..")3.')  in.;  the  muzzle  velocily  is  202011.  per  .srr.;  the  velocity  of  recoil, 
10}  ft.  per  src;  the  gun  is  rilhd  so  that  the  projectile  makes  one  turn  in  40  calibret. 

Total  energy  of  explosion  =  energy  of  shot  +  energy  of  recoil; 
Energy  o/s/io/ =  energy  of  translation +eiiergy  i)f  rotation 

rm    ^2020)'^  500    1     /_rU    2020\'/3..'J35\' 
~  32.2        2  32.2  2    l   tV     40.^ '    ^    12   ' 

=  310SO]24.2  4-'»77-">-S,r) 
-31777SS2.8ft.-lb8.; 


EXAMPLES. 


179 


,         .,     36X2240(16})'    oqnfitoiftih- 
Emrgy  of  recoil- — 5^-^ —  -    *     -330d5<J.1  U.-1D8. 


HMice,  if  C  be  the  energy  of  1  lb.  of  powder, 

€'■300-31777882.8  +  330652.1 

-32108534.9  ft.-lbs., 

and  hence  C  - 107028.45  ft.-lbs.  -  47.7  f t.-tons. 

Kx.  13.  Let  W  he  the  weight  of  a  fly-wheel  in  pounds,  and  let  its  maximum 
1,11(1  itiiiiimum  angular  velocities  be  <o,,  wi,  respectively.  The  motion  beingone  of 
miatiim  only,  the  energy  stored  up  when  the  velocity  rises  from  102  to  o»i,  or  given 
„ut  iflun  it  falls  from  wi  to  ut,,  is 

i!,  1 .  l)ping  the  linear  velocities  corresponding  to  01,,  m,,  and  k  being  taken 
<'(|ual  to  the  mean  radius  of  the  wheel. 

It  is  usual  to  specify  that  the  variation  of  velocity  is  not  to  exceed  a  cer- 
tain fractional  part  of  the  mean  velocity. 

Let  V  be  the  mean  velocity,  and  —  the  fraction.    Then 


V 

V%  —  V2-—] 


v,+v,-2V. 


Therefore 


vr  —  vi 


Ilenre  the  work  stored  or  given  out- . 

V.K.  14.  An  engine  weighing  64  tons  travels  round  a  curve  of  1000  ft.  radius 
.It  thf  rntf  of  45  miles  per  hour.  Find  the  horizontal  thrust  on  the  rails,  and  also 
fiiiii  the  direction  and  magnitude  of  the  residtant  thrust. 

....     ,..,,...  ,      64X2000    1     /45X5280\'     ,_.„. 

1 1„.  honzontal thrust  m  pounds  »  —^^  --—  [  (=5x60")       ^^'^^^' 

The  rrrtical  weijtht  in  pounds -64X2000  =  128,000. 

Up  refore  the  resultant  thrust  is  inclined  to  the  vertical  at  an  angle  whose 
17424      l.OSO 


tansreiit  is 


■-.136125. 


IL'NOOO        8 
I'hc  resultant  thrust  in  pounds 


-  V' (17,424)'  +  (128000)'  - 120.181. 

F.x.  1").  Find  the  total  kinetic  energy  of  a  system  of  rigidly  connected  heavy 
1>art . ..  s  revolving  about  a  fixed  axis  with  a  uniform  angular  velocity  id. 


ISO 


THEORY  OF  STRUCTURES. 


Let  \Vx,  15%,  Wi...\W  Ix'  the  weights  of  fhi-  particles  in  pounds,  and 
lot  I,  Xi,  Xi  .  .  .  Xn  Ih'  the  distances,  respectively,  of  the  particles  from  the 
fixe<l  axis. 

The  kinetic  energies  of  I  he  several  weifjhts  are 


(Ji"')%  --         (XimY, 

1     (I  -     '' 


1  w. 


{Xntny. 


Therefore  the  total  kinetic  enerfjy 


or-  /ir,    , 
=  — I  —xr 
-'  \'J 


9 


Xn'l 

(J  I 


1 


-  g 


I  twing  the  moment  of  inertia  of  the  system,  and  k  the  radius  of  gyration. 

10.  Inertia — Balancing. — Newton's  First  I,a\v  of  Motion,  called 
also  tlic  Laic  of  Inertia,  states  that  "a  body  will  continue  in  a  state 
of  n>st  or  of  iniiforin  motion  in  a  straight  line  unless  it  is  made  to 
change  tliat  state  by  external  forces." 

This  jiroperty  of  resisting  a  ciiange  of  state  is  termed  inertia, 
and  in  dynamics  is  always  employed  to  measure  the  quantity  of 
matter  contained  in  a  l)ody,  i.e.,  its  mass,  to  which  the  inertia  must 
he  necessarily  ])ro])ortional.  Thus,  to  induce  motion  in  a  body, 
energy  must  b(>  exiK>nded,  and  must  again  be  alvsorbed  before  the  body 
can  be  l)rought  to  rest.  Tlie  inertia  of  the  reciprocating  parts  of  a 
n:achine  may  therefore  heavily  strain  tiie  framework,  which  should 
be  bolted  to  a  firm  fouiulatioii,  or  nmst  be  sufficiently  massive  to 
counteract  by  its  weight  tlio  otiierwise  unbalanced  forces. 

Ex.  10.  Considcrthe  case  of  a  direct-acting  horizontal  steam-engine,  Fig.  2S9. 
^.  •'    ~~^  ,^1   At  any  given  instant  let  the  crank  OP  and 

the  connecting-rod  VP  make  angles  0  and 
<1>,  resiK'ctively,  with  the  line  of  stroke  AB. 
__^  ,,  Let    ('  be  the  velocity  of  Ihc   crank-pin 

"^  •j'Y  centre    P,   and   let    u    be   the   corresponding 

piston    velocity,    which    must    evidently    l)e 
the  same  as  that  of  the  end  C  of  the  con- 
V\c,.  2S0.  nectiti;:-rod. 

Ix't  OP  pnidiircd  meet  the  vcrlic-al  through  V  in  /. 

At  the  moment  u:ider  consideration  the  jwints  C  and  Pare  turning  about 
/  as  an  instaidaucom  centre.    Therefore 

u  ^^^sin  ("4-^ 
V      IP         cos  <}> 


:v^ 


IS'ERTIA     BALASCISG. 


181 


Let  W  be  the  weight  of  the  reciprocating  parts,  i.e.,  the  piston-head,  piston- 
rod,  Cross-head  (or  motion-block),  and  a  portion  of  the  connecting-rod. 
\ssume  (1)  that  the  motion  of  the  crani<-pin  centre  is  uniform; 

(2)  that  the  obliquity  of  the  coiinci  ting-rod  may  be  disregarded 
without  sensible  error,  and  hence  0=0. 
Draw  PN  perpendicular  to  AB,  and  let  O.V  -jc;  ON  is  equal  to  the  distance 
„f  tlic  piston  from  the  centre  of  the  stroke,  corresponding  to  the  position  OP 
(,f  the  crank. 

rhe  kinetic  energy  of  the  reciprocating  parts 


'  9  -^ 


W  v' 


9 


9 


K'-t?' 


r  bt^ing  the  radius  OP. 

Ihcreforo  the  change  of  kinetic  energy,  or  work  done, correspondmg  to  the 

values  X,,  x-o{  X, 


'  9   2\     r'     r 


Let  R  be  the  mean  pressure  which,  acting  during  the  same  interval,  would 
do  the  same  work.    Then 


W  v'  x,^-x,' 

r 


K{xx-Xt), 

g  2       '' 


and  therefore 


D     W  »'x, +Xf 
9  2      r' 


Hence,  in  the  limit,  when  the  interval  is  indefinitely  small,  xi-Xi-x,  and  the 
pressure  corresponding  to  x  becomes 


R- .X. 

9  r' 


This  is  the  pressure  due  to  inertia,  and  may  be  written  in  me  form 

R-C-, 

r 

(■  /  -=E  JL  J  being  the  centrifugal  force  of  W  assumed  concentrated  at  the 

(■rink-pin  centre.  /?  is  a  maxim-m  -ad  equal  to  C  when  x=-r,  i.e.,  at  the 
poihis  .1,  B,  and  its  value  at  intermediate  points  may  be  represented  by  the 
viriiial  ordinates  to  AB  from  the  straight  line  EOF  drawn  so  that  AE-BF 
"''.  In  low-speed  engines  C  may  l)e  so  small  that  the  effect  of  inertia  may 
I"'  disregarded,  but  in  quick-running  engines  C  may  become  very  large  and 
till  inertia  of  the  reciprocating  parts  mav  give  rise  to  excessive  strains. 

Aiioihcr  force  acling  upon  the  crank-sh.ift  is  the  centrifugal  force  of  the 
(Tiiik,  crank-pin,  and  of  that  portion  of  the  connecting-rod  which  may  be 
suuased  to  rotate  with  the  crank-pin. 


182 


THEORY  OF  STRUCTURES. 


I*t  u>  be  the  weight  of  .he  mass  coneentratcd  at  the  crank-pin  centre  which 
will  produce  the  same  centrifugal  force  as  these  rotating  pieces  (i.e.,  ur-sum 
of  products  of  the  weights  of  the  several  pieces  into  the  distances  of  their 
centres  of  gravity  from  O). 

The  centrifugal  force  of  «•=——. 

'J   r 
Thus  the  total  maximum  pressure  on  the  crank  shaft 

-  C  +  i^  il'  =  il'(  ir  +  uO  -  r ( JF  +  u')-, 
9    r       gr  '  g' 

10  being  the  uniform  angular  velocity  of  the  crank-pin. 

This  pressure  may  l)o  counteracted  l)y  placing  a  suitable  balance-weight 
(or  weights)  in  such  a  position  as  to  develo|)c  in  the  ()|)posite  direction  a  cen- 
trifugal force  of  equal  magnitude. 

Let  U'l  be  such  a  weight,  and  R  its  distance  from  0.    Then 


h^r(W  +  xi\ 


m 


from  which,  if  /?  is  given.  Tl',  may  he  obtained. 

During  the  first  half  of  the  stroke  an  amount  of  energy  represented  by 
the  triangl."  AEO  is  oh.iorb,;!  iti  accelerating  the  n-ciprocating  parts,  and  the 
■s.im,  amount,  n't-rcsciitcd  i)y  the  triangle  i:Oh\  is  given  out  during  tiie  si-coiid 
h:ilf  of  the  stroke  when  the  reciprocating  part      ire  Ix-ing  retarded. 

During  th..  u|)-str..kc  „f  a  irr//,ri/  engine  the  w.'ights  of  the  reciprocating 
parts  act  in  a  direction  opposite  to  the  motion  of  the  piston,  while  during  the 
down-stroke  they  act  in  the  same  direction. 

In  .If;  j)r.Kluccd  (I'ig.  J'M),  ,,,ke  EE'  to  represent  the  weight  of  the  recip- 
rocating parts  on  the  same  scale  as  .1  A' rc'presents  the 
pressure  due  to  inertia.  Draw  E'O'F'  parallel  to  EOF. 
During  the  up-stn>ke  the  ordi nates  of /fO' represent 
the  pressures  re<|uircil  to  accelerate  the  reciprociitiii^ 
p;iris,  ihc  pressures  while  t hey  are  retarded  being  repre- 
sented by  the  ordinales  of  O'F'. 
The  ejis*'  is  exactly  reversed  in  the  <lown-stroke. 

A'.«.— The  formula  /I'-Cy  may  1k>  easily  deduced  as  follow*: 


rhcrefore 


u-v  sin  11;  the  acceleration     ''"  =-  i,  eos  O-'f  -ilV 

ill  dt      r' 


g  di 


accelerating  force -force  due  to  inertia 


f  Jf 


g  r'  r 


INERTIA.    BALANCING. 


183 


F.x.  1 7.  Consider  a  double-cylinder  engine  with  two  cranks  at  right  angles 
anil  let  d  be  the  distance  between  the  centre  lines  of  the  cylinders  (Fig.  292). 


T  otcjit. 


.Centre  lilM 


Fio.  291. 


tC.Binrt      ofcjir. 
Pio.  292. 


The  pressures  due  to  inertia  transmitted  to  the  crank-pins  when  one  of  the 
cranks  makes  an  angle  0  with  the  line  of  stroke  are 

P,  =  C  cos  0     and    Ps  -  C  sin  9. 

Tlicse  are  equivalent  to  a  single  alternating  force 
P=C  (cos  d±9\nO) 
acting  half-way  between  the  lines  of  stroke,  together  with  a  couple  of  moment 


Tlic  fore*'  and  couple  are  twice  reversed  in  each  revolution,  and  their  maxi- 
iiiuiii  values  a."e 

Pmax. -Cv'2     and     .Um.x. -yV'2. 

In  order  to  avoid  the  evils  that  might  result  from  the  action  of  the  force 
111  muple  at  high  8|)eods,  suitable  weights  are  introduced  in  such  positions 
iliii  the  centrifugal  fonvs  due  to  their  rotation  tend  to  fca/o/icc  both  the  force 
ml  tlie  couple.  For  example,  the  weights  may  be 
li  vil  \\\Mm  the  lly-whcel  of  a  stationary  engine,  or, 
,:im.  uiH)t\  the  drivitig-whcels  of  a  locomotive, 

I. it  a  balance-weight  Q  Ix'  placed  nearly  diamet- 
n  illv  opposite  to  the  p««ntre  of  each  crank-pin  (Fig.  293), 
III  lit  H  Ih!  the  distance  from  the  axis  to  the  centre  of 
i:r,i\ity  of  Q. 

\a\  f  \tv  the  horizontal  distance  between  the  balance- 
«iii:lits,  Fio.  293. 

I  ho  centrifugal  force  F  due  to  the  rotation  of  Q 

Q  (velocity  of  Q)'    «l^*i-^^ii« 
"fl  «  alt  r'        gr*   ' 


184 


THEORY   OF  STRUCTURES 


and  this  force  F  is  equivalent  to  a  single  force  F  acting  half-way  between 

the  weights  and  to  a  couple  of  moment  F-.    Let  ^  be 

the  angle  between  the  radius  to  a  balance-weight,  and  the 
common  bisector  of  the  angle  between  the  two  cranks 
(Fig.2i)4). 

Since  there  are  two  weights  Q,  there  will  be  two  couples 

each  of  moment  F— ,and  two  forces  each  equal  to  F  acting 

half-way  between  the  weights,  the  angle  between  the  axes 
of  the  couples  being  I80°-2./.  and  that  between  the  forces  being  2</..  The 
moment  of  the  resultant  couple  is  Fe  sin  <j>,  and  its  axis  bisects  the  angle 
between  the  axes  of  the  separate  couples;  the  resultant  force  parallel  to  the 
line  of  stroke  -2F  cos  ^. 

Q  and  <t>  may  now  be  chosen  so  that 


and 


Then 


and 


and  therefore 


2F  cos  (}>  -maximum  alternating  force  -Cn/2, 
Fe  sin  4>  -maximum  alternating  couple  -—v^ J. 

tan  ^  -— , 

e 

g  r*         g    r   e  \j     2     ' 

o  --  lc!±il 


Ex.  IS.  Again,  the  prrmire  C  iil  n  <lFnil-p„inl  may  be  balanced  by  a  uriglit  Q 
ilitimrlriralhi  apposite. 

If  A  is  the  radius  of  the  weight-circle,  then 

g  r  gr'    ' 


and  therefore 


Q-wr.. 


The  weight  Q  may  1*  replnowl  by  a  weight  q'^^^I  on  the  near  and  a  weight 


,.e-d 


Q  -^   on  the  far  wheel.    Thus,  since  the  cranks  are  at  right  onglea,  there  will 
be  two  weiehto  90»  apart  on  each  wheel,  vii..  Q'^^  )„  line  with  the  crank  and 


CURVES  OF  PISTON  VELOCITY. 


185 


y— — .    These  two  weights,  again,  may  be  replaced  by  a  single  weight  B  whose 

I'.ntiifugal  force  is  the  resultant  of  the  centrifugal  forces  of  the  two  weights. 

Tims 


\g  h)   '\g   2e   rJ       \g   2e  R  }  ' 


v'  being  the  lineal  velocity  at  the  circumference  of  the  weight-circle. 
Therefore 


B'-Q'-^-f. 


B 


q    \e'+(P 
'eNj     2    ■ 


rje+d 


If  Of  is  the  angle  between  the  radius  to  the  greater  weight  Q-;r—  and  the 

2e 


crank  radius, 


tan  a  - 


Qe-dv^ 
g  2e  R     e-d 
Qe+dv^~e+d' 
g    2e  R 


Note. —  In   outBide-cylinder    engines    e-d    is    approximately   nil,    and 

B-Q-W^. 

II.  CurvM  of  Piston  Velocity. — Consider  the  engine  in  Ex.  16. 
s 


Fm.  295. 

T,pt  CP  produced  intersect  the  vertical  through  0  in  7*,  and 
ii!  or  tiikc  OT'-OT. 

The  piston  velocity  u  and  the  velocity  v  of  the  crank-pin  centre 
are  ronnecte*!  by  the  relation 


u_    8in(0f^)    OT    or 
V  "     co8^     'op'  OP' 


(1) 


186 


THEORY  OF  STRUCTURES. 


If  the  velocity  r  is  assumed  constant,  and  if  it  is  represented  by  OP, 
then  on  the  same  scale  OT'  will  represent  the  piston  velocity  u. 
Drawing  similar  lines  to  represent  the  value  of  ii  for  every  position 
of  the  crank,  the  locus  of  T'  will  be  found  to  consist  of  two  closed 
curves  OGS,  OHT,  called  the  polar  curi'es  of  piston  velocity.  They 
pass  through  the  point  0  and  through  the  ends  S  and  T  of  the  ver- 
ti  >al  diameter.  On  the  side  towards  the  cylinder  they  lie  outside 
the  circles  having  OS  and  OT  as  diameters,  while  on  the  side  away 
from  the  cylinder  they  lie  inside  the  circles.  If  the  connecting-r  ' 
is  so  long  that  its  obliquity  may  be  disregarded, 

^  =  0,     u  =  D  sin  ^, 

and  the  curves  coincide  with  the  circles. 

A  rectangular  diagram  of  velocity  may  be  drawn  as  followp. 


C     N 
Fio.  296. 


C     M 


Tpon  the  vertical  through  C,  Fig.  206,  take  CL  =  OT;  the  locus 
of  L  is  the  curve  required  for  one  stroke.  A  similar  curve  may 
be  drawn  for  the  return-stroke  either  Ix^low  MN  or  upon  the  pro- 
longation Mi{"M.\)  of  MX. 

If  the  ol)li(]uity  nf  the  connecting-rod  is  neglected,  the  curves 
evidently  coincide  with  the  semicircles  uiwm  MX  and  XR,  MX 
(  =  .\7?)  defining  the  extreme  positions  of  C.  The  obliquity,  how- 
ever, caus(>s  the  actual  curve  to  fall  above  the  semicircle  during 
the  first  half  of  the  stroke,  and  Iwlow  during  the  second  half. 

Again,  let  the  connecting-rod  (/)-n  cranks  (r).    Then 

sin  0     I 


sin  (/>    r  "   ' 


and  by  eq.  (1), 

/  •    />  ,        nL      JN      /  •    /.     sin  ^coa^  \ 
u - v(sm  0-^cosO  tan  c^)  - r{  sin  0+  -  —  1. 

\  Vn-'-am^Oj 


.    .    (2) 


CURVE  OF  CRANK-EFFORT.  187 

If  the  obliquity  is  very  small, 

,       ...     sin  ^  .        , 

tan  9 -am  9  -  — - — ,  approximately, 

ft' 

...       .                    /  .     -     sin  5  cos  <?\       /  .    ^    sin  25\ 
and  therefore       u-v(sm<?+ \=vysmd+    - — j. 

12.  Curve  of  Crank-effort. — The  crank-effort  F  for  any  position 
op  of  the  crank  is  the  component  along  the  tangent  at  P  of  the 
tlirust  along  the  connecting-rod. 


therefore 


This  thrust  =• 


cos  (p' 

sin  (d  +  4)) 


cos  <t>    ' 


If  tlie  pressure  P  upon  the  piston  is  constant,  and  if  it  is  rejv 
nsciitpd  by  OP,  then,  on  the  same  scale,  OT,  Fig.  297,  will  rep- 
irsint  the  crank-effort.  Thus  the  curves  of  piston  velocity  already 
drawn  may  also  be  taken  to  represent 
cuives  of  crank-effort.  If  the  pres- 
siiH!  P  is  variable,  as  is  usually  the 
o:is«>,  let  OP,  the  crank  radius  repre- 
stiif  the  initial  value  of  P.  After 
cvjiaiision  has  begun,  take  OP'  in  OP, 
f(ir  any  position  OP  of  the  crank,  to 
riprcsont  the  correspon<ling  pressure 
which  may  be  directly  obtained  from 
thi'  iiidicator-<liagram.  Draw  P'T'  parallel  to  PT,  and  take 
(IT"  =0T'.  Then  07"' will  represent  the  required  crank-effort,  and 
the  linear  and  jjolar  diagrams  may  be  drawn  as  already  described. 

13.  Curves  of  Energy —Fluctuation  of  Energy. — In  the  curve  of 
(T.ink-effort  as  usually  drawn,  the  crank-effort  for  any  position  OP 
'I  ilu'  crank  is  the  ordinate  57/,  the  abscissa  DH  being  equal  to 
ilii-  arc  AP,  i.e.,  to  the  distance  traversed  by  the  point  of  applica- 
ti  u  of  the  crank-effort.    Thu.s,  DSE  m<\  EVG  being  the  curves, 


Fia.  297. 


DE  •'EG-  semicircumference  of  crank-circle  -  w. 


J8S 


THEORY  OF  STRUCTURES. 


If  the  obliquity  is  neglected,  the  curves  of  crank-efTort  are  the 
two  curves  of  sines  shown  by  the  dotted  lines. 

The  area  /AS'//  also  e\idently  represents  the  work  done  as  the 


crank  moves  from  OA  to  OP,  and  the  total  work  done  is  represented 
by  the  area  DSE  in  the  forward  and  by  EVG  in  the  return  stroke. 
Let  Fq  be  the  tnean  crank-effort.    Then 

FoX27:r  =  2Px2r, 
assuming  P  to  be  constant. 

Therefore  Fn  =  ^^—. 


np 

Draw  the  horizontal  line  1234567  at  the  distance  —  from  DEG 

t:  ' 

and  intersecting  the  verticals  through  D,  E,  and  G  in  1,  4,  and  7, 
and  the  curves  in  2,  3.  5,  and  6.  The  engine  may  be  supposed  to 
work  against  a  constant  resistance  R  equal  and  opposite  to  the  mean 
crank-effort  Fq. 

From  /;  to  2,  /?>  crank-effort ,  and  the  speed    must    therefore 
continually  ilitninish. 

From  2  to  3,  /e<  crank-effort,  and  the  speed  must  continually 
increase. 

Thus  2  is  a  point  of  min.  velocity,  and  therefore  also  of  niin. 
kinetic  energy. 

From  3  to  E,  ft >  crank-effort,  and  the  speed  niast  continually 
diminish. 

Thus  3  is  a  point  of  max.  velocity,  and  therefore  also  of  max. 
kinetic  energy. 

Similarly,  in  the  return-stroke,  5  and  6  are  points  of  min.  and 
max.  velocity  resjx'Ctively. 


Ci'RVES  OF  ENERGY 


1S9 


The  change  or  fluctuation  of  kinetic  energy  from  2  to  3  =  area  283, 
l)ouiiiled  by  the  curve  anil  by  23. 

llic  fluctuation    from  3  to  5  =  area  3Eo,  bounded  b\    35  and 

liv  the  curve. 

F    Fr 
Affiiin,  since  7^  =  7r»  the  ordinates  of  the  curves  mav  be  taken 

r-  p       pf.> 

t(i  ri  present  the  moments  of  crank-oflfort,  and  the  abscissa)  are  then 
the  corresponding  values  of  0. 

The  work  done  between  ^1  and  any  other  position  P  of  the  crank- 

=  /  FrdO  =  Pr         sm  d  +  -,-  ]dO 

Jo  Jo   \  \/n2_  sn2(9/ 


=  Pr(l-  cos  e  +  n-Vn^-  »\n^0). 

If  there  are  two  or  more  cranks,  the  ordinates  of  the  crank- 
(tfiirt  curve  will  be  equal  to  the  algebraic  sums  of  the  several  crank- 
efforts.  For  example,  if  the  two  cranks  are  at  right  angles,  and  if 
Fi.  /'\.  are  the  crank-efforts  when  one  of  the  cranks  (F'l)  makes  an 
;iiij:!('  0  with  the  line  of  stroke, 

sin  2b\ 


^.-/■(si...'^) 


ainl 


2« 
sin  2e\ 


^      r./       .    sin20\ 
F.  =  P{cosO—^y 

Therefore      Fi  +  F2  =  P(sin  d  +  cos  6)  =  combined  crank-efTort, 

/'  iieiiig  supposed  constant. 

Xiite. — In  the  case  of  the  polar  curves  of  crank-effort,  if  a  circle 

2P 

\-  Icscriljcd  with  O  as  centre  and  a  radius  =  mean  crank-effort  = — , 

;: 

ii   will  inters(>ct  the  curves  in  four  points,  which  are  necessarily 
I II  III  Its  of  max.  and  min.  velocity. 


EXAMPLES. 

I.  A  stone  wpiphinn  S  oz.  falls  for  5  seconds.  What  is  its  momentum,  and 
wl,  ,t  fcirco  will  stop  it  in  3scconds?  Ans.  80;   H  lbs. 

'.  What  force,  acting  fur  0  seconds  on  a  mass  of  12  lbs.,  will  change  its 
velu.  ity  from  200  to  320  ft.  per  second?  Ana.  7^  lbs. 


190 


THEORY  OF  STRUCTURES. 


3.  The  velocity  of  a  body  is  observed  to  increase  by  four  miles  per  hour  in 
every  minute  of  its  motion.  Compare  the  force  acting  on  it  with  the  force  of 
^"^''^y-      .,  Am.  Ilto3fi00. 

4.  A  railway  tram  whose  mass  is  100  tons,  moving  at  the  rate  of  a  mile  a 
mmute,  is  brought  to  rest  in  10  second.-*  by  the  action  of  a  uniform  force 
Find  how  far  the  train  runs  during  the  time  for  which  the  force  is  applied 
Also  determine  the  force,  stating  the  units  employed. 

„    ^        .  Ans.  440  ft.;  27i  tons. 

5.  If  the  unit  of  mass  is  the  mass  of  12  lbs.,  and  the  units  of  length  and 
time  are  1  \  ft.  and  12  seconds  respectively,  find  the  measures  of  the  mass 
velocity,  and  momentum  of  a  body  which  weighs  1  cwt.  and  is  moving  with  a 
velocity  of  35  ft.  per  second. 

Aw).  mass=9J;  velocity -30;  momentum -280 

6.  A  goods  truck  of  6  tons,  travelling  at  3  miles  an  hour,  collides  with  another 
truck  at  rest,  and  both  move  on  together  at  2  miles  an  hour.  Find  the 
mass  of  the  second  truck.  ^^    3  ^^j^ 

7.  Find  the  average  force  which  will  bring  to  rest,  in  2  ft.,  an  ounce  bullet 
moving  at  the  rate  of  1500  ft.  per  second.  How  long  will  it  take  to  bring  it  to 
rest? 

Ans   35,156i  poundals;  ,fj  sec. 

8.  A  man  of  12  stone  weight  climbs  up  a  mine-shaft  800  ft.  deep  by  a  lad- 
der.    What  work  does  he  do?    If  he  exerts  J  horse-power,  how  long  will  he  be? 

Ans.  134,400  ft.-lbs.;   20 ,«r  minutes. 

9.  A  shot  IS  fired  from  a  gun,  which  is  fixed,  with  a  certain  charge  of  powder. 
If  the  quantity  of  powder  be  quadrupled,  in  what  proportion  will  the  velocity 
of  the  shot  be  iiureasod?  ^„.,    Doubled. 

10.  A  body  falling  from  a  mast  took  .25  second  to  fall  from  the  hatchway 
to  the  bottom  of  the  hold,  a  distance  of  20  ft.  From  what  height  did  it  fall 
and  with  what  velocity  did  it  strike?  .I,,,,,  lioj^ft.-  84  f/s. 

ti.  A  tr.iin  going  at  the  rate  of  45  miles  an  hour  takes  half  a  minute  in 
passing  another  train,  230  yards  long,  going  in  the  same  direction  at  the  rate 
of  lo  miles  an  hour.    What  is  the  length  of  the  first  train?    Ans.  210  yards. 

iJ.  A  train  215  yards  long,  going  at  the  rate  of  .')5  miles  an  hour,  takes 
10  seconds  in  passing  another  train  going  in  the  opposite  direction,  at  the 
rate  of  35  miles  an  hour.     What  is  the  length  of  the  second  train? 

Ans.  225  yards. 

13.  .lust  .as  a  tramcar  roaches  a  man  stamliiig  l)y  thi^  tramway  it  has  a 
veloeity  of  SJ  ft.  per  second;  the  man  takes  hold  of  and  mounts  the  car.  What 
change  of  velocity  takes  place,  the  weights  of  the  car  and  man  being  1  ton 
and  10  stone  respectively?  .j,,,,    g  f /(, 

14.  One  hundred  and  fifty  pounds  is  drawn  up  the  shaft  of  a  coal-pit, 
and,  starting  from  rest,  acquires  a  velocity  of  U  miles  an  hour  in  the  first 
minute.  Assutninz  that  the  acceleration  is  uniform,  find  how  heavy  the  mass 
appears  to  one  drawing  it  up.  ^t^,.  i'jqjj  lbs. 

15-  A  train  of  100  tons,  running  on  a  level  line,  is  kept  going  by  the  loco- 
motive at  a  uniform  pace  of  .jO  miles  per  hour;  tlie  steam  is  suddenly  shut 
off,  and  the  train  comes  to  rest  after  it  has  travelled  2  miles  farther.    What 


EXAMPLES. 


191 


was  the  force  applied  by  the  locomotive  to  the  train,  supposing  the  resistance 
(if  the  rail  and  air  to  be  constant?  Ans.  A  weight  of  lUi  ton. 

1 6.  If  a  700-lb.  shot  be  fired  from  a  75-ton  gun  with  a  speed  of  1200  feet 
jicr  second,  find  the  speed  of  recoil  of  the  gun.  Ans.  5  f /s. 

17.  What  force  must  be  applied  for  one  tenth  of  a  second  to  a  mass  of 
10  tons  in  order  to  produce  in  it  a  velocity  of  3840  ft.  per  minute?  What 
would  be  the  energy  of  the  mass  so  moving? 

Ans.  A  weight  of  200  tons;  640ft.-tons/sec. 

18.  A  bullet  moving  at  the  rate  of  1100  ft.  per  second  passes  through 
a  thin  plank,  and  comes  out  with  a  velocity  of  1000  ft.  per  second.  If  it  then 
passes  through  another  plank  exactly  like  the  former,  with  what_velocity 
will  it  come  out  of  this  second  plank?  Ans.  100V79f/s. 

19.  A  shot  of  600  Ibe.  is  fired  from  a  10-ton  gun  with  a  velocity  of  1000 
ft.  per  second.  If  the  mass  of  the  powder  be  neglected,  find  the  velocity  of 
recoil.  Ans.  26.8f/s. 

ao.  A  bullet  whose  mass  is  1  oz.  is  fired  with  a  velocity  of  1210  ft.  per 
second  into  a  mass  of  1  cwt.  of  wood  at  rest.  What  is  the  velocity  with  which 
the  wood  begins  to  move?  Ans.  8.098  in./s. 

21.  An  1800-lb.  shot  moving  with  a  velocity  of  2000  ft.  per  second  im- 
pinges on  a  plate  of  10  tons,  passes  through  it,  and  goes  on  with  a  velocity 
of  4()0  ft.  per  second.     If  the  plate  be  free  to  move,  find  its  velocity. 

Ans.  128.57  f/s. 

22.  To  find  the  velocity  of  an  8-lb.  shot  that  will  just  penetrate  an  armor- 
plate  10  ins.  thick,  the  resistance  being  84  tons.  If  the  velocity  of  the  shot 
be  doubled,  what  must  be  the  thickness  of  the  plate  in  order  that  the  shot 
may  only  just  penetrate  it?  Ans.  1120  f/s;    40  ins. 

23.  The  velocity  of  flow  of  water  in  a  service-pipe  48  ft.  long  is  64  ft.  per 
second.  If  the  stop-valve  is  closed  in  i  of  a  second,  find  the  increase  of  pres- 
sure near  the  valve.  Ans.  375  lbs.  per  sq.  i.i. 

24.  A  body  approaches  an  observer  with  a  velocity  due  east.  If  the 
observer  moves  due  north  with  an  equal  speed,  in  what  direction  will  the 
body  appear  to  move?  Ann.  South-east. 

25.  A  horseman  at  full  gallop  fires  at  a  stationary  animal.  Show  that 
he  must  aim  behind  the  animal. 

26.  If  the  animal  is  also  running  in  a  parallel  direction,  show  that  the 
horacman  must  aim  in  front  or  behind  according  as  his  speed  is  less  or  greater 
than  that  of  the  animal. 

27.  Rjun  is  falling  vertically,  aud  it  is  observed  that  the  splashes  mad© 
I'v  the  drops  on  the  window  of  a  moving  railway  carriage  are  i  iclined  to  the 
virfical.  Explain  this,  »nd  point  out  in  which  direction  the  splashes  are 
'i^^'i'"''!-  Ans.  At  30"  with  vertical. 

If  the  train  is  also  travelling  at  60  miles  per  hour,  and  the  inclination  of 
thi  .phishcs  to  the  vertical  is  30°,  what  is  the  velocity  of  the  faUing  drops? 

Ans.  88V3  f/s. 
38.  A  steamboat  ia  going  north  at  15  miles  per  h(  ur  while  an  cast  wind 


^1 


I: 


mm4 


192 


THEORY  OF  STRUCTURES. 


is  blowing  at  5  miles  per  liour.    Find  the  angle  the  direction  of  the  smoke 
makes  with  the  ship's  keel.  Ans.  W.  by  S.  at  cot-'  3. 

jQ.  Knowing  the  direction  of  the  true  wind  and  the  velocity  and  direc- 
tion of  the  apparent  wind  on  a  ship,  aa  shown  liy  the  direction  of  the  vane 
on  the  mast,  determine  the  velocity  of  the  ship,  supposing  there  is  no  leeway. 

30.  To  determine  (a)  the  direction  taken  by  tlie  smoke  of  a  steamer,  (b) 
tho  direction  and  velocity  with  which  the  wind  appears  to  blow  to  a  passenger 
on   board.     (N.B. — The  smoke  is  carried  along  with  the  wind.) 

31.  A  train  travels  at  the  rate  of  45  miles  an  hour.  Rain  b  falling  verti- 
cally, but  owing  to  the  motion  of  the  train,  the  drops  appear  to  fall  past  the 
window  at  an  angle  tan~'  1.5  with  the  vertical.  Find  the  velocity  of  the 
rain  drops.  Ans.  44  f/s. 

32.  The  rim  of  the  wheel  of  a  centrifuRal  pump  moves  at  30  ft.  per  second; 
water  flows  radially  at  5  ft.  per  second;  the  vanes  are  inclined  backward  at 
an  angle  of  35°  to  the  rim.  What  is  the  absolute  velocity  of  the  water?  What 
is  the  component  of  this  parallel  to  the  rim?  Ans.  23.4  f/s;   22.8  f/s. 

33-  Find  the  time  in  which  it  is  possible  to  cross  a  road,  of  breadth  100  ft., 
in  a  straight  line  with  the  least  velocity,  between  a  stream  of  vehicles  of 
breadth  40  ft.,  following  at  intervals  of  20  ft.,  with  velocity  5  f/s. 

Ans.  50  sees. 

34.  A  man  stands  on  a  platform  which  is  ascending  with  a  uniform  acceler- 
ation of  G  ft.  per  second  per  second,  and  at  the  end  of  four  seconds  after 
the  platform  has  begun  to  move  he  drops  a  stone.  Find  the  velocity  of  the 
stone  after  three  more  seconds.  Ana.  72  f/s. 

35.  A  man  in  a  lift  throws  up  a  ball  vertically  with  a  velocity  of  vis 
and  catches  it  again  after  (  seconds;   prove  that  the  vertical  acceleration  of 

the  lift  is  y-2— . 


36.  A  railway  train  is  going  at  30  ft.  per  second;  how  must  a  man  throw 
a  stone  from  the  window  so  that  it  shall  leave  the  train  laterally  at  1  fi  ,)er 
second,  but  have  no  velocity  in  the  direction  of  the  train's  motion? 

Ans.  30.03  f  s  at  178°  C  wiljh  direction  of  train's  motion. 

37-  If  a  bucket  of  water,  weighing  20  lbs.,  is  pulled  up  from  a  well  with 
an  acceleration  of  8  ft.  per  second  per  second,  find,  in  pounds  we'ght,  ♦he 
force  which  must  be  applied  to  the  rope.  Ans.  25  lbs. 

38.  Find  the  maximum  velocity  of  an  ice-yacht  sailing  at  right  ■ire!'"' 
to  a  wind  of  10  miles  an  hour  if  the  angle  between  the  sail  and  the  keel  is 
sin-'  .0  (neglecting  all  resistances  to  motion  along  the  keel). 

An.".  13if'»- 

39.  Find  the  angle  between  the  sail  and  the  keel  of  the  ice-yacht  if,  when 
the  wind  is  perpendicular  to  the  keel  the  maximum  velocity  of  the  yacht 
is  four  times  the  velocity  of  the  wind.  Ans,  tan-'  25. 

40.  An  ice-yacht  is  sailing  at  right  angles  to  the  wind  at  the  rate  of  20  miles 


EXAMPLE.^. 


193 


m  hour.  Its  sail  makes  with  the  keel  an  angle  of  30°.  Find  the  least  pos- 
sihlo  velocity  of  the  wind.  Ans.  6|\/3  f/s. 

41.  Show  that  it  is  possible  with  an  ice-boat  to  beat  to  windward  faster 
tliiii  the  wind  is  blowing  in  an  opposite  direction. 

Show  also  that  it  is  quicker  to  beat  to  leeward  instead  of  saiUng  dead 
!■•  fore  the  wind. 

4-.  A  steamer  takes  3  minutes  on  the  measured  sea-mile  with  the  tide 
aiid  4  minutes  against  the  tide;  find  the  speed  of  the  tide  and  the  speed  of 
t!iu  vessel  through  the  water.  Ans.  17J  knots;   2i  knots. 

43.  A  ship  sailing  north-east  by  (•oiiiii:i>-  tiirough  a  tide  running  4  knots 
timis  lifter  2  hours  she  h'ls  made  good  4  miles  south-cast;  determine  the  direc- 
tion of  the  current  and  the  speed  of  the  ship. 

44.  A  ship  is  sailing  at  OJ  knots  directly  towards  a  battery  where  artil- 
lery [iractice  is  going  on.  The  man  at  the  helm  observes  the  flash  of  a  gun 
and  hears  the  report  15  seconds  afterwards;  five  minutes  after  the  first  flash 
he  observes  a  second  flash,  and  hears  the  report  12  seconds  afterwards.  Assuni- 
ini:  that  a  sea-mile  is  COSO  ft.,  find  the  velocity  of  sound.       Ans.  1140  f/s. 


45.  Two  particles  are  started  simultaneously  from  the  points  A  and  B, 
■  ft.  apart,  one  from  A  towards  li  with  a  velocity  which  would  cause  it  to 
ri.i' h  li  in  3  seconds,  and  the  other  nt  right  angles  to  the  former  and  with 
three  fourths  of  its  velocity.  Find  their  relative  velocity  in  magnitude  and 
direction,  the  shortest  distance  between  them,  and  the  time  at  which  they 
are  nearest  to  one  another.  A  \  f/s;  3  ft.;  1.92  see. 

4'i.  I  rom  the  edge  of  a  cliff  two  stones  are  tu.  at  the  same  time,  one 
vertiKilly  downwards  with  a  velocity  of  30  ft.  per  .  ond,  the  other  vertically 
upwanls  with  the  same  velocity.  The  first  stone  reaches  the  ground  in  7J 
sediids.     How  much  longer  will  the  other  be  in  the  air?    /Ins.  li  sees. 

4 '.  A  steam-engine  moves  a  train  of  mass  CO  tons  on  a  level  road  fr'  m  rest, 
aiiil  iii'(|uire3  a  speed  of  5  miles  an  hour  in  5  minutes.  If  the  same  engine 
ii"  .1  fiiiother  train  and  give  it  a  speed  of  7  miles  an  hour  in  10  minutes,  find 
ih"  iii:i^3  of  the  second  train.  (The  mass  of  the  engine  is  included  in  that 
if  '!i'  irain,  and  the  forces  exerted  by  it  are  the  same  in  both  eases.) 

.fins.  85?  tons. 
43.  A  force  which  can  just  support  200  lbs.  acts  for  one  minute  on  6000  lbs. 
•'  !  I  \elo>ity  does  v  produce  in  it?  Ans.  64  f/s. 

4).  In  a  ship  sailing  at  IG  miles  per  hour  it  is  observed  that  the  direction 
f  til.  wind  is  apparently  30°  to  the  line  of  keel  and  from  the  bows;  its  velo- 
;>  IS  :i|iparently  4  miles  per  hour.     What  is  its  true  direction  and  magnitude? 

ylns.  Cosec-'6.34;    12.69  m/h. 
51.  The  velocity  of  a  ship  in  a  straight  course  on  an  even  keel  is  8^  miles 
11  h  ur;  a  ball  is  howled  across  the  deck,  pcrpondic ular  to  the  ship's  length, 
with  !k  uniform  velocity  of  3  yards  in  a  second.    Describe  the  true  path  of 
the  hull  in  space,  and  show  that  it  will  pass  over  45  ft.  in  3  seconds  nearly. 


i 


194 


rilKORY  OF  STRUCTURES. 


51.  A  train  passes  two  men  walking  beside  a  railway  in  3i  seconds  and 
3f  seconds  respectively;  a  second  train  passes  the  men  in  4j  seconds  and  4J 
seconds  respectively.  Show  that  this  train  will  overtake  the  first  train  and, 
if  on  a  different  line,  could  pass  it  completely  in  30  seconds. 

52.  Two  ships,  A  and  B,  of  which  H  bears  from  A  35°  E.  of  N.,  lie  east- 
wards of  a  coast  which  presents  seawards  a  long  vertical  precipice  running 
N.  and  S.  A  cannon  is  fired  from  13,  and  3  seconds  after  the  flash  is  seen  from 
A  the  report  is  heard,  and  7  seconds  later  the  echo  from  the  precipice.  Find 
the  distance  of  A  from  the  coast,  assuming  1100  f/s  as  the  velocity  of  sound. 

Ans.  4(i'.'.")  ft. 

53.  A  goods  train  is  running  uniformly  at  15  m/h ;  a  passenger  train 
runs  on  a  parallel  line  with  an  average  speed  of  25  m/h,  but  stops  at  a  station 
every  mile  for  1  minute.  The  goods  train  has  a  start  of  1  minute.  When 
and  where  will  the  trains  pass?    Solve  this  problem  graphically. 

Ans.  i  mile  Ix'VoikI  l>t  stjitioii  in  .">  minutes. 

54.  A  man  rows  a  boat  through  the  water  at  the  rate  of  3  miles  an  hour 
in  a  direction  (10°  oast  of  north,  in  a  current  flowing  southwards  at  the  rate 
of  1^  miles  an  hour.  Show  that  the  boat  will  travel  due  eastwards^nd  find 
rate  of  progress.  Ana.  jV3  m/h. 

55.  AB  is  a  given  straight  line,  and  P  a  given  fixed  point  without  it;  a 
particle  Q  moves  along  AB  with  a  given  constant  velocity  When  Q  is  in 
any  assigned  position,  find  its  angular  velocity  with  respect  to  P. 

-4ns.  pv/PQ',  V  being  the  constant  velocity  and  p  the  perpendicular  from 
Pto  AB. 

56.  A  particle  under  the  action  of  a  number  of  forces  moves  with  a  uniform 
velocity  in  a  straight  line.    What  condition  must  the  forces  fulfil? 

57.  A  lift  is  descending  and  coming  to  rest  with  a  uniform  retardation 
of  4  F.P.S.  units.  A  man  in  the  lift  weighs  out  a  pound  of  tea  with  an  ordinary 
balance,  and  a  pound  of  sugar  with  a  spring  balance.  How  many  pounds  of 
each  does  he  really  obtain?  .4ns.  1  lb.  of  tea;  I  lb.  of  sugar. 

58.  A  ;5-toii  cage,  desieiiding  a  shaft  with  a  speed  of  &  yards  a  second,  is 
broii<;ht  to  a  stop  by  a  uniform  force  in  the  space  of  18  ft.     What  is  the  ten- 


sion in  the  rope  while  the  stoppage  is  occurring? 


Ans.  4\'i  tons. 


59.  A  1-oz.  bullet  is  fired  with  a  velocity  of  1000  ft.  per  second.  Find 
the  velocity  with  which  a  2-oz.  bullet  could  be  fired  from  the  same  rifle  with 
treble  the  charfre  of  powder.  Ans.  122.5  f  s 

60.  If  a  14-lb.  shot  leave  the  muzzle  of  a  2-ton  gun  with  a  relative  speed 
of  .540  ft.  per  secoiul,  find  the  speed  of  recoil.  ^Ins.  1.GS2  f,  s. 

6i.  A  pun  weisrhing  .">  tons  is  charged  with  a  shot  weighing  28  lbs.  If 
the  pun  he  free  to  move,  with  what  velocity  will  it  recoil  when  the  ball  leaves 
it  with  a  velocity  of  1(X)  ft.  jht  second?  Ans.  J  f  s. 

62.  A  railway  train  travels  \  niile  on  a  smooth  level  line,  while  its  speed 
increases  unifoimly  from  1.")  to  20  miles  an  hour.  What  proportion  does 
the  pull  of  the  engine  bear  to  the  weight  of  the  train?  Ans.  itVti. 


EXAMPLES. 


195 


63.  A  13-ton  gun  reooils  on  Ijeing  fired  with  a  velocity  of  10  ft.  per  second, 
ar.il  i-i  brought  to  rest  by  a  uniform  fricti(>n  equal  to  the  weight  of  A  ton. 
lliiw  far  does  it  recoil?  Ann.  !>  ft. 

64.  Determine  the  charge  of  powder  required  to  send  a  32-lb.  shot  to  a 
raiij'  of  L'.")00  yards  with  an  olpvation  of  1')°,  supposing  the  initial  velocity 
>  1(100  ft.  a  second  when  the  charge  is  half  the  weight  of  the  shot,  and  that 
the  iiiitiid  energy  of  the  shot  is  projKirtional  to  the  charge  of  powder. 

A  ns.  3  lbs. 

65.  A  ,5-02.  ball  moving  at  the  rate  of  1000  f  s  pierces  a  shield  and  moves 
iin  with  a  velocity  of  400  f  s;   what  energy-  is  lost  in  piercing  the  shield? 

.4«.s.  4076  ft.-lbs. 

66.  .\  half-ton  shot  is  discharged  from  an  81-ton  gun  with  a  velocity  of 
Iti.'O  ft  |)er  second.  What  will  be  the  velocity  with  which  the  gun  will  begin 
to  n  cull  if  the  mass  of  the  powder  be  neglected?  Will  the  gun  or  the  shot 
Ik>  utile  to  do  more  work  before  coming  to  rest,  and  in  what  proportion? 

Arts.  10  f/s;  shot  does  162  times  that  of  gun. 

67.  .\  ball  weighing  12  lbs.  leaves  the  mouth  of  a  cannon  horizontally 
with  a  \eIocity  of  1000  ft.  per  .second;  the  gun  and  carriage,  together  weigh- 
ing 12  cwt.,  slide  upon  a  smooth  plane  whose  inclination  to  the  horizon  is 
30^  Find  the  distance  through  which  the  gun  and  carriage  will  be  driven  up 
tho  [ilane  by  the  recoil.  Ann.  4.982  ft. 

68.  The  pressure  of  water  in  a  hydraulic  company's  main  is  750  lbs.  per 
S'liuirc  inch,  and  the  average  flow  is  25  cubic  feet  pet  muiuie.  What  horse- 
power dut's  this  reprc.-<tiif.'  If  the  charge  for  the  water  is  two  cents  per  100 
gal^,  what  is  the  cot  j    r  horse-pfiwer  hour?        .4ns.  88.2  H.P.;  2j'^  cents. 


6«. 
a  niiii 
I'aii  111' 

70. 
per  s.i 

illlW    II 

71. 

isit<  h 

7-'. 
of  1   ill 

7'- 

Jhafi  I 

>'<■'■)  i  . 

T.t. 

nf  I'.lill 


75- 
fatiiii'i 


An  accumulator  loaded  to  a  pressure  of  750  lbs.  per  square  inch  has 
of  21  ins.  diameter,  with  a  stroke  of  24  ft.  How  much  horse-power 
obtained  for  a  period  of  50  seconds?  Ans.  226.8. 

I'iiid  the  weight  which  will  give  an  average  fluid  pressure  of  750  lbs. 
lare  inch  in  an  accumulator  with  a  14-in.  ram  and  a  stroke  of  16  ft. 
inch  energy  can  be  stored  up? 

Ans.  115..")00  1bs.;    1,848,000  ft.-lbs. 
A  steam-pump  raises  11  tons  of  water  15  ft.  high  every  minute.    What 
.rsc-jM)wor?  A71S.  10. 

1  intl  the  horse-power  required  to  raise  a  weight  of  10  tons  up  a  grade 
I  •_'  at  a  speed  of  6  miles  per  hour  against  a  resistance  of  9  lbs.  per  ton. 

Ans.  31.3. 
I'iiid  the  horse-power  of  an  engine  that   would  empty  a  cylindrical 
ill  of  water  in  32  hours  if  the  diameter  of  the  shaft  lx»  8  ft.  and  its  depth 
ih<'  weight  of  a  cubic  foot  of  water  being  t)2.5  lt)s.  Ans.  8||. 

I  i.termine  the  horse-power  transmitted  by  a  In-lt  moving  with  a  velocity 
fl.  per  minute,  passing  round  twt)  pulleys,  supposing  the  difference 
nil  of  the  two  \niTts  is  the  weight  of  ICi'A)  \hg.  Ans.  30. 

A  cylindrical  shaft  of  10  ft.  diameter  has  to  be  sunk  to  a  depth  of  100 
s  through  (halk,  the  weight  of  the  chalk  being  143^  lbs.  per  cubic  foot 


i 


1% 


THEORY  OF  STRUCTlHES 


\Vh:it  horse-power  is  required  to  lift  out  the  material  in  12  working  days  of  8 
hours  each?  Ans.  10  7. 

76.  An  engine  is  retjuired  to  raise  in  3  minutes  a  weipht  of  13  ewt.  from  a 
pit  whose  depth  I*  >4()  ft.     Find  the  horse-power  of  th<' engine.     Aim.  12.3."). 

77.  .\n  aeeunuilator-ram  is  S.s  ins.  in  diameter  an<i  has  a  stroke  of  21  ft. 
Find  the  store  of  energy  in  fipot-i>ounds  when  the  ram  is  at  the  top  of  the 
stroke  and  is  loaded  until  the  ])ressure  is  7")()  ihs.  per  square  inch. 

.1;(.s-.  !).-)S,320  ft.-lbs. 

78.  In  a  dilTerential  aiciunulator  the  diameters  of  the  spindle  are  7  ins. 
Tud  .">  ins.:  the  stroke  is  10  ft.  Find  the  store  of  energy  when  full  and  loaded 
10  20«)()  Ihs.  per  s(|uare  iiiih.  .l//.s.  37.-(,0O0  ft. -Mm. 

79.  In  a  dilTerential  pres>  the  diameters  of  the  u])per  and  lower  portions 
of  the  ram  are  riins.  and  Sins,  respectively.  The  pressure  is  lOOOlhs.  per  square 
inch.,  and  the  stroke  is  10  ft.  Find  the  load  on  the  aeeumulator,  the  ma.ximum 
store  of  energy,  and  the  store  of  water. 

.W/.v.  22,(K)0lbs.;  220.(MM)  ft.-lba.;   Ij;  eu.  ft. 

80.  Find  the  horse-power  of  a  fall  where  18,000,000  eu.  ft.  passes  per 
minute,  falling  Hi2  ft.  An.s.  .'),,-)22,727. 

81.  .\  hoii/oiil;il  axle  10  \n>.  (liatncter  lias  a  vertical  load  upon  it  of  20 
tons,  and  a  hori/<intal  pull  of  }  tons.  The  coefFicient  of  friction  is  0.02.  Find 
tlie  hi'at  generated  per  ininut",  and  the  horse-[H)W('r  wasted  in  friction,  when 
making  "0  revolutions  per  minute.  .1/(n.   ITm  thermal  units;  .'J.d.'J  ll.P. 

8j.  .\  .'Ml.l'.  steam-crane  is  found  to  raise  a  weiidit  of  10  tons  to  a  height 
of  .'Oft.  in  20  minutes;  wh.it  pait  of  the  work  is  done  .against  friction?  If  the 
erani'  is  kept  at  similar  work  for  S  Iiours,  Jiow  many  foot-pounds  of  the  work 
arc  wa-ted  ,  i,  fii,  ii,>n'.'  .1/,.s-.  S,i\  20,()40.(KX)  ft.-llis. 

83.  Find  the  energy  per  .second  of  a  waterfall  .'!()  y<ls.  high  and  1  mile  hroad 
where  tlie  tiiass  of  w;itcr  is  20  ft.  deep  and  h:is  a  velocity  if  7J  miles  an  hour 
when  it  arrives  at  the  fall.     Tlie  weiL'ht  of  water  is  1021  oz.  per  cubic  f(M>f. 

.1//.S.     hl.">-4(il    ft  .-tons  .see. 

84.  A  shaft  .">C)0  ft.  il.cp  .and  "  ft.  in  diameter  is  full  of  water:  how  many 
foot-pounds  of  work  .'lie  ie(|uired  \>i  I'lnpty  it,  and  how  long  would  it  take  an 
engine  iif -{J  hor<e-|Knver  to  ijo  t  he  vMirk'.'  .N.H.  Of  course  it  is  to  be  assuriH'<l 
liial  there  i<  no  llo'-  01'  water  into  1  he  sli.ift.) 

A/IS.   IM2,'.(H).(KM)  ft.-l'os.:   ll'itlCf;  minutes. 

85.  lind  the  work  e\|iended  in  raising  the  materials  ilI2  lbs.  per  ciihic 
.f.  (ii  lor  ,1  biiik  toner  I2."(  It.  in  height  and  of  21  ft.  external  anil  Hi  ft.  itit(Miial 
diaiiieicr.     Ill  whiil  lime  could  the  materials  1k'  rais«'d  by  u  21-11.1'.  engine? 

. !/..«.   220.(KH1,0(K)  ib.-ft.:    14J  hours. 

86.  A  unfiirm  he.ini  wcitrlis  KKK)  His.  .•iml  is  20  ft.  long.  It  hangs  by  one 
end.  luuiiil  \viiii  h  it  c.'in  turn  freely.  Ilnu  inany  fnot-pounds  of  work  mti-t  1n' 
done  to  r.aise  it  from  it-  jouc^st  to  itx  hiuhcsl  position'       .l/,.v.  20.0<K»  fl.-ilis 

S7  l!ie(iric  l.iinps  g  ve  I  caiidle-p'iuer  for  !  wjilts;  liow  many  10-  or 
lio'v  inaiiv  picandle  l.itrips  may  tie  worked  per  electric  horse-jHiwcr?  Tlie 
ciiniliined  I'tiiciency  o|  eng:iie  dyn.'ilno,  and  gearing  U'liig  7il  |)er  cent,  wimi 
\n  the  candle-power  avadable  for  every  indieatod  horse-jiower? 

Ans.    IS;   11;   130.a5. 


EXAMPLES. 


197 


S8.  A  cast-iron  fly-whoel  of  36  sq.  ins.  section  and  120  in?,  mean  diameter 
v.vAf<  'iO  revolutions  per  minute.  Find,  approximately,  the  mean  energy 
nf  VI nation.  Also  find  the  number  of  revolutions  per  minute  after  losing  800 
fi-lln.  of  energy.  Ans.  54,424  ft.-lb.s. ,  59i. 

89.  If  the  earth  be  assumed  to  be  spherical,  how  much  heat  would  be 
il(vi''(i|H'd  if  its  axial  rotation  wore  suddenly  .stopped,  a  unit  of  heat  corre- 
spMiuliiij:  to  778  ft.-lbs.?  Weight  of  mass  of  earth  =  10"  XG.029  tons;  diam- 
ti,r  (.f  earth  =8000  miles.  Ans.  10"  X63,()3.5. 


go.  \  fly-wheel  supported  on  a  horizontal  axle  2  ins.  in  diameter  is  pulled 
round  by  a  cord  wo'ind  round  the  axle  carrying  a  weight.  It  is  found  that  a 
wi  i.'ht  of  4  ll)S.  is  just  sufficient  to  overcome  the  friction.  A  further  weight 
,f  Ki  ll)a.,  making  20  in  all,  is  applied,  and  after  two  seconds  starting  from 
rrst  it  is  found  that  the  weight  has  gone  down  12  ft.  Find  the  moment  of 
iiirrtia  of  the  wheel.  Ans.  .014. 

Qi.  Find  the  horse-power  of  an  engine  which  pumps  up  wal-'r  from  a  depth 
vi  .(I  ft.  and  delivers  it  at  the  rate  of  1000  gals,  per  minute  through  a  pipe 
will  -(-  cross-section  is  1  sq.  ft.  Ans.  15iV. 

02.  A  fly-wheel  with  a  rim  of  uniform  axial  thickness  weighs  1000  lbs., 
ha<  a  (i(t-in.  external  and  a  48-in.  internal  diameter,  and  makes  W)  revolutions 
pr  minute.  The  greatest  fiuctuation  of  energy  is  1000  ft.-lbs.  Find  the  varia- 
iiiiM  111  s|K'ed. 

Ans.  9.43  revolutioiiS  per  minute. 
g3.  .\  beam  will  safely  carry  1  ton  with  a  deflection  of  1  in.     From  whnt 
h.  it'lit  may  a  weight  of  100  lbs.  drop  without  injuring  it,  neglecting  the  effect 
i.t  i!ii  rliu?  Auk.  10.2  in. 

94.  A  cut  of  .Ofi  in.  depth  is  being  made  on  a  4-in.  wrought -iron  shaft 
rcvdlviiig  nt  10  revolutions  j)er  minute;  the  traverse  feed  is  .3  in.  per  revolution; 
ihi  pn-ssureon  the  tool  is  found  to  be  435  lbs.  What  is  the  horse-jmwer  ex- 
[K  iiilrd  on  the  tool?    How  much  metal  is  removed  per  hour  per  horse-power? 

.-In.*.  .1381;    08.84  cu.  in. 
<i-,.  The   travel  of  the  table  of  a  planing-niachine  whi(  h  cuts   Ixith  waj-s 
1^  'I  It  :   taking  the  resistance  to  Ije  overcome  .'it  400  lbs.  and  the  number  of 
'1'  ul.li  strokes  per  hour  at  80,  find  the  horse-iM)wer  ab8orlK>d  in  cutting. 

•Ins.  .29. 
q''.   I  he  lly-wheel  c.f  a  3-H.P.  riveting-machine  fluctuates  iK'tween  SO  and 
I'Vdlutions  per  minute;  every  two   seconds  an   oiM-ration  (it(U!s  wlii.  h 
•'  s'vcn  eighths  of  all  the  energy  supply  for  two  seconds.     Find  the 
-  Muiment  of  inertia.  .■\u».  f'"<>. 

\  lly-wheel  weighing  5  tons  has  a  mean  radius  of  gyration  of  It)  ft. 

Ill  1 1  is  carried  on  a  shaft  of  12  ins.  diameter,  and  is  running  at  C'l  rcvolu- 

!«  r  minute.     Mow  m.'iny  revolutions  will  the  wheel  make  iM'forc  -"niiping, 

■  orlTiiienl  of  friction  of  the  shaft  in  its  Uarinir  is  ll.d'l." .'     Otliir  u'sist- 

iii.iy  lie  neglected.)  Ans.  354  0. 

■'.  Ill  .•!  gas-engine,  using  thi*  Otto  cycle,  the  I. II. P.  is  8,  and  tlie  sjieed  is 

w.lniiiiiis   [XT  miiiiitv.    Ti<'Miiii»i  ciirii  fiiiiiih  niiiglo  stroiic-  ri.?  rfff< tive 

;ii.'   nsistiince  as  unifortn,  find  how  many  f(M)t-lbs.  of  energy  must  be 


l.'fl  I 

tr.|ir 


if  ill, 

aim  ■ 


108 


THEORY  OF  STRUCTCRKS. 


ston'd  in  a  fly-whool  in  onlcr  that  tho  sjx'cd  shall  not  vary  by  more  than  nno- 
fortieth  abovo  or  hclnw  its  mean  value.  Ans.  29,255  ft.-lbs. 

99.  A  sphorp,  mas.s  10  lbs.,  at  rest  '  •  struck  by  another,  ma.sa  8  lbs.,  moving 
with  a  velocity  of  20  miles  [xt  hour  in  a  direction  making  an  angle  of  45°  with 
the  line  of  centres  at  the  ni nierit  of  impact;  the  coefiicient  of  rebound  is 
J.     Determine  the  subsequent  motion.        Ans.  cot  fl=9;  vel.  --''^V^l  f/s. 

100.  A  steamer  of  SOCK)  t/>ns  displacement  sailing  due  east  at  16  knots 
an  hour  collides  with  a  steamer  of  .")0(X)  ton.s  displacement  sailing  at  10  knots 
an  hour.  Find  ♦he  energy  of  collision  if  the  latter  at  the  moment  of  collision 
is  going  (1)  due  west;    (2)  northwest;    (3)  northeast. 

Ans.  92,823;  79,951;  17,820  ton-ft. 

loi.  A  fly-wheel  weighs  10,0(X)  lbs.,  and  is  of  such  a  size  that  the  matter 
composing  it  may  be  treated  as  if  concentrated  on  the  circumference  of  a  cir- 
rle  12  ft.  in  radius;  what  is  its  kinetic  energy  when  moving  at  the  rate  of 
15  revolutions  a  minute? 

How  many  turns  would  it  make  before  coming  to  rest,  if  the  steam  were 
cut  off  and  it  moved  against  a  friction  of  400  lbs.  exerted  on  the  circumference 
of  an  axle  1  ft.  in  diameter?  Ans.  55,,561.2  ft.-lbs.;  44.2  tun    . 

lo:  Two  inelastic  bodies,  the  one  of  100  lbs.  moving  due  W.  at  I  t/i, 
the  other  of  50  lbs.  moving  due  K.  at  10  f  s,  collide.  Find  the  energy  oi  col- 
lision. What  will  Ix?  the  energ>-  of  collision  if  they  move  in  the  same  direction, 
i.e., due  K.  ordue  W?  ,1,,.,.  408}  ft.-lbs.;  52t',  ft  -lbs. 

103.  Prove  that  for  a  rope  round  pulleys,  running  at  3000  ft.  a  minute, 
to  tnuuMiiit  40  horse-power,  the  tension  on  the  driving  side  must  be  880  lb8., 
supposing  it  double  the  tension  on  the  slack  side. 

104.  A  fly-wheel  weighs  20  tons  and  its  .adius  of  gjTstion  is  5  ft.  How 
mill  h  work  is  given  out  while  the  six-ed  falls  from  60  to  .50  revolutions  |)er 
"'ii'ute?  Ann.  94.3  ft. -tons. 

105.  A  tower  is  to  be  built  of  biickwork,  and  the  ba.so  is  a  rectangle  22  ft. 
by  9,  and  the  height  is  00  ft.,  the  walls  Uing  2  ft.  thick.  Find  the  nunik-r 
of  units  of  v.'irk  ex|K-nded  on  raising  the  bricks  from  the  ground,  and  the  nunr 
Iht  of  hours  in  which  an  engine  (tf  3  horse-power  would  raise  them,  a  ubic 
foot  of  brickwork  weif>hing  1  cwt.    Ana.  20,.345,ON.S  ft.-lbs.;  5.42(t8  hours 

106.  A  rifle-bullet  .45  m.  in  diameter  weighs  1  oz. ;  the  charge  of  pf)wder 
wciL'hs  K")  grains;  the  muzzle-velocity  is  Mi.M)  ft.  per  second;  the  weight  of 
the  rille  is  9  lbs.  Xriihrtinii  the  ^r/.v,  determine  the  energy  of  1  lb.  of  |)owder. 
If  the  bullet  loses  \  of  its  velocity  in  its  j.assage  through  the  air,  find  the  aver- 
age force  of  the  blow  on  the  target  into  which  the  bullet  sinks  i  in. 

If  there  is  a  twi.st  of  1  in  in  20  in.,  find  the  cuurge  to  give  the  same  muzzle 
velocity    the  length  of  the  barrel  In-ing  3o  in. 

Ana.  140,570  ft.-lbs  :  1S!(S4J  lbs.;  8.5.10<13  gniins. 

107.  The  table  of  a  small  pluning-machine,  which  weijchs  1  rwt.,  mnke 
six  double  strokes  of  4}  ft.  each  (kt  minute.  The  coefficient  of  friction  Ix-tx.ceii 
the  sliding  BurfucvH  io  0.07.  X^hul  w  the  work  performed  in  fo<n-jH)uniis  |Kr 
minute   in  moving  the   table?  Ant.  423.3 


EXAMPLES. 


199 


1 08.  The  fly-wheel  of  a  40-H.P.  engine,  making  50  revolutions  per 
niiiiiitf,  is  20  ft.  in  diameter  and   weighs   12,000  lbs.     What  is  its  kinetic 

It  the  wheel  gives  out  work  equivalent  to  that  done  in  raising  5000  lbs. 
tlitcugli  a  height  of  4  ft.,  how  much  velocity  does  it  lose? 

Ihc  axle  of  the  fly-wheel  is  12  in.  in  diameter.  What  proportion  of  the 
|i.r-.-| lower  is  re(iuired  to  turn  the  wheel,  the  coefficient  of  friction  being  .08? 

If  the  liy-wheel  is  disconnected  from  the  engine  when  it  is  making  50  revo- 
hiiiiiiis  i)er  minute,  how  many  revolutions  will  it  make  before  it  comes  to  rest? 
Auk  511,2(10.4  ft.-lbs.;  1.04  ft.  jier  stjc;  ,*5;  160.4. 

109.  A  uniform  circular  disc,  whose  radius  is  4  ins.  and  weight  2  lbs.,  can 
turn  about  a  horizontal  axis  through  its  centre  at  right  angles  to  its  plane. 
Will  II  started  at  the  rate  of  200  revolutions  ])er  minute,  and  left  to  itself, 
11  is  observed  to  come  to  rest  in  one  minute. 

I'rove  that  the  retarding  couple  (supposed  constant)  has  the  .same  moment 
;,s  ii  weight  of  25J  grains  attached  to  the  rim  at  the  extremity  of  a  horizontal 
r:i('iiis.     .\lso  find  how  many  revolutions  it  makes. 


1 10.  A  particle  is  placed  on  a  rough  horizontal  plate  at  a  distance  of  9  ins. 
frniM  a  vertical  axis  about  which  the  plate  can  turn.  Find  the  greatest  num- 
l»r  uf  revolutions  per  minute  the  plate  can  make  without  causing  the  particle 
to  Mi|.  upon  it,  the  cfK'fficie.it  of  friction  being  J.  Ans.  .50i?. 

111.  A  wrought -iron  (ly-wheel  10  ft.  in  diameter  makes  M  revolutions 
|.i  r  iiiiiiute.  Find  the  intensity  of  stress  on  a  transverse  section  of  the  rim, 
(li^iv.'MKliiig  the  influence  of  the  arms.  If  the  wheel,  which  weighs  W  lbs. 
pvr-  (lilt  work  equivalent  to  that  done  in  raising  W  through  a  height  of 
.■il  tt.  ill  1  sec,  what  velocity  will  it  lose?  If  the  axle  of  the  wheel  is  10  in. 
ill    iliaiiieter   and    if  .08   is   the    coefficient    of    friction,  show   that    it    will 

ir 


l;,k 


III 


.'.'.(H) 


ii.r.  to 


rnthewhwl.     (</-32.2.) 

Ann.  10,33.j  lbs.;  5.6  ft.  per  sec. 
in.  An  engine  of  400  horse-power  can  draw  a  train  of  200  tons  gross 
,11  incline  of  1  in  2S()  at  M  miles  an  hour.     Determine  the  resistance  of 
M.id  in  pounds  per  ton.  •4«s.  17  llw. 

113.  IVove  that  a  train  going  45  miles  an  hour  will  lie  brought  to  rest 
iliniit  :{7H  yards  by  the  brakes,  supixtsing  them  to  press  with  two-thirds 

iif  I  lie  weight  on  the  wheels  of  the  engine  and  brake-vans,  which  are  half  the 
v,(  L'lii  (if  the  train,  taking  a  c(H'fficient  of  fri<'tion  0.18. 

I'liivc  that  an  engine  ca(»able  of  exerting  a  u.  iform  pull  of  .3  tons  can  take 
till-  triiiii,  weighing  120  tmis,  on  the  level  fn>m  one  station  to  stop  at  the  next, 
:■  Mill  s  dlT,  ill  alMiut  :\  inins.  38J  .sees.,  the  six>ed  being  kept  uniform  when  it 
hi-  Il  ached  45  iiiilcs  an  hour. 

114.  Determine  the  constant  effort  exerted  by  a  horse  which  does  1,650,000 
ft. -II 11.  of  work  in  one  hour  when  walking  at  the  rate  of  2i  miles  per  hour. 

Ann.   125  lbs. 

115.  A  train  is  drawn  by  a  locomotive  of  100  H.P.  at  the  rate  of  GO  milca 


'm 


200 


THEORY  OF  STRUCTURES 


an  hour  against  a  resistance  (if  20  lbs.  per  ton.      What  is  the  gross  weight 
..f  the  train?  Ans.  50  tons. 

ii6.  A  train  of  202 J  tons  is  drawn  up  an  incline  of  1  in  75,  5J  miles  long, 
asainst  a  resistance  of  10  lbs.  |)er  ton,  in  ten  minutes.  P'ind  the  H.P.  of  the 
engine.  The  sjK'ed  on  the  level,  the  engine  exerting  769.42  H.P.,  is  43.4  miles 
jK-r  hour.     What  is  the  resistance  in  pounds  |)er  t(m? 

.l/(.s-.   1027  H.P.;  22.7  lbs.  per  ton. 

117.  An  engine  with  its  tender  weighs  SO  tons.  It  is  moving  uniformly 
at  the  rate  of  20  miles  an  hour,  against  a  resistance  of  7  llw.  a  ton.  At  what 
horse-power  is  it  working?  Ans.  170J. 

If  it  drew  .ifter  it  a  train  of  12  carriages,  each  weighing  10  tons,  at  the 
rate  of  40  miles  an  hour,  against  a  resistance  of  8  lbs.  a  ton,  at  what  horse- 
power would  it  now  '.k^  working?  Ans.  29p-  170?. 

118.  (iiven  a  locomotive  with  two  lS"x20"  cylinders,  the  connecting- 
rod  -  0  ft.,  the  boiler-i)ressure  =  140  lbs.,  and  driving-wheels  of  7'  0"  diameter, 

calculate  the  adhesion-f  ietion,  i.e.,  the  ratio  ^"rcc  at  periphery 

weight  on  drivers' 
Ana.  ftsec  0sin(0  +  <l>),  where  fiW  ■=■  14,040  lbs. 

119.  A  railway  wagon  weighing  20  tons,  with  two  pairs  of  wheels  S'  0" 
centre  to  centre,  and  with  its  centre  of  inertia  7'  0"  above  top  of  rails,  has 
its  whe<-ls  skidded  while  running.  Take  ;<=0.15.  Required  the  total  retard- 
ing force  and  pressure  of  each  wheel. 

Ans.  .3  t<  li^;  7.375,  12.625  tons. 

120.  Prove  that  if  a  motor  car  going  at  100  km  /h.  can  be  stopped  in 
2(X)  m.,  the  brakes  can  hold  the  car  on  an  incline  of  1  in  5,  and  determine  the 
time  required  to  stop. 


121.  What  horse-power  is  given  up  in  lowering  by  2  ft.  the  level  of  the 
surface  of  a  lake  2  sf|uare  miles  in  area  in  3(X)  hours,  the  water  Ix'ing  lifted 
to  an  average  height  of  .">  ft.?  Ait.s.  29.33 

122.  .\  cistern  at  a  height  of  2.")0  ft.  above  the  river  is  to  Ije  filled  with 
2().(KM)  gals,  of  water  every  24  hours  by  pun)ps  worked  by  a  turbine  with  a 
fall  (.f  4  ft.  Determine  how  much  water  the  turbine  will  require,  and  its 
liursc-power  with  an  efficiency  of  70  jmt  cent. 

.!/(.<.  S7-"),(MM)  gals,  per  1:4  hours:   l.O'i. . 
12V  Takitii  the  average  power  of  a  man  as  one  tenth  of  a  horse-jwwer, 
nii.l  the  clficiency  of  ijic  pump  used  as  ().(,  in  what  time  will   ten  men  empty 
.Ml  Ilk  of  .Vl'  ■  .lO'  ■  (i'  lilli.d  with  water,  the  lift  l)eing  an  average  height  of 
•'"  "■ '  Aii.i.  1278  minutes. 

124.  A  lin-«  ngine  pump  is  provided  with  a  nozzle  the  sectional  area  of 
^'ll).ll  I-  I  s, I, ;,:■,.  ;.,,.[,  ,,,,,1  til,,  vv;it(r  is  |in.ject(>(l  througli  the  nozzle  with 
an  avi  race  n,  imal  \.lucily  of  l.JO  ft.  jmt  second;  inid  il)  the  nun/lier  of 
cubic  feel  ,11-  Iimi-.mI  |.t  seromi,  ,2'  the  wcJL'hl  <if  water  discharged  jht  inin- 
"''■  '  ''"■  '  i'"'ti  •  ciicrirv  ..f  ca.h  pound  of  wnler  as  it  leaves  the  nozzle.  (4) 
the  lior^'-pow.r  ..f  thc<  iigine  rcciuired  to  drive  the  pump,  assuming  the  effi 
'■"'"'>■  '"  '»<■  •'<'  I"!-  "lit.  .l„,v.  ,<|  ,.,1.  ft.:    |..-,1  tons;   2r.2.;t  fl.-lbs.-;  .38.3 

125.  I  iiiil  tlu   sli,,iu-t  iitiic  from  rest  to  rest  in  which  a  chain  capable  of 


EXAMPLES 


201 


bcMriiig  a  safe  load  of  25  tons  can  raise  a  weight  of  10  tons  out  of  a  hold  15  ft. 
c|pp|i:  also  find  the  greatest  load  which  can  be  raised  or  lowered  in  2i  seconds. 

Ans.  IJ  seconds;  21 J  tons. 

126.  A  heavy  vertical  chain  is  drawn  upwards  by  a  given  force  of  P  Ibs- 
wciirlit,  which  exceeds  its  weight  ir.  Find  its  acceleration  and  its  tension 
ill  ;iiiy  assigned  point.     Show  that  the  tension  at  its  middle  ix)iiit  is  iP. 

Alls.    ~~,    g;  nP,  n  being  the  fraction  of  the  chain  below  the  point  con- 

sidcred. 

127.  A  rojie  mX)  ft.  long,  and  weighing  2  lbs.  a  foot,  is  wound  on  a  roller. 
What  is  the  difference  of  its  potential  energy  in  this  position  and  in  its  position 
when  200  ft.  f)f  the  rope  have  rolled  out,  neglecting  friction  and  the  weight 
(if  I  lie  roller  and  supjX)sing  that  no  part  of  the  rope  touches  the  ground? 

Ans.  1,280,000  ft.-poundals. 

128.  A  chain  hanging  vertically,  520  ft.  long,  weighing  20  lbs.  per  foot, 
is  wound  up;  what  work  is  done?  Aiis.  1352  ft. -tons. 

129.  What  is  the  kinetic  energj'  of  a  tramcar  moving  at  6  miles  per  hour, 
l.idcn  with  thirty-six  passengers,  each  of  the  average  weight  of  11  stones, 
wcijilit  of  car  2i  tons?  What  is  its  momentum?  If  stopped  in  two  seconds, 
\vh;it  is  the  average  force?  If  the  force  is  constant,  this  must  also  be  the 
spaic  average  force;   find  the  distance  of  stopping  if  the  force  is  constant. 

Ans.  13,400  ft.-lbs.;    3045.565;    1522.8  lbs.;    88  ft. 

!3o.  Find  the  horse-jxiwer  of  a  locomotive  which  moves  a  train  of  mass 
.)()  tuns  at  the  rate  of  30  miles  an  hour  along  a  level  railroad,  the  resistance 
from  friction  and  the  air  Ixjing  16  lbs.  weifiht  per  ton.  Ans.  64. 

131  A  cylinder  and  a  ball,  each  of  radius  /^  start  from  rc-it  and  roll  down 
all  inclined  plane  without  slipping.  If  V  is  tV-c  velocity  of  translation  after 
(iisdiidiug  through  a  vertical  distance  h,  show  that 


uiid 


V'-f(2jf/i)  in  the  case  of  the  cj'linder 
V*-^(2gh)  in  the  case  of  the  ball. 


■•fe 


It 


132.  A  wheel  having  an  initial  velocity  of  10  ft.  per  second  ascends  an 
im  iini'  of  1  in  100.  How  far  will  the  wheel  run  along  the  incline,  neglecting 
fn.  iioir,'    )';=.32.2.  ^".1.  232.0  f I 

133.  The  lly-whecl  of  an  ejigine  of  4  H.P.  running  at  75  revolutions  per 
ihi;  itc  is  ciiuivalent  to  a  heavy  rim  of  45  ins.  mean  diameter,  weighing  500  lbs. 
lilt,  'iiiinc  the  ratio  of  the  kinetic  energ}'  in  the  fly-whcil  to  the  energj-  de- 
vil..pid  in  n  revolution,  and  also  find  the  maximum  and  mininiuni  speeds 
"f  '.  iiilinn  when  the  llucluatioii  of  cnerg}'  is  one  fourth  of  the  cncr.'ry  of  a 
n'\i'i'iiion.  An.s.  1;    l.'.>4;  7.").2    74. S. 

I  u.  It  was  foutid  that  when  a  length  of  12  ins.  was  cut  off  the  muzzle  of 
:i  I'  ni.  cuit.  the  velocity  fell  from  1400  to  1330  f/s.  If  the  weight  of  the  shot 
«:i-  MHi  lbs.,  calcuJate  the  pressure  on  its  base  as  it  left  the  muzzle. 

Ans.  11.13  tons. 

US.  Three  gtwds  trucks,  weighing  respectively  5  tons,  7  tons,  and  8  tons. 


202 


THEORY  OF  STRUCTURES. 


are  placed  on  the  same  line  of  rails.  The  first  is  made  to  impinge  on  the 
second  with  a  velocity  of  00  ft.  jx-r  second  without  rebounding.  The  first 
and  .'second  together  impin-ie  in  the  same  way  on  the  third.  Find  the  final 
^■'•'"f'ty.  Ans.  15  f/s. 

i3f>.  A  steel  punch  J  in.  in  diameter  is  employed  to  punch  a  hole  in  a  plate 
i  in.  in  thickness;  what  will  l)e  the  least  pressure  necessary  to  drive  a  putuh 
through  the  jJate  when  the  shearing  strength  of  the  material  is  35  tons  j)er 
square  inch?  Ans.  51.56  tons. 

137.  Find  the  weight  of  rim  required  for  the  fly-wheels  of  a  punching- 
machine,  intended  to  punch  holes  li  in.  diameter  through  li-in.  plates;  speed 
of  rim  30  ft.  jx-r  .second.— .1/i.s.  :\>l  tens,  35  tons/!>q.  in.  being  the  shearing 
strength  !ind  ;is.suming  "jOf;  of  the  cncigy  to  lie  absorbed  by  the  punching. 

138.  In  a  fly-pre.ss  for  punching  holes  in  iron  plates,  the  two  balls  weigh 
30  lbs.  each,  and  are  placed  at  a  radius  of  :«)  ins.  from  the  a.xis  of  the  screw, 
the  siTcw  itself  having  a  pit(  h  of  1  in.  What  diameter  of  hole  could  be  punched 
by  such  a  press  in  a  wrought-iron  plate  of  i  in.  thickness,  the  shearing  strength 
of  which  is  '.'.■j.2  tons  [K-r  .square  inch?  -Awsunie  that  the  balls  are  revolving 
at  the  rate  of  60  revolutions  |)er  minute  when  the  punch  comes  into  contact 
with  (he  plate,  and  that  the  resistance  is  overcome  in  the  first  sixteenth  of 
an  inch  of  the  thickness  of  the  plate.  Ans.  1.136  in. 

130.  Find  the  stress  due  to  centrifugal  force  in  the  rim  of  a  cast-iron  wheel 
8  ft.  diameter,  running  at  160  revolutions  per  minute. 

Ans.  567.55  lbs.  /.sq.  in. 

140.  Find  in  pounds  the  horizontal  thrust  on  the  rails,  of  an  engine  weigh- 
ing 20  tons,  going  round  a  cur\-e  of  600  yds.  radius  at  30  miles  an  hour. 

An,s.  Weight  of  133  cwts. 

141.  A  leather  ht-lt  runs  at  2400  ft.  per  minute.  Find  how  much  its  tension 
is  increased  by  centrifugal  action,  the  weight  of  leather  being  taken  at  60  lbs. 
I)er  cubic  foot.  .4,,,..  -jO;!  lbs.  per  square  in<h. 

MJ.  Find  the  centrifugal  force  arising  from  a  cylindrical  crank-pin  6  in. 
long  and  3i  in.  in  diameter,  the  axis  of  the  pin  being  12  in.  from  the  axis  of 
the  engine-shaft,  which  makes  100  revolutions  per  minute.  How  would  you 
balance  such  a  pin?  Aiik.  .'jo.02  lbs. 

143.  A  railway  carriage  of  mass  1000  lbs.  is  travelling  with  velocity  .W 
I  s  round  a  curve.  If  the  radius  of  the  curve  be  lOOO'  ft.,  find  the  magnitude 
and  direction  of  the  resultant  thrust  on  the  raiks.  If  the  rails  are  4  ft.  SJ  ins. 
iipait,  find  hf)w  much  the  outer  rail  must  Ixj  rais<>d  so  that  the  carriage  may 
])res.s  fHTiH'ndiiMilarly  on  the  rails. 

Ans.  1003.04  lbs.;  tan-'  A;  4i  ins. 

144-  I"  a  fly-wheel  weighmg  12,000  lbs.  and  making  .'>0  revolutions  per 
minute,  the  ccnt'-c  nf  gravity  is  <mr  srveritnitth  of  an  inch  out  of  the  centre. 
Find  the  rent lifugal  force.  .l/i.s.  .~)0.4  lbs. 

145-  III  the  preceding  (piestion,  if  the  axis  of  rotation  is  incHned  to  the 
plane  of  the  wheel  at  an  angle  cot  '.001.  (ind  the  centrifugal  couple,  the 
radius  of  gyration  In'ing   10  ft.  Ann.  102.S.9  ft.-lbe. 

146.  A  rublMT  tire  weighing  2  lbs.  imt  foot  is  stretched  over  the  circum- 
ference of  a  wheel  3  ft.  in  diameter,  the  tangential  pull  in  the  rubber  being 


EXAMPLES. 


203 


10  Ihs.  Find  the  radial  pressure  exerted  by  the  tire  on  the  circumference  of 
the  wheel  per  inch  of  length  when  the  wheel  is  at  rest,  and  the  speed  at  which 
the  wheel  must  revolve  to  make  the  tire  cease  to  exert  any  radial  pressure. 

Ans.  IS  lbs.;  angular  velocity  =  j n/ 10. 

147.  An  engine  of  mass  1  ton  is  travelling  round  a  curve  at  the  rate  of 
HO  inilesan  hour.  If  the  curve  is  an  arc  of  a  circle  whose  radius  is  1210  ft., 
determine  the  horizontal  thrust  between  the  engine  and  the  rails. 

Ann.  .05  ton. 

148.  Prove  that  the  india-rubber  band  of  a  bicycle  will  become  slack  when 
running  at  more  than  (ngdT/W)*  f/s,  where  W  denotes  the  weight  of  the 
hand  in  pounds,  T  the  tension  in  pounds,  t  nd  d  the  diameter  of  the  wheel  in 

149.  What  is  the  angular  velocity  of  the  4-ft.  wheel  of  a  car  which  is  travel- 
ling at  the  rate  of  30  miles  per  hour?  What  is  the  relative  velocity  of  the 
.n't  ic  and  the  highest  point  of  the  wheel?  Aiw.  22 ;  30  m/h. 

150.  \  locomotive  engine  weighing  9  tons  passes  round  a  curve  f500  ft. 
in  nidius  with  a  velocity  of  30  miles  an  hour.  What  force  tending  towards 
till  ((-litre  of  the  curve  must  be  exerted  b''  the  rails  so  that  the  engine  may 
innvc  on  this  curve?  ^'">'-  Weight  of  18.15  cwts. 

151.  If  trains  are  to  run  at  30  miles  an  hour,  find  how  much  the  outer 
niil  should  be  raised  on  a  curve  of  half  a  mile  radius,  the  gauge  being  4  ft., 
so  tiiat  ihero  shall  be  no  side  thrust  on  the  flange.  Ans.  1.1  in. 

152.  A  railway  train  is  running  smoothly  along  a  curve  at  the  rate  of  (iO 
miles  an  hour,  and  in  one  of  the  cars  a  pendulum  which  would  ordinarily 
o-  illato  seconds  is  observed  to  oscillate  121  times  in  two  minutes.  Show 
that  the  radius  of  the  curve  is  =  1050  ft.  nearly. 

153.  If  a  railway  carriage  without  flanges  tc  its  wheels  moves  on  a  cir- 
<ul  11  curve,  show  how  the  effect  of  the  centrifugal  force  may  be  counteracted 
liy  I  ris'  of  the  outer  rail,  and  find  what  the  rise  of  the  outer  above  the  inner 
rail  sliould  be  if  the  radius  of  the  circle  Ix;  1320  ft.,  with  a  velocity  of  the  train 
:i()  miles  an  hour,  and  the  breadth  of  the  track  5  ft.  Ans.  2J  ins. 

154.  .\  train  weighing  4000  lbs.  per  lineal  ft.  has  a  speed  of  30  miles  jK?r 
Imui  arotnid  a  10-degrce  curve  on  a  steel  viaduct.  Find  (o)  the  transverse 
|e,  ~-inv  iH-r  lineal  ft.  of  structure  due  to  the  centrifugal  load;  (/>)  the  suiK>r- 
e|.v,ii.)ti  of  the  outer  rail.  Ann.  (a)  420  lbs.;  (h)  5.05  inches. 

155.  The  fly-wheel  of  an  engine  makes  80  revolutions  jx-r  minute,  and  the 
renpn.ial  of  the  coefficient  of  fluctuation  of  the  velocity  is  not  to  exceed  40. 
Deienuine  the  least  moment  of  energy  required,  the  fluctuation  of  energy 
p.  see,„ul  l)eing  8000  Ibs.  If  the  weight  of  the  wheel  is  4000  lbs.,  find  the 
la  Inis  of  gvration.  •■^"s-  4500;  6  ft. 

156.  A  circular  disk  of  cast  iron  (sn.  g. -7.1)  10  ins.  in  diameter  and  1  in 
tiii.  k  acts  as  a  pulley  for  a  cord  carrying  10  lbs.  on  one  end  and  5  lbs.  on  the 

-tie T.  Find  the  angular  velocity  of  rotation  of  the  pulley,  .ind  the  linejir 
veloeities  of  the  weights  .50  second"  after  starting  from  rest,  disregarding  the 
tl  I  I  nation  of  the  shaft  and  its  inertia.        Ana.  770;   320  ft.  per  second. 


I      'l 


204 


THEORY  OF  STRUCTURES 


157.  A  fly-wheel  of  a  shcaring-maphino  has  l.-iO.OOO  ft.-lbs.  of  kinetic  enerev 
Btorod  in  It  when  its  speed  is  2oO  revolutions  per  minute;  what  energy  do/s 
It  part  with  during  a  reduction  of  speed  to  200  revolutions  per  minute'  If 
82  per  cent  of  this  encrprv  given  out  is  imparted  to  the  shears  durin.^ 
a  stroke  of  2  ins.,  what  is  the  average  force  due  to  this  on  the  blade  of  the 
®^'''*'"^'  A  m.  54 .0(K)  ft .  lbs. :  53 1  3(i0  lbs. 

158.  A  4  in.v.3  i„.  diameter  crank-pin  is  to  be  balanced  by  two  weights 
on  the  samf  side  of  the  crank;  the  length  of  the  crank  is  12  ins.;  the  engine 
makes  100  revolutions  {)cr  minute:  the  distance  of  the  C.  of  G.  of  each 
weight  from  the  axis  of  the  shaft  is  6  in.     Find  the  weights. 

159.  A  heavy  ball  attached  by  a  string  to  a  fixed  point  O  revolves  in  a 
horizontal  circle  with  a  given  uniform  angular  velocity  to.  Find  the  vertical 
depth  of  the  centre  of  the  ball  below  the  point  of  attachment. 

If  a  uniform  rod  be  substituted  for  the  ball  and  string,  find  its  position 

Also  find  the  rmsition  when  the  ball  is  attached  to  the  fixed  point  by  a 

uniform  rod;   r  being  the  ratio  of  the  weight  of  the  rod  to  the  weight  of  the 

It' 


Ans,  h  —■ 


.,-1 


2  w'n+3 


160.  The  deflection  of  a  truss  of  /  ft.  span  is  /x.OOl  under  a  stationary 
load  W.  What  will  be  the  increased  pressure  due  to  centrifugal  force  when 
W  crosses  the  bridge  at  the  rate  of  60  miles  an  hour?  ^.^  „r 

Ans.  l^^~ 

161.  A  fly-wheel  20  ft.  in  diameter  revolves  at  30  revolutions  per  minute. 
Assuming  weight  of  iron  450  lbs.  per  cubic  foot,  find  ihe  intensity  of  the 
stress  on  the  transverse  section  of  the  rim,  assuming  it  unaffec+ed  by  the 
'^^^^-  Ans.  9»)  lbs.  per  square  inch. 

162.  Diameter  of  a  pij>e  is  IS  in.;  at  one  iwint  it  is  cur\-ed  to  an  arc  of 
0  ft.  radius.  Water  flows  round  the  curve  with  a  velocity  of  0  ft.  per  second. 
Dctcrmiiie  the  centrifugal  force  jx'r  foot  of  length  of  bend  measured  along 
^'"■••i^'^-  yl ns.  20.717  lbs. 

163.  A  disk  of  weight  n  and  area  .1  square  feet  makes  n  -evolutions  per 
s.i<nu\  about  an  axis  through  its  c(>ntre,  inclined  at  an  angle  0  to  the  normal 
to  the  plane  of  the  disk.     Find  the  centrifugal  couple. 


An.f. 


"''-"-  tan  0  ft.-lbs. 
5.12 


164.  Assuming  15,000  lbs.  per  square  inch  as  the  tensile  strength  of  cast 
iron,  Mti.l  tMkiiiu'  5  as  a  factor  of  safety,  find  the  riiaximuni  working  speed  and 
the  hursiiu-  spee.l  for  ,•!  c.ist-iron  tlv-wheel  of  L>()  ft.  mean  diameter  and  weigh- 
ing'J  I, (lOO  ll,s.,  ih"  section  of  the  rim  Ih'Iii^'  l<i()s,|.  in. 

Ans.   175;{  f/s;  ;j!n  ,S4  f/s. 
A  t',0-iii.  (Iriviiic-wlieel  weighs  ;rj  tons,  .-ind  its  C.  of  G.  is  1  in.  out  ol 
liiid  the  greatest  and  the  least  pressure  on  the  rails. 

Am.  .3i(i±— ) 
166.  A  wheel  (,f  V,,  i.'lii  ir.  radius  of  -vi-ition  /,-,  and  making  n  revolutions 


"is- 
centre. 


If  ■■* 


EXAMPLES. 


205 


per  second  on  an  axle  of  radius  R,  comes  to  rest  after  having  made  N  revo- 
lutions.    Find  the  coefficient  of  friction. 


Ans.  sin  1,6 


,  und  coeff.  of  fric.  =tan^. 


167.  The  maximum  variation  in  the  enerpcy  of  a  fly-wheel  whose  mean 
<IKOil  is  101  revolutions  per  minute  is  lO.oOO  ft.-lbs.  Find  the  weight  and 
ipprnximate  section  of  the  rim  of  the  wheel,  the  mean  diameter  being  10  ft., 

nil  the  greatest  variation  of  speed  being  2i  per  cent  from  its  mean  value. 

168.  A  hammer  weighing  2  lbs.  strikes  a  nail  with  a  velocity  of  l.j  ft.  per 
-r  ond,  driving  it  in  J  in.     What  is  the  mean  pressure  overcome  by  the  nail? 

Ans.  67.J  lbs. 
i6g.  A  mass  of  .50  lbs.  falls  from  a  height  of  50  ft.  and  penetrates  2  ft. 
inti)  loose  sand.    To  find  the  resistance  of  the  sand  in  pounds  weight. 

Ans.  1.300  Ihs. 

170.  A  pile-driver  of  300  lbs.  falls  20  ft.,  and  is  stopped  in  ,'0  second.  What 
IS  the  average  force  exerted  on  the  pile?  Ans.  .3344  lbs. 

171.  Determine  in  tons  the  mean  thrust  on  the  terminus  buffers,  which 
.-•top  in  G  ft.  a  train  of  200  tons  going  at  (5  m/h;  also  find  the  time  it  brakes 
in  seconds.  Ans.  40J  tons;     l/r  seconds. 

172.  Show  that  a  brake  resistance  of  38.5  lbs. /ton  will  destroy  a  velocity 
uf  CiO  m./h.  in  704  ft.  and  in  16  seconds. 

173.  With  full  brake  power  it  is  possible  to  pull  up  a  train  in  225  ft.  when 
riiriiiiiig  at  30  m/h,  and  in  flOO  ft.  at  fiO  m/h.  IIow  many  seconds  does  it 
1  >kr  to  pull  up,  and  what  will  be  the  distance  overshot  by  the  delay  of  a  second 
III  tlic  action  of  the  brakes?  Ans.  10;\  .sees.:  88  ft. 

174.  The  brakes  of  a  train  reduce  its  sjx-ed  3J  m/h  every  second,  and 
I  ilvi'  one  second  to  Ik"  applied.  In  what  distance  can  a  train  Ijc  stop]iod  when 
L-iiig  at  30  m/h  and  at  60  m/h? 

175.  Determine  the  length  and  time  in  which  a  l)arge  of  ,50  tons  moving 
!'  L'  tii/h  can  b«-  brought  up  by  a  rope  round  a  post,  su])posing  the  breaking 
i'  ill  of  the  roiie  is  1  ton.  Ans.  13J  ft.;  9'  sees. 

176.  An  express  train,  timed  to  run  at  the  full  speed  of  00  m/h,  is  checked 
Iv.  :i  siijnal  to  20  m/h  over  a  mile  of  road  under  repair.  The  train  takes 
"!  '  mile  from  rest  to  get  u;)  full  speed,  and  half  a  mile  to  pull  up.  Show  that 
ill''  train  will  1k>  2  minuJc;-  40  seconds  late. 

177.  A  weight  f.alls  16  ft.  am'  does  2.")(K)  ft.-lbs.  of  work  upon  a  pile  which 
it  lirivcs  4  ins.  against  a  uniform  resistance.  Find  the  weight  of  the  ram, 
:iii'l  the  resistance.  Ans.  160  lbs. ;  7680  lbs. 

178.  When  a  nail  is  driven  into  wood,  why  do  the  blows  swm  to  have 
lit  ill'  if  anv  effect  unless  the  wood  is  backed  up  by  a  piece  of  metal  or  stone? 


ST! 


I 


179.  A  hammer  weighing  2  lbs.  strikes  a  steel  plate  with  a  velocity  of  10 


203 


THEORY'  OF  STRUCTURES. 


ft.  per  second,  and  is  brought  to  rest  in  .000^  second.     What  is  the  average 
force  on  the  steel?  Ans.  6250  Ibe. 

i8o.  A  hammer  weiphinp  10  lbs.  strikes  a  blow  of  10  ft.-lbs.  and  drives  a 
nail  .5  in.  into  a  piece  of  timber.  Find  the  velocity  of  the  hammer  at  the 
moment  of  contact,  and  the  mean  resistance  to  entry.  Also  find  the  steady 
pressure  that  will  i)roduce  the  same  effect  as  the  haninier. 

i4n.s\  Sf/s;  240  lbs.;  480  lbs. 

i8i.  In  Ex.  li>0,  taking  the  weipht  of  the  nail  to  be  4  oz.  and  the  weight 
of  the  piece  of  timber  to  Ije  100  lbs.,  find  the  depth  and  timy  of  the  penetration 
(a)  when  the  timlxT  is  fixed;  (h)  when  the  timber  is  free  to  move. 

Also  in  case  {/>)  find  the  distance  through  which  the  timber  moves. 

-Iw.s.  (a)  fi'  in.;   ,',,  second;   (h)  .4424r)  in.;     .009448  second;   .04113  in. 

182.  An  inelastic  pile  weighing  half  a  ton  is  driven  12  ft.  into  the  ground 
by  30  blows  of  a  hammer  weighing  2  tons  falling  30  ft. 

Prove  that  it  would  require  120  tons  in  addition  to  the  hammer  to  be  super- 
posed  on  the  pile  to  drive  it  down  slowly,  supposing  the  resistance  of  the  ground 
uniform;   but  240  tons  if  the  resistance  increases  as  the  penetration. 

Prove  that,  with  uniform  resistance,  each  movement  of  the  pile  takes 
0.0228  -iecond. 

183.  If  the  resistance  of  the  ground  to  the  penetration  of  an  inelastic  pile 
is  00  tons,  prove  that  15  blow?  of  a  hammer  weighing  1  ton  falling  20  ft.  will 
drive  the  pile  4  ft.  into  the  ground,  the  pile  weighing  J  ton. 

Prove  also  that  the  time  of  each  movement  of  the  pile  is  0.01863  second. 

184.  An  inclagtic  pile  weighing  78S  lbs.  is  driven  3i  ft.  into  the  ground  by 
120  blows  from  a  weight  of  112  lbs.  falling  m  ft  Find  the  steady  load  upon 
the  pile  which  will  produce  the  same  effect,  assuming  the  ground-resistance 
to  be  ((I)  uniform;  (h)  proportional  to  the  depth  of  penetration.  If  the  re- 
sistance is  uniform,  how  long  (c)  dcH's  each  movement  of  the  pile  last?  How^ 
many  blows  {<!)  are  required  to  drive  the  pile  the  first  half  of  the  depth,  viz., 
1?  ft.,  the  ground-resistance  being  7168  lbs.?  How  far  (c)  does  the  pile  sink 
under  the  last  blow'.' 

Aiis.  (a)   I4,:«C.ll)s.:  (/.)  28,672  lbs.;  (c)  .0107  second;  (rf)  30;  (e)  .0146  ft. 

185.  A  pile  is  driven  <i  ft.  vertically  into  the  ground  by  u  blows  of  a  steam- 
hammer  fastened  to  the  li.'ad  of  the  pile.  Prove  that,  if  p  is  the  mean  pres- 
sure of  the  steam  in  pounds  per  s(iuure  inclr,  d  the  diameter  of  piston  in  inches, 
I  \\h'  lcii>rth  of  .troke  in  feet,  II'  the  weight  in  pounds  of  the  moving  parts 
of  the  hainnuT,  -/■  lbs.  the  weight  of  the  pile  and  the  fixed  parts  of  the  hammer 
attached  to  it,  then  the  mean  resistance  of  the  ground  in  pounds  is 


n  +w\        1      '  /  a 


186.  Prove  that  if  a  hammer-head  weiirhing  2  lbs.,  striking  with  a  velocity 
of  ,-)0  f/s  a  nail  ,'„  in.  in  diameter  and  wciirhiii!:  1  oz..  drives  the  nail  1  in. 
into  a  plank  t.f  wood,  thrn  a  bull.t  0..')  in.  in  diaineicr  and  weighing  1  oz.,  strik- 
ing with  a  velocity  1.1(M>  f  s,  will  penetrate  1.1(1  in.  of  the  wood;  supposing 
the  resistance  uiiifortn  and  proportional  to  the  sectional  area  of  the  hole. 


EXAMPLES. 


207 


Determine  also  the  penetration  of  the  bullet,  supposing  the  resistance  pro- 
portional to  the  penetration.  Ans.  2.2  ins. 
Determine  also  the  time  of  penetration  in  each  case. 

Ans.  .00344,  .00013  and  .000244  sees. 

187.  Prove  that  the  mean  resistance  of  the  wood  is  204  lbs.  to  a  nail  wcigh- 
ina  1  "z.,  supposing  a  hammer  weighing  1  lb.,  striking  it  with  a  velocity  of 
34  f  s  drives  the  nail  1  in.  into  a  fixed  block  of  wood. 

If  the  block  is  free  to  move  and  weighs  68  lbs.,  prove  that  the  hammer  will 
<lriv('  the  nail  only  {J  in. 

Trove  that  the  nail  is  0.0052  and  0.005128  second  in  penetrating  the  wood 
in  till  two  cases,  during  which  the  block  if  free  will  move  0.015  in. 

188.  Show  that  the  total  work  done  in  raising  a  number  of  weights  to  a 
pivcM  level  is  the  product  of  the  sum  of  the  weights  and  the  vertical  displace- 
iiit'iit  of  their  centre  of  gravity. 

189.  An  engine  has  to  raise  4000  lbs.  1000  ft.  in  5  minutes.  \Vhat  is  its 
hiirsc-iiower?    How  long  will  the  engine  take  to  raise  10,000  lbs.  100  ft.? 

Ans.  24,  J  H.P. ;  1 J  minutes. 
iQo.  How  many  men  will  do  the  same  work  as  the  engine  in  the  preceding 
question,  assuming  that  a  man  can  do  900,000  ft. -lbs.  of  work  in  a  day  of 
(•hours?  AriK.  480  men. 

191.  Determine  the  horse-power  which  will  be  required  to  drag  a  heavy 
roik  wcifihing  10  tons  at  the  rate  of  10  miles  an  hour  on  a  level  road,  the  co- 
pffirii'iit  of  friction  being  0.8.  What  will  be  the  speed  up  a  gradient  of  1  in  50, 
thr  same  power  being  exerted?  Ans.  477!  s;   Sfi  miles  per  hour. 

192.  Two  horses  draw  a  load  of  4000  lbs.  up  an  incline  of  1  in  25  and  1000 
ft.  iniig.     Determine  the  work  done.  Am.  160,000  ft.-lbs. 

193.  .\t  what  speed  do  the  horses  walk  if  each  horse  does  16,000  ft.-lbs.  of 
wnrk  [ler  minute?  Ans.  2i'r  miles  per  hour. 

194.  It  is  said  that  a  horse  can  do  about  13,200,000  ft.-lbs.  of  work  in  a 
day  of  >i  hours,  walking  at  the  rate  of  2 J  miles  per  hour.  What  pull  (in  pounds) 
fdul.l  .such  a  horse  exert  continuously  during  the  working  day?  How  many 
sii(  ii  liorses  would  be  required  to  do  as  much  work  as  an  engine  of  10  H.P., 
w(nkm<:day  and  night?  Ans.  125  lbs.;  36. 

195.  F"ind  the  shortest  distance  in  which  a  train  going  at  30  m/h  can 
Iw  l>roui;ht  up  by  continuous  brakes  pressing  on  the  wheels  with  three  fourths 
f>f  till'  weight  of  the  train.    Take  .16  as  the  coefficient  of  friction. 

Ans.  About  252  ft. 

196.  A  train  of  60  tons  travels  20  miles  in  1  hour  with  nine  intermediate 
st(,pp  ,.r,.s,  each  of  2  minutes,  at  intervals  of  2  miles.  The  resistance  of  the 
rn;ii|  is  10  lbs.  per  ton, and  the  brake-power  of  the  engine  and  brake-van,  half 
til  •  weiffht  of  the  train,  is  inic  sixtli  of  its  weight.  Find  the  pull  of  th  ■  (Migine 
:i!  I  Its  horse-power  when  running  at  full  speed.    Ans.  2184  lbs.;  332.8  H.P. 


Ji!..'l. 

iSii 

if'S 

1 

I     ■? 

*  n 

,'    -J  ■'; 


si 


107.  An  express  train  reduced  speed  from  60  to  20  miles  an  hour  in  SOO 
yds.,  the  distance  between  the  distant  and  home  signals.     How  nmch  farther 


208 


TIIEOHY  OF  STRUCTUlfES. 


out  should  the  distant  signal  be  placed,  or  how  much  should  the  brake-power 
be  increased?  Ans.  900  yds.;  12J%. 

198.  A  train  of  100  tons  fitted  with  continuous  brakes  is  to  be  run  on  a 
level  line  between  stations  one  third  of  a  mile  apart  at  an  averape  speed  of 
12  m/h.  incliidinn  two  thirds  (  f  a  minute  stop  at  each  station.  Prove  that 
the  wei:iht  on  the  driving-wheels  must  exceed  22i  tons,  with  an  adhesion  of 
one  sixth. 

Trove  that  this  line  can  he  worked  principally  by  gravity  if  the  road  i.s 
curved  down  iM'tween  the  stations  to  a  radius  of  11^740  ft.,  implying  a  dip 
of  ;?:5  ft.  between  the  stations,  a  gradient  at  a  station  of  1  in  13,  and  a  maxi- 
mum running  velocity  of  31  m/h. 

199.  Show  that  the  time  lost  by  a  train  with  continuous  brakes, over- 
shooting a  station  a  ft.  and  backing  in,  is (>/«+!+ 1)^  —    seconds  at  least 

\."9 

the  coefficient  of  adhesion  being  n  and  the  weight  en  the  driving-wheels  -  of 

n 

the  wcij;ht  of  the  train. 

200.  (a)  A  train  weighing  If.O  tons  (of  2240  lbs.)  travels  at  30  miles  an 
hour  n,-^;iinst  a  resistance  of  H)  Ihs.  jht  ton.     What  horsc-|X)wer  is  exerted? 

('")   With  the  same  horse-power  what  will  Ix-  the  speed  up  a  gradient  of 

1  in   !(»)? 

I'  1  If  the  steam  is  shut  ofT,  how  far  will  the  train  run  before  stopping  (1) 
0:1  til  ■  in-line;    (2)  on  the  level? 

('/I  If  th"  ilriiw-har  suddenly  breaks,  m  what  distance  would  the  carriages 
ri(M)  tons  in  weight)  he  sto()|>ed  if  the  brakes  are  applied  immediately  the 
fnK  ture  occurs,  the  wei^dit  of  the  Ijrake-van  Ix'ing  20  tons  and  the  coefficient 
of  friction  .2? 

(1  If  the  eiigino  (weight  =  f •>  t*,  -  c.)ntin!i(-(!  la  exert  the  same  power 
after  the  fracture,  what  would  be  its  ultimate  s|.eed? 

('*  What  resistance  would  l)e  reciuired  to  stop  the  whole  train,  after  steam 
is  siiiit  olf,  in  KKXJ  yds.  on  the  level? 

AH>i.  (u)  J2S;  (h)  '.)■-  miles  per  hour;  (r)  (1)  199.2  ft.,  (2)  6776  ft.;  ((/) 
(iSO.8  ft.  on  the  level,  .■,2.0  ft.  on  the  incline;  (c)  80  miles  an 
hiiur  on  the  level,  24.0  miles  on  the  inchne;  (/)  22.58  lbs.  jier 
ton. 

201.  Prove  that  a  train  going  00  miles  an  hour  can  Ix;  brought  to  rest  in 
al)out  :!i:i  yils.  hy  the  brakes,  suiiposin::  tlieni  to  jm-ss  on  the  wheels  with 
twc,  thirds  ,,f  the  weislit  of  tile  train,  taking  a  coefficient  of  friction  O.IS  in 
adiiitiiwi  to  n  passive  resistance  of  20  li,s.  a  ton. 

Prove  tint  the  mean  uniform  pull  to  Ik;  exerted  by  an  engine  to  take  this 
train,  wei;;iiing  l(K)  tons,  on  the  run  from  one  station  to  a  stop  at  the  next, 

2  miles  off,  in  4  minutes,  is  aljoui  2.1,')  tons. 

202.  A  train  of  120  tons  is  to  Ix;  taken  from  one  station  to  the  next,  a 


EXAMPLES. 


209 


milr  off,  up  an  incline  of  1  in  SO,  in  4  minutes  without  using  the  brakes.  Prove 
that  with  no  road  resistance  the  enpne  must  exert  a  pull,  until  steam  is  turned 
oil.  of  about  6203  lbs.,  and  the  weight  on  the  drivers  must  be  37,218  lbs. 
=  1(1.6  tons,  with  an  adhesion  of  one  sixth. 

203.  With  a  coefficient  of  adhesion  /t  a  motor  car  actuated  and  braked 
mi  the  hind  axle  cnn  get  up  a  speed  v  f/s  in  x  ft.,  or  be  brought  to  rest  again 
in  V  ft.  given  by  g:^ax=-v'[a- ith)  and  giia\j-'V^(a  + nh),  a  being  the  distance 
Iriuccii  the  axles  and  h  the  height  from  the  ground  of  the  C.  G.  midway 
lift  ween  the  wheels. 

204.  Determine  in  tons  the  greatest  train  ai.  engine  capable  of  exerting 
a  uniform  pull  of  3  tons  can  take  on  the  level,  from  one  station  to  the  next,  a 
milt'  off,  in  4  minutes,  supposing  the  resistance  of  the  road  estimated  at  20 
ilis.  ii  ton,  and  the  brake  power  at  400  lbs.  a  ton  in  addition. 

Ans.  202  tons. 

205.  Supposing  that  1  in  m  is  the  steepest  incline  a  train  can  crawl  up 
witli  uniform  velocity,  and  1  in  n  is  the  steepest  incline  on  which  the  brakes 
ran  hold  the  train,  prove  that  the  quickest  run  up  an  incHne  of  1  in  p,  from 
0111'  station  to  stop  at  the  next,  a  distance  of  a  ft.,  which  can  be  made  is 

;    lajm  +n)p' 
\,7i/)-m)(p-:     ) 

must  be  carrifd  by  the  driving-wheels  of  the  engine  with  a  coefficient  of  ad- 
hesion /(,  assuming  the  resistance  on  the  level  to  be  equivalent  to  an  incline 
of  1  in  q. 

206.  If  the  end  of  a  railway  wagon  exposes  a  surface  of  6X4  ft.  to  the 
"iiiil,  what  is  the  greatest  gradient  up  which  a  20  lb.  to  the  square  foot  gale 
will  drive  it?    Take  the  weight  at  10  tons,  the  friction  10  lbs.  per  ton. 

Ans.  1  in  59. 

207.  A  locomotive  and  tender  weigh  70  tons,  of  which  26  tons  are  carried, 
l\v  the  driving-wheels.  Taking  the  adhesion  at  i,  friction  10  lbs.  i)er  ton- 
wlrit  maximum  gradient  can  tl  •"  ••tigine  ascend?  Ans.  1  in  16. 

208.  Determine  the  pull  of  an  engine  and  the  weight  on  the  drivers,  in 
uh  rh  it=20,  1  =24,  D=60,  for  a  mean  effective  pressure  p  =  100  lbs,  on  the 
sipiaro  inch,  taking  an  adhesion  of  one  sixth. 


seconds.     Also  determine  the  proportion  of  the  weight  which 


I 


i 


•if 


i  1' 


s 


2og.  The  weight  upon  the  driving-wheels  (D  in.  in  diameter)  of  a  loco- 
iiMiivc  is  IF  tons;  the  adhesion  =  one  fifth;  the  cylinders  have  a  diameter  of 
'/  111.  and  a  stroke  of  I  in.  Find  the  steam-pressure  required  to  skid  the  wheels, 
I  Ml  lieing  2000  lbs  .         ,„„  WD  , 


Ans 


400  '-^  lbs.  |)er  square  inch. 


210.  In  an  express  engine  the  driving-wheels  are  8  ft.  in  diameter,  and 
the  load  on  them  is  1.5  tons;  the  cylinders  are  18"X28".  Find  the  pressure 
'  f  -'iiin  which  will  skid  the  wheels  with  an  adhesion  of  one  sixth.  Determine 
ilu'  ratio  of  the  velocity  of  the  engine  to  the  velocity  of  the  piston  at  any 
|iiiiiit  of  the  stroke.  Ans.  .W  Ibs./sq.  in.,  a  ton  being  2000  lbs. 


<     'I 


.t^I 


210 


THEORY  OF  STRUCTURES. 


211.  A  loromotive  capable  of  exerting  a  uniform  pull  of  2  tons,  with  a 
24-it..  stroke,  20-in.  cylinder,  and  60-in.  drivinjr-wheels,  hauls  a  train  l)et\vcen 
two  stations  ;}  miles  apart.  The  pross  \veip:ht  of  the  train  and  locomotive  = 
200  tons;  the  road  resi.stance  12  li)s.  ;  cr  ton  (of  20(X)  lbs.);  the  brakes, 
when  .ipplied,  press  with  two  thirds  of  the  wi'i<:ht  on  the  wheels  of  the  enfjine 
and  iirake-vaii,  viz.,  00  tons,  the  coefficient  of  friction  being  .18.  Find  («) 
the  least  time  In'twet'ii  th<'  statio'is;  (/))  the  distance  in  which  the  train  is 
brought  to  rest:  (<)  the  maximum  speed  attained;  (d)  the  pressure  of  the 
steam;    to  the  weight  ui>on  the  driving-wheels. 

Alts,  (a)  .")i;j.,S  seconds;    (//)  '.WO  ft.;    (r)  42  miles  per  hour;    (rf)  2.'>  lbs. 
IKT  square  inch.;    (c)  111  tons. 

212.  If  the  siwed  in  the  last  question  is  limited  to  30  miles  an  hour,  find 
(a)  the  time  between  the  stations;  (/i)  the  distance  in  which  the  train  is  brought 
to  rest ;   (r)  the  disttnice  traversed  at  30  miles  an  hour. 

Alls.  (ri).-.4;{i  seconds:  (/))  50-ii  ft. ;  (O  7773Ht. 

213.  If  the  steam-pressure  in  tlie  above  locomotivi  is  increased  to  .")0  lbs. 
per  square  inch,  find  (u)  the  weight  of  the  heaviest  train  which  can  b"  hauled 
in'twceu  the  stations  in  10  minutes,  the  road-resistance  being  20  lbs.  per  ton 
(of  2000  lbs.),  and  the  braking-power  being  sufficient  to  bring  the  train  to 
rest  in  a  distance  of  720  ft. 

Also  find  (6)  the  braking-power;  (r)  the  weight  thrown  upon  the  drivers, 
the  coefficient  of  friction  In'ing  J;   (</)  tlie  maximum  s(K'ed  attained. 

.\i,s.  in)  3IOi  tons;    (/>)   l.').t)  tons;    (r)  24  tor>s;    (d)  30  miles  per  hour. 

214.  Two  trains,  each  with  a  brake-power  of  I'.H)  lbs.  jwr  ton  (of  2000  lbs.), 

run  U'tween  Montreal  and  Toronto,  a  distance  of  333J  miles,  agi  inst  an  average 

resistance  of  10  lbs.  |>er  ton.     One  train  runs  through,  and  the  other  stops  at 

O.V 
A"  intermediate  stations.    Show  th.'t  the  .saving  of  fi;.'l  in  the  former  is 

jxr  cent ;   the  speed  is  not  to  exceed  M  muos  f)er  hour. 

215.  Prove  that  the  horse-power  of  an  engine  drawinj?  a  train  of  120  tons 
up  an  incline  of  1  in  224  at  :«)  miles  an  hour  is  3.3(1,  taking  the  resistance  of 
the  road  on  the  level  at  this  speed  at  2.')  lbs.  a  ton. 

Deternnne  also  the  horse-i)ower  of  an  engine  drawing  a  train  of  2tK)  tons 
up  an  incline  of  1  in  140  at  30  miles  iii  hour,  resistance  of  road  IS  lbs.  a  ton; 
or  of  the  engine  drawing  the  train  down  an  incUne  of  I  in  140  at  .'•0  miles  an 
hour,  resistance  at  this  s|M'cd  ,')0  lbs.  a  ton.  Aiis.  OOtiJ  HP.;  r)44  l\V. 

Dcl.Tiniiic  the  pull  of  the  <'npne,  ami  suppo.sing  it  constant,  find  how 
far  the  train  would  go  uji  an  incline  of  1  in  UK),  Ix'fore  the  velocity  dropjied 
from  .'lO  to  30  miles  an  hour,  taking  an  average  uniform  resistance  of  the  road 
of  20  lbs.  a  ton.  •I"-''-  -ii  <""■•*;  ^'^-*''  f- 


216.  A  locotnotive  exerting  a  uniform  pull  of  4  tons  hauls  a  train  of  2fKl 
tons  u[>  ati  mduie  of  I  in  2(H),  iM'lwceu  two  sttitions  2  miles  aj.art,  the  greatest 
allowable  s|M'cd  being  M  miles  an  hour.  If  t  li"  ro.id-rcsislaiice  is  10  lbs.  per 
t,,!i  |.,f  jiMM!  I!h,.>  H'ld  if  the  Iirukes  'Te  capab!''  of  evert inti:  ii  pressure  of  l(Ml 
tons,  the  adhesion  Uing  I'ln   fijlh,  find  lui  the  time  U'tween  the  stations;  \i) 


EXAMPLES. 


21t 


thi'  (iistanre  in  which  the  train  is  brought  to  rest;   (c)  the  distance  traversed 
at  ".(I  miles. 

Also,  if  the  speed  is  not  limited  to  30  miles,  find  (d)  the  least  time  in  which 
thi'  distance  can  lx>  accomplishfd ;  (<•)  the  maximum  speed  attained;  (/)  th& 
ilisi:iniv  in  which  the  train  is  hroupht  to  rest. 

A>h<.  (a)  5}  minutes;  (b)  27")  ft.;  (c)  7260  ft.;  (d)  4.47  minutes;  (,■)  -).3.g 
miles  per  hour;   (/)  S.SO  ft. 

217.  With  the  same  lirake-pnwer,  adhesion,  and  road-resistance,  fin<l  the 
wiiir!;'  r  the  heaviest  train  which  the  hMomotive  in  the  precedinj;  question, 
\irtMi!;  the  uniform  pull  of  4  tons,  can  haul  between  the  two  stations  in  & 
i'l""**.  Ans.  2y.S  tons. 

:tS.  If  the  locomotive  has  fiO-in.  drivers  and  24"X20"  diameter  cylinders 
f..  I  he  wcjirht  required  upon  the  drivers  when  the  steam-pressure  is  .")0  lbs 
per  sriuare  inch.  Ans.  20  tons. 

Jig.  Prove  that  the  loss  of  time  in  going  from  A  to  C,  two  points  on  a 
railway  at  the  same  level  8  miles  apart,  due  to  an  incline  of  1  in  100  from  A 
up  to  /{,  and  an  incline  of  1  in  ;}00  from  li  down  to  (',  instead  of  going  on  a 
li  vcl  line  from  A  to  C  at  a  uniform  velocity  of  4.')  miles  an  hour,  is  2  minutes 
■JO  ~.M  .)nds,  equivalent  in  time  to  a  detour  on  the  level  of  1  mile  (iO  chains. 

It  is  supposed  that,  with  full  steam  on,  the  velocity  drops  from  4.">  to  15 
iiiil.s  .III  hour  at  the  summit  B,  and  that  in  descending  the  incline  full  steam 
is  kipt  on  till  the  velocity  is  again  4."),  after  which  the  velocity  is  kept  uniform 
liv  p  irtlv  shutting  off  steam ;  and  prove  that  this  happens  at  a  point  Q,  distant 
fnitn  li  about  I  mile  S92  yards. 

I'mvc  that  the  train  would  l)e  "stalled  on  the  grade"  if  the  incline  fromi 
.1  III  li  was  a  quarter  of  a  mile  longer. 


120.  Prove  that  if  the  weight  of  the  train  is  200  tons,  and  the  resistance 
of  thr  road  is  14  lbs.  a  ton,  the  pull  of  the  engine  from  /I  to  (^  is  2v\  tons, 
aihl  from  Q  to  C  is  T,  ton;  and  determitie  whether  there  is  any  extra  exjiendi- 
tun  of  work  due  to  the  inclines  on  this  and  the  return  journey. 

iM.rniirie  the  reijuisite  weight  on  the  driving-wheels,  taking  an  adhesion 
"f  '""■  ^i^'h.  Alts,  i,  ton;  n..;  1J|  tons. 

221.  The  distance  lietween  two  stations  is  10.")  miles,  and  there  are  27 
imriiiiidi.ite  .stations.  The  average  resistance  of  a  parliamentary  train  stop- 
piii'  I  ,ill  stations  is  taken  us  S  lbs.  a  ton,  while  the  resistance  to  an  evfiress 
tri  •,  ■\lii(||  runs  through  without  stopi.ing  is  taken  at  10  lbs.  a  ton,  the  brake- 
1"     I   I'l  i"ich  viifr  l)eing  taken  at  SO  lbs.  a  ton  additional. 

>  ippoving  the  sp«H'd  to  l)e  kept  constant  by  reducing  steam  when  if  has 
rri  I1..I  ;j()  miles  an  hour,  find  which  train  is  most  expensive  in  fuel,  and  by 
li   i  tiiuih  |)cr  coiit.  Anif.  Parliamentjiry  exceeds  that  of  express  by  0}%. 

W   rk  out  the  same  problem  with  the  resistances  and  full  sjieed  donbliHl. 

-J/M.  ;«',. 

221.  Uctermine  the  loss  of  time  and  the  extra  e\|¥'nditun'  of  work— if 
■ii.v-  .iuc  lo  rro8.sing  a  jmss  by  a  railway  having  an  incline  of  1  it,  m  uq,  nn<l 
1  111  /  down,  instead  of  going  in  a  k-vel  f uiniel  of  length  /  ft   through  the  pjisa; 


>12 


TUEnRY  OF  STRUCTURES. 


supposing  that  V  f/s  is  the  maxiinutn  spoed  allowed  on  the  line,  and  that 
the  volocity  of  the  train  drops  from  V  to  v  f/s  at  the  summit  of  the  pass; 
dotprniine  the  pull  of  the  enpine  P  in  tons  for  a  train  of  IV'  tons  weight,  taking 
the  resislance  of  the  road  at  r  lb   per  ton. 

Determine  the  lenpth  of  detour  of  equal  time. 

223.  The  rrank  of  a  horizontal  engine  is  3  ft.  6  ins.  and  the  connect- 
inp-md  "•  ft.  long.  At  half-stroke  the  pressure  in  the  connecting-rod  is  500 
ll)s      What  is  the  corresponding  twisting  moment  on  the  crank-shaft? 

Alls.  1716i   ft.-lbs. 

224.  A  piston  and  rod  and  cross-heail  weigh  330  lbs.  At  a  certain  instant 
whfii  the  resultant  total  forces  due  to  sti  im-pressure  is  3  tons,  the  piston  has 
an  acceleration  of  370  ft.  jier  second  in  the  same  direction.  What  is  the  actual 
force  acting  at  the  cros.s-head?        Am.  290*1  lbs.,  the  ton  Iwing  2240  lbs. 

225.  If  the  connecting-rod  is  5  ft.  long  and  the  crank  is  1  ft.  in  an  engine 
making  -HH)  revolutions  per  minute,  what  are  the  accelerations  of  the  piston 
when  it  is  farthest  from  and  nearest  to  the  crank?  The  piston  and  rod  and 
cros!i-h(;ul  weigh  330  lbs. ;  the  area  of  piston  =  120  sq.  ins. ;  at  the  beginning  of 
either  the  in-  or  the  out-stroke  the  pressure  is  SO  lbs,  per  square  inch  on  one  side 
in  excess  of  what  it  is  on  the  other.  Find  the  total  forces  on  the  cross-head  in 
these  two  cases. 

226.  A  revolving  weight  W  is  at  a  distance  r  from  the  axis  of  rotation 
and  is  to  Ik-  balanced  by  two  weights  each  r,  from  the  axis,  the  one  being  0 
on  the  right  and  the  other  b  on  the  left  of  the  weight  to  be  balanced.    Find, 

the  weights.  .^,,,  n--^;  n-  \. 

riu  +  h  n  a  +  6 
227  There  is  a  balance-weight  of  ISO  lbs.  at  a  distance  of  3.4  ft.  from  the 
rentn  ,  and  another  weight  of  l.">0  His.  at  a  .listance  of  2..">0  ft.  from  the  centre, 
in  a  direitinn  at  right  angles  to  th«>  first,  both  on  the  same  driving  wheel  of 
a  locomotive.  l-"ind  the  amount  and  position  rf  any  single  weight  which 
would  have  the  same  l)alancing  effect  :is  these  two. 

.1/i.v  Wt.  •  cniTik  r;idius™722}  ft.-lbs. 
228.  Kach  jiistiiM  of  !i  locomotive  weighs  IMH)  ilis  Wh;it  balance-weights 
will  l)alanee  o^ie  piston  so  that  the  max.  alternating  c(iu|)le  and  horiwuital 
force  niiiv  !«'  nil?  Stroke  =24  in.;  distance  C.  to  C.  of  cylinders -42  ins.; 
railiid  ilistiirue  of  i)alan(c  weights  -ij'.t  nis. ;  distance  U'tween  centres  of  gravity 
of  biiLinee-weiiihts  -  .")7  itis.  .\iin.   12.1.")   Ii>s.;    SO.l.')  lbs. 

2 JO.  ihe  piston  and  rod  and  cross-heaii  weigh  .'{:«)  lbs.;  the  connectinft- 
rod  3iK)  ll)s. ;  the  12-in.  crank  is  eqiiivaleiit  to  270  lbs.  at  crank-pin;  the  C.  "f 
Ci.  of  the  >  imnecting-rod  is  0.,3S  of  its  lingth  from  the  crank-pin.  If  the  centre 
I  ties  of  the  cylinders  art>  27  ins.  apart,  and  the  middles  of  the  wheels  arc  .')9 
ins.  iipart,  negld-ting  valve-motions,  etc.,  calculate  the  balance-weights  Ht>d 
their  [idsitions.  Place  their  centres  of  mass  3  ft.  from  the  centres  of  the  wheels 
Is  the  locomotive  in  perfect  balance?    What  is  the  nature  of  the  balance." 

Anx  Wm  lbs.  and  <,^-24*3(>',  8.ssuming  §  of  total  reeip.  wis  at  crunk 
nidius 

230.  I  iiid  Ihe  acicleralioii-presmire  at  each  end  of  the  stroke  of  a  wrtical 


BXA.VPLES 


213 


iiiv.rtpd  high-speed  steam-engine  when  running  at  oOO  revolutions  per  minute 
stroke  !»  ins.,  weight  of  reciprocating  parts  110  lbs.,  diameter  of  cyhnder 
s  ins.,  length  of  connecting-rod  1.5  ft.  (3=32.2). 

Am.  54.8  Ibs./sq.  in.  at  bottom ;  85.5  Ibs./sq.  in.  at  top. 

231.  In  a  horizontal  marine  engine  with  two  cranks  at  right  angles  distant 
V  fi.  from  one  another,  weight  of  reciprocating  parts  attached  to  each  crank 
,  111  tons,  revolutions  75  per  minute,  stroke  4  ft.,  find  the  alternating  force 
',;A  ,,mple  due  to  inertia.  Ans.  54.2  tons;    218.20  ft.-tons. 

-32.  An  inside-<vlinder  locomotive  is  running  at  50  miles  an  hour;  the 
arivin^-wheels  are  6  ft.  in  diameter;  the  distance  between  the  centre  lines 
„f  ihc  cylinders  is  30  in.,  the  stroke  24  in.,  the  weight  of  one  piston  and  rod 
3U()  ll'-^  and  the  horizontal  distance  between  the  balance-weights  4?  ft.;  the 
.iKuiHlcr  of  the  weight-circle  is  4i  ft.  Find  the  alternating  force  and  couple, 
an<l  also  the  magnitude  and  position  of  suitable  balance -weights. 

Ans.  78711b8.;  9839ft.-lbs.;   100.5  lbs.;  27?°. 

233.  The  weight  of  the  reciprocating  parts  of  an  engine  is  3  lbs.  per  square 
inch  of  piston  area;  the  stroke  is  36  in. ;  the  number  of  revolutions  per  minute 
is  4');  the  back-pressure  is  2  lbs.  per  square  inch;  the  ubst)lute,  initial  steam- 
pr.v.siire  is  60  lbs.  per  square  inch ;  the  ratio  of  expan.sion  is  3.  Find  the 
prt-s^ure  necessary  to  start  the  pi.ston,  and  also  the  effective  pressure  at  each 
J  of  the  stroke.  Ans.  3.1  Ibs./sq.  in.;  57.2,  20,  20.1  Ibs./sq.  in. 

234.  .\n  engine  with  a  24-in.  cylinder  and  a  connecting-rod  =  six  cranks - 
fi  ft,,  makes  60  revolutions  per  minute.  Show  tht  ^c  pressure  required  to 
start  and  stop  the  engine  at  the  dead-points  =  Ji  of  t;      veight  of  reciprocating 

parts. 

235.  Find  the  ratio  of  thrust  at  cross-head  to  tangential  effort  on  crank- 
pii,  when  the  cra>ik  is  45°  from  the  line  of  stroke,  tb-  connecting-rod  being 
■  four  rninks.  .ln«.  6  to  5. 

236.  Draw  the  linear  diagram  of  erank-effort  in  the  ea«e  of  amgle  crank, 
th.  ...tmecting-nxl  being -four  cranks.  Assume  the  resistance  uniform  and 
a  n.iistant  pressure  of  <)000  lbs.  on  the  piston,  the  stroke  being  4  ft.  and  the 
nmulN  r  of  revt.lutions  per  minute  55.  Also  find  the  fluctuation  of  energy  in 
fi.  t-|Hiiitid?<  for  one  revolution. 

237.  An  engine  with  a  connecting-rod -six  rrankfl-6  ft.  receives  steam 
at  Til  11>«.  precsun-  iht  square  inch,  and  cuts  off  at  me-quartcr  stroke.  Fmd 
th.  ,  rank-effnrt  when  the  piston  has  travelled  our  third  of  its  forward  stroke. 
ji  itn.tcr  of  Diston  -2  ft.    Also  find  the  position  of  the  piston  where  its  velocity 

An»   23.231  lbs.;  9-80°  53'. 
1^   1  maximum.  "'**•  *'^<'""^  '""  < 

238.  Data:  Stroke -3  ft.;  number  of  revolutions  per  minute -60;  cut-off 
,1  lie-half  stroke;  initial  pressure  -56  lbs.  per  square  inch  absolute;  diameter 
.f  i.-tcn  -10  in.;  weight  of  n-ciprocating  parts-.VK)  llw.;  bac k -pressure - 
i  li...  iK-r  «,uure  inch  absolute.  Find  the  effertive  pr«anre  at  each  fourth 
..f  ih,  stroke,  taking  account  of  the  inertia  of  the  piston.  Abo  End  the  pr«». 
iun-  ((luivalent  to  inertia  at  eommenecment  of  stroke. 

An*.  52,  58.5,  44.5,  41.4  lb8./s(|.  in.;  12.95  ll«i./sq.  in. 


214 


THEORY  OF  STRUCTURES. 


239-  A  pair  of  2oO-H.P.  engines,  with  cranks  at  00°,  and  working  against 
a  uniform  resistance  and  under  a  uniform  steam-jiressure,  arc  runnii  .  at  60 
revolutions  per  minute.  Assuming  an  indefinitely  long  eonneeting-rod,  find 
the   maximum  and    minimum   moment-  rank-effort,  the   liuctuation  of 

energy,  and  the  cwfReient  of  energy. 

Arts.  28,200  ft.-ll).v  .   17,.").'0  ft.-ll)s.;  Sr^K)  ft.-lbs.;  .04. 

240.  An  iiiside-eylinder  locomotive  runs  2.5  miles  i>er  hour;  its  drivers 
are  tJO  in.  in  dianuser;  ilie  stroke  is  24  ins.;  the  distance  betwwn  the  centre 
lines  of  the  cylinders  =;50  in.;  weight  of  reciprocating  paris-oOO  lbs.;  hori- 
zontal distance  U'twceii  balance-weights  =  .V,»  in.;    diameter  of  weighi-<'ircle 

=  12  in       Find  the  alternating  force,  alternating  couple,  and  the  magnitude 
iiiiil  i>usition  of  suitable  balance-weights. 

.l/w.  22f).8  lbs.;   41I3.Sft.-lbs.;    ?S  =  26°. 

241.  A  box  of  sand  weighing  I  ton  is  suspended  by  parallel  cords,  each 
8  ft.  long,  and  is  found  to  recoil  through  an  arc  whose  chord  is  0  ft.  when  struck 
horizontally  through  the  ("C.by  a  l)all  weighing  .0!  ton,  and  the  penetration 
of  the  ball  is  2  ft.  Find  the  resistance,  the  time  of  ixMietration,  and  the  dis- 
tance through  which  the  box  moves  {ballistic  jundulum). 

-Ihs.  \\:\\  tons;     ,', ,  sec:  .237fi  in. 
Also,  what  IS  the  striking  velocity  of  the  ball'  Ans.   1212  f/s. 

242.  If  a  thill  disc  weighing  10  lbs.  and  of  10  ins.  diameter  rotate  at  1000 
revolutions  [ler  minute  about  an  axis  through  its  centre,  and  if,  instead  of 
iH'ing  iierpendiculfir  to  the  shaft,  it  is  out  of  truth  by  ,,',„  of  its  radius  find 
the  couple  ou  the  shaft  in  inch-pounds.  Ans.  12.8  in.-lbs. 


CHAPTER  IV. 
PTRIuSS;   STRAIN;   ELASTICITY;  OSCILLATION;  THIN  CYLINDER. 


I.  Stress  and  Strain;  Resilience. — The  science  relating  to  the 
,<tnn>ith  of  niatpfials  is  partly  theoretical,  partly  practical.  Its 
primary  object  is  to  investigate  the  forces  developed  within  a  body, 
mill  to  determine  the  most  economical  dimensions  and  form,  con- 
-i-tmt  with  stability,  of  that  body.  Certain  hypotheses  have  to 
1»  made,  but  they  are  of  such  a  nature  as  al'vays  to  be  in  accord 
with  the  results  of  direct  observation. 

11  ic  materials  in  ordinary  use  for  structural  purposes  may  be 
tcrnuMl,  generally,  solid  bodies,  i.e.,  bodies  which  offer  an  appreciable 
rc^^istance  to  a  change  of  form. 

A  body  acted  upon  by  external  forces  is  said  to  be  strained  or 
ik'jormed,  and  the  straining  or  deformation  intluces  stress  amongst 
thi    |iarticles  of  the  body. 

The  state  of  strain  is  simple  when  the  stress  acts  in  one  direc- 
tion (iiily.  anil  the  strain  itself  is  measured  by  the  ratio  of  the  defor- 
mation to  the  original  length. 

Tlic  ."^tate  of  strain  is  comimund  when  two  (or  more)  stresses  act 
siiiiultaiicously  in  different  directions. 

A  strained  body  tends  to  assunie  its  natural  state  when  the  strain- 
iim  Inrccs  are  removed;  this  tendency  is  called  its  elcu^tirity.  A 
tl  ihihikIi  knowledge  of  the  laws  of  elasticity,  i.e.,  of  the  laws  which 
riimiici  ili(.  external  forces  with  the  internal  stresses,  is  absolutely 
Urn  say  for  the  projKT  comprehensiofi  of  the  strength  of  mate- 
ri:il-  This  pro|X'rty  of  elasticity  is  not  |»oss<>ssed  to  the  same  degree 
liy  ill  JMxlit's.  It  may  Ix^  almost  jM-rfecf  or  almost  zero,  but  in 
till'  majority  of  cases  it  has  a  mean  value.  Hence  it  naturally  fol- 
lows that  solid  bodies  may  be  classified  Iwtween  two  extreme,  though 
idoal,  states,  viz.,  a  perfectly  elastic  state  and  a  perfectly  soft  state* 

215 


216 


THKORY  OF  STRICTURES. 


Perfectly  elastic  bodies  which  have  been  strained  resume  their 
original  forms  exactly  when  the  straining  forces  are  removed.  Per- 
fectly soft  bodies  are  wholly  devoitl  of  elasticity  and  offer  no  resist- 
ance to  a  change  of  form. 

Bodies  capable  of  undergoing  an  indefinitely  large  deforma- 
tion under  stress  are  said  lo  he  plastic. 

Every  botly  may  be  subjected  to  five  distinct  kinds  of  stresses, 
viz. : 

(a)  A  longitudinal  jmll,  or  tension. 

(6)  A  longitudinal  thrust,  or  compression. 

(c)  A  shear,  or  tangential  stress,  which  may  be  defined  as  a 
stress  tending  to  make  one  surface  slide  over  another  with  which 
it  is  in  contact. 

(d)  A  transverse  stress. 

(e)  A  twist  or  torsion. 

Under  any  one  of  these  stresses  a  body  may  suffer  either  an 
elastic  deformation  of  a  tenijwrary  character  or  a  plastic  deforma- 
tion of  a  permanent  character. 

Let  a  wire  or  a  bar,  of  length  L  and  uniform  sectional  area  A, 
be  fixeil  at  the  upixT  end  and  hang  vertically.  A  load  P  uniformly 
distributed  over  the  lower  end  will  stretch  the  bar  by  an  amount  I 
which  can  V)e  readily  measured  by  an  extensometer  or  by  the  aid 
of  a  telescope  ami  a  vertical  scale.  If  this  bar  is  now  replaced  by 
one  of  the  .same  length  but  of  twice  the  sectional  area,  it  is  found 
that  it  requires  twice  the  load  to  stretch  it  by  the  same  amount. 
Also,  the  load  upon  a  column  or  strut  producing  a  certain  .shorten- 
ing 01  th?  length  is  twice  as  great  as  that  recjuired  to  produce  the 
same  amount  of  shortening  in  a  strut  of  the  same  length  but  one 
half  the  sectional  area. 

The  ratio  y,  i.e.,  the  chamje  nf  lemjth    (or  deformation)   per  mil 

oj  loKjth,  is  called  the  strain. 

Again,  the  load  acts  at  every  cros.s-sect ion  in  a  similar  manner, 
and  it  is  wi-ll  to  rememlK-r  that,  according  to  St.  Venant,  the  actual 
distril)Uti()n  of  loail  on  a  small  area  is  not  of  much  im|K)rtance. 
WliatiVi-r  tilt-  iiistribution  may  \n\  the  "train  at  a  jxiint  not  vpry 
near  is  the  same.  To  eliminate  any  difference  which  might  exist 
at  a  point  where  the  bar  lia.s  a  gn-ater  or  less  sectional  area,  it  is 


STRESS  AND  STRAIN. 


217 


r 


advisable  to  use  the  load  per  unit  of  area,  viz.,  -j,  instead  of  the 

total  load  P. 

The  load  per  unit  of  area  is  called  the  stress. 

Exix^riment  then  shows  that,  within  certain  limits,  the  stress  is 
jinidically  proportional  to  the  strain. 

This  relation  between  stress  and  strain  was  enunciated  by  Hooka 
aiul  is  known  as  Hooke's  Law. 

It  may  be  expressed  by  the  equation 

f-£e, 

where  /-• ^-the  load  per  unit  of  area,  or  stress,  and 


e  =  j  =the  strain. 

The  multiplier  E  ia  a  number  whose  value  depends  upon  the 
character  of  the  material.  It  is  calleil  Young's  coefjicient  or  modu- 
hi.s  o'l  elasticity,  and  tables  at  the  end  of  the  chapter  give  the  value 
of  R  for  different  materials.  It  nmst  not  be  forgotten  that  the 
iiuiro  homogeneous  a  material  is  the  more  accurate  is  the  relation 

The  longitudinal  strain  is  accompanied  by  an  alteration  in  the 
tnmsviTse  tlimensioiis,  the  lateral  unit  strain  being  T-,  where-   la 

a  cdcliiciont  which  usually  varies  from  ^  to  i  for  solid  bodies  and 
i-  ;  pproximately  i  for  the  metals  of  construction.     In  the  case  of 

India-rubber,  if  the  deformation  is  small,  -  is  about  i. 

Generally. the  deforma  ion  may  be  calculated  per  unit  of  original 
l.Mi|;th  without  sensible  error,  but  for  India-rubber  it  is  more  accurate 

to  make  the  calculation  per  unit  of  stretched  length  (  -  j^)- 

The  ratio  of  the  lateral  to  the  longitudinal  strain  is  called  Poisson's 

ratio. 

Kx.  I.  A  cable  1000  ft.  long  stretihes  6  ins.  under  a  |iv«n  load;  what 
is  the  strain? 

The.trsin-^-^2-.0005. 


21S 


THEORY  OF  STRUCTURES. 


Ex.  2.  .4  short  cast-iron  strut  24  ins.  long  and  of  8  sq.  ins.  sectimal  area 
bears  a  load  of  32  ton.s.  If  E=  SOOO  tons/sq.  in.,  how  much  is  the  strut  short- 
ened?   What  is  the  stress f     What  the  strain? 


and  therefore 


The  stress  =  V  =4  tons/sq.  in. 
4  =8000  X the  strain, 

the  strain  =ff 0*0 ff  =.0005. 


The  diminution  of  length  =24 Xthe  strain  =.012  in. 

Ex.  3.  .1  mild-steel  bar  100  ft.  long  and  of  2  sq.  ins.  sectional  area  carries 
a  load  uhirh  deirlojies  in  the  materud  a  stresn  of  Hi,(XX)  Ihs./sq.  in.  If  E^ 
30,000,000 /fc.s. /w/.  in.,  by  how  much  is  the  bar  lengthened?  What  is  the  strain? 
What  is  the  total  load,' 

The  tot.al  load  =  1  (UXK)  X  2  =32(K)0  lbs. 
10000  =30000000  -the  strain. 


Therefore 
and 


the  strain  = 


KMKM) 


1 


30(X)0000      1875' 


the  stretch  =  100  X— |_-  =—  ft.  =.64  in. 


Again,  the  change  of  length  /  is  produced  by  a  load  which  gradually 
increases  from  0  to  P,  and  therefore 

p  j^     IT  » 

the  work  done  in  changing  the  length  =  tt-  /  =  -;7  tt  =  .te,  AL, 

and  this  is  the  energy  stored  up  in  the  strained  bar.  It  is  called  the 
resilience  of  the  bar  if  the  load  is  only  a  little  less  than  that  which 
produces  a  jm-nianent  change  of  length,  i.e..  a  permanent  deformation 
or  set.  Such  a  load  is  the  proof  load,  and  the  stres.s  /  becomes  the 
jrroof  stres.^.  Also,  since  AL  is  the  volume,  the  resilience  per  unit  of 
volume  is  j'^/'lE. 

Thu.s  tlie  resilience  is  the  greatest  amount  of  energy  which  can 
be  stored  by  a  material  without  becoming  i)ermi»nently  deformed. 
If  the  stress  is  variable,  the  resilience  is  necessarily  less,  while  a  blow 
or  a  siiock  may  devclope  in  tlie  material  a  larger  amount  of  strain 
en(  rgy  than  it  can  bear,  so  that  a  local  deformation  (or  seO  and  a 
plastic  yielding  may  occur. 

Exs.  4  and  5.  Find  the  work  done  in  producing  the  changes  of  length  in 
Ex.1.  2  and  3. 

32 
In  Ex.  2,  the  work  in  in.-tons-  ;-x.012-.192. 


STRESS  AND  STRAIN. 
32000 


219 


In  Ex.  3,  the  work  in  in.-lbs. 


X  .64  =  10240, 


or  aS'i'ii 


..           ,   .    .     ,,         (16000)' X2X 100X12     ,-„._ 
the  work  m  mAh..^^       ,x30000000 ^"^^^ 


i;x.  6.  A  bar  of  weight  Wi,  length  L,  and  uniform  sectional  area    ^^^^^ 
.1  I.N  fixed  at  its  upper  end,  hangs  vertically,  and  carries  a  weight  Wi        T 
lit  its  lower  end.    Determine  the  stretch  of  the  bar  and  the  work  of 
ftn  telling. 

Consider  a  slice  of  thickness  dx  at  the  distance  x  from  the 
fi\('(l  end. 


W 

rhe  weight  on  the  slice  =-7-'(/^-x)  +lFj,  and  if  dl  is  the  ex- 


tiiision  of  dx  under  this  weight, 


A  dx' 


Fia.  299. 


Therefore,  l'^^£  {  x^^'"^)  +"''  }  ^^^'iTU^'  ^"'')- 

The  work  done  in  stretching  dx  by  the  amount  dl,  under  a  weight  which 
gradually  increases  from  0  to  -^(L—x)  +Wt, 

and  therefore  the  work  done  in  stretching  the  whole  bar 

-5irrir'"-"^''-r-bf.(^'-'-----)- 

l!x.  7.  A  vertical  steel  bar  of  200  ft.  length  is  fixed  at  it.t  upper  end,  hangs 
vertically,  and  carries  a  load  of  100  tons  at  the  lower  end.  The 
stress  in  the  material  is  not  to  exceed  7  tons/sq.  in.  and  is  to  be 
the  same  at  every  point  of  the  bar.     Determine  the  form  of  the  bar. 

Consider  a  slice  of  thickness  dx  and  sectional  area  -4  at  a 
distance  x  from  the  fixed  end. 

At  x+dx  the  area  is  A-dA.  the  amount  d.i  being  the 
additional  area  required  at  x  to  support  the  weight  of  the  slice 
dx.     Therefore 

-fd.i  —wAdx, 

w  being  the  specific  weight,  and  /  the  stress  developed  in  the 
material. 


220  THEORY  OF  STRUCTURES. 

This  equation  may  be  written 

w,      dA 


Integrating, 


-(L-X)=l0ge-^, 


.4i  bt>ing  the  sectional  area  at  the  lower  end  and  L  being  the  length  of  the  bar. 
Therefore  the  form  of  the  bar  is  given  by 


.1. 


=c' 


^(t-x) 


If  A2  is  the  area  at  the  upper  end,  i.e.,  when  x=0, 


-i^=> 


■  —€• 


.ll 


and 


Such  a  bar  is  called  a  bar  of  uniform  strength. 

100 
In  the  present  case  A,= —  =l4f  sq.  ins. 

49()\200 


Fig.  301. 


or 


,4,  =  100^7,^^;"^,l_W  m^i^^l  05„j5  ^q  i„^ 


100. 
7 


E.\.  8.  A  pier  of  height  L,  with  its  axis  vertical  and 
carn/itig  a  weight  on  the  top,  is  to  have  the  ame  stress 
f  per  unit  of  area  at  every  pirint,  i.e.,  is  to  l>c  of  uni- 
form strength.  Determine  the  form  of  the  pier,  w 
being  its  speeific  weight. 

If  .4  is  the  area  of  the  pier  at  x  from  the  base, 
then  -dA  is  the  diminution  of  area  at  x+dx,  cor- 
res(x>n(liiig  to  the  weight  ;,i  the  slice  dx  by  which 
the  load  upon  .4  is  reduced.    Therefore 


wAdx  -  -jdA, 


w      dA 


Integrating, 


w      ,     A 


Ai  being  the  area  of  the  base  of  the  pier. 
Thu.s  the  form  of  the  pier  i.s  given  by 


A 

A,'" 


K<i 


THEORY  OF  VIBRATIOSS. 


221 


Adx  —  Ai  I     e     '   dx 
Also,  if  A   is  the  area  of  the  top  of  the  pier, 

a  111  1  the  surcharge  =jAt=jAie    '    . 

If  Zy  =30  ,  i.e.,  if  the  pier  is  infinitely  high,  • 

-7'' 
e     '     -0. 

Tlirrefore  At  anu  the  surcharge  are  each  nil,  and  the  volume  of  the 

IlilT  --.It—. 
W 

V.\.  9.  ,1  steel  bar  of  2  sq.  ins.  sectional  area  has  its  ends  fixed  between  two 
i'limnrnble  blocks  when  the  temperature  is  at  40°  F.  What  pressure  will  be  exerted 
iiji'iii  tlic  blocks  if  the  temperature  of  the  bar  is  raised  to  100°  F.,  the  coefficient 
1 .1 )  r>/  linear  dilatation  per  degree  being  .00108^  1 80?    E -=  30,000,000  lb.  /sq.  in. 

If  tht  length  of  the  bar  is  L,  and  if  it  were  allowed  to  lengthen  freely 
miller  an  increasing  temperature,  then,  when  the  temperature  is  f  F., 

the  additional  length  =Lat. 

Hence  the  stress  developed  when  the  extension  is  prevented 

Ij 

an<l,  in  the  present  case,  the  pressure  on  the  blocks 

AniAQ 

-  2  X  30000000  X  "^4^  X  60 


180 


=21600  lbs. 


2.  Theory  of  Vibrations. — Lot  a  particle  of  weight  W  be  acted 
upon  l)y  a  force  which  is  proportional  and  acts  directly  opposite  to 
its  displacement  from  a  fixed  point  0,  Fig.  302.    Then 


W(Px 


T 


T 


■ing  some  constant. 


Fio.  302. 


THEORY  OF  STRUCTURES. 


(L 


Multiplying  each  side  by  l>~~  and  integrating, 


where  -^  =  0,  when  x  =  o  and  the  velocity  of  the  particle  at  any  distance 
X  from  0  is  given  by 

f2  =  |-!(a2-x2). 


The  equation  may  be  written 


Integrating  again, 


V'l 


sm' 


■:-=#«, 


c  being  a  constant  of  integration. 

But  x  =  a,  when  t  =  0,  and  therefore  c  =  -. 

mamams  .       x        jau      tt 

Hc-nce  sm-  =^?^,+| 

or  x  =  acos>J^f, 

and   therefore  the  time  of  a  complete  oscillation  ia 

Ex.  10.  A  spiral  spring,  Fi<,.  30.3,  whose  coefficient  of  stiff- 

i;r  •?<>•?         "?","■  ''. "  '"'"'"'  "■''''  "  "''f''"  "'  '""^  ^^'^  depressed  a  distance 

•     "■        -r  ''<'<"'•  !'••>'  poxitioii  of  ((piilibrium. 
Disregarding  tlic  weight  of  the  spring, 

W,P.T        , 

"T  -Jn  ='he  "restoring"  force  of  the  spring 


=  -cx, 


Jll 


THEORY  OF  VIBRATIONS.  228 

:i:m1  therefore  the  spring  will  oscillate,  the  time  of  complete  oscillation  being 


Kx.  11.  A  straight  spring  AB  is  loaded  in  the     i i 


;m  W/c  irith   a  tmight  W.     It   is  then  deflected  stilt 
i  nil, IT  hi)  an  antount  x. 

I,i't  the  "restoring"  force  F  due  to  this  addi- 
tioiKil  (leflcftion  be  such  that 

FP 


Fio.  304. 


x  =  n 


El' 


I  Ixing  the  length  of  the  beam.    (See  Chap.  VII.) 
Then 


K'!!£=-F"-—x 
nP 


g  dp 

Thus,  the  weight  will  oscillate  about  its  position  of  equilibrium,  and  the 

IWnP 
time  of  a  complete  oscillation— 2n.j — =^. 

Kx.  12.  A  revolving  shaft  ia  supported  at  the  point  B  of  a  cast-iron  standard 
ABD 


--a -»(< o — *iD 

Fig.  3ft5. 

Lot  W  be  the  weight  carried  at  B.    Each  rod  supports  in  the  direction 

W  I 
f  Its  length  a  force  (either  a  tension  or  a  compression)  -—  -r-. 

Let  B  Ix!  displaced  to  n  and  draw  nk  perpendicular  to  AB.    Then  AB  is 
^llu^tened  by  the  length  Bk.     Also, 

Bfc      AC     a 
Bn  'aB  ~l  ' 


Tl'.rrefore 


Bk^-x-'jBn 


.    ■      t        ■    V  iWld^x 

aiiil  tile  rcstonng  force  is  fix  —  —  "TT  wT' 


2J4 


THEORY  OF  STRUCTURES. 


E  Ix'ing  Young's  modulus  of  elasticity  for  the  materiaj  of  the  bars.      Hence 
the  period  of  a  complete  vibration  of  the  rod 

\-2,ihE 

General  Equation  of  Motion. — The  procoding  examples  seem  to 
indicate  that  if  vibrations  are  onee  started  they  will  never  die  out. 
They  are  destroyed,  however,  sooner  or  later  by  other  forces,  some 

of  tiie  nature  of  friction  which  are  constant,  and  others  of  the  nature 

(J  \ 
It)- 


Fig.  .-jog. 


The  complete  equation  of  motion  may  be  writ  ten  in  the  form 

d'x      ,  (Ix 


Put 


j=  —  +  ije' 


Then  the  etjuation  becomes 


7l+f'/-/'=').v-o, 


from  which  y  ^  I)  cos  \'q  -  jr't, 

l>  hfiiijr  a  coetficient  whose  value  i-<  to  Im-  found. 


!Ieiu-e 


/=.--  + 1)0-1*  cos  \  7  -  ji^t, 


and  the  extent  of  the  vibrations  diminishes  as  t  increases,  i.e.,  they 
die  out. 


THEORY  OF   VIBRATIONS. 


225 


Forced  OsriUntion. — The   term   jree  oscillation   is  applied   to   a 

cdinlition  of  vibration  in  which  the  forces  acting  upon     „„„„ 

the  IkmIv  il('])en(l  only  on  the  displacements  of  the  sev- 

( ral    particles   from    the    jKisition  of  eciuiiilmum.      If 

niln  r  forces  which  are  due  to  external  causes  and  arc 

tiiintions  of  the  time  also  act,  the  oscillation  is  called 

;i    forced    oscillation.      For    example,    let    the   spiral 

^juiii;:  he  loaded  with  a  weight  U'  and  then  depressed     '  y  I 

thmu^ii  a  distance  x  l)el<)w  its  neutral  position,  the  re-      Jt.""~— 

-inriiig  force  being  ex.     Let  a  force  q  sin  nt,  which  is  a      ^'°"  **'' 

tuiiction  of  the  time,  also  act  uyion  tiie  weight.     Then  its  equation  of 


notion  l)ecomes 


Integrating, 


fPx 
dP 


■■  -  TTrCX  +  7  sin  nt. 
^.lcos>J^<-^ 


sinrU. 


.1  being  a  constant  whose  value  is  to  be  determined. 

Thus  the  oscillation  is  compounded  of  a  free  and  a  forced  oscilla- 


tidii.  the  iK'rio<ls  being 


7T 

and  — ,  resjiectively. 


3.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End  of  a  Ver- 
j  tical  Elastic  Rod.— An  ehvstic  rod  of  natural  length  L{OA) 

i" '      and  sectional  area  A  is  su:. tended  from  0,  and  carries  a 
I  weight  P  at  its  lower  ead,  which  elongates  the  rod  until 

its  length  is  OB^L  +  l. 

Assume  that  the  mass  of  the  rod  a.s  compared  with  P 
is  sufficiently  small  to  be  disregarded,  then 

/ 


/•  =-  EA 


1/ 


i 


If  the  weight  is  made  to  descend  to  a  point  C,  and 
is  then  left   free  to  return  to  its  state  of  equilibrium,  it 
'■^         nnist  necessarily  de.scrilM'  a  series  of  vertical  oscillations 
*-         alKHit  li  as  centre. 
;,„^  Take  li  as  tlie  origin,  :uid  at  any  time  t  let  the  weight 

be  nt  .U  distant  j  from  li;   also  let  liC^r. 
Two  ca.<w>s  may  Ik>  consi<lered. 

/'VaV,  suppose  the  end  of  the  rod  to  be  gradmUy  forced  down  to 
iiid  then  suddenlv  n-Ieased. 


'     i  I 


220  THEORY  OF  STIiUCTUIiES. 

According  to  the  principle  of  the  conservation  of  energy, 

—  o  I  j7  )  =  the  work  done  between  C  and  M 
g  2\dt/ 


or 

and  hence 


~  L  \2     2/' 
ij2\dtl       I  2^       '  '' 


r,  the  velocity  of  the  weight  at  M 


,-4(^- 


x»)*. 


Now  r  is  zero  when  r=  ±r,  so  that  the  weight  will  rise  above 
B  to  a  point  Ci  where  BCi=c  =  IiC. 
Again,  from  the  last  ccjuation, 


*J 


dx 


n        (f2-x2)»' 

and  integrating  between  the  limits  0  and  x, 


4 


■sm 


-i_ 


/  c' 

and  the  oscillations  are  therefore  isochronous. 

r     IT 
WTienj-f,  '"">>"» 

and  the  time  of  a  complete  oscillation  is 

4 

Xcit,  suppose  the  oscillatory  motion  to  be  caused  by  a  weight  P 
falling  without  friction  from  a  |)oint  I),  and  being  suddenly  checked 
and  held  by  a  catch  at  the  lower  end  of  the  rod. 

Take  thf  same  origin  and  data  as  lK«fore,  and  let  AD-h, 
The  elastic  resistance  of  the  nnl  at  tlie  time  (  is 

/  +  i 


EA 


L   ' 


*!f; 


THEORY'  CF  VIBRATIONS 


227 


and  the  equation  of  motion  of  the  weight  is 
gdfi 


l^J.,-E^^.P-l,..,.-.l 


or 


Integrating, 

|  — I  --yx^+ci,  ci  being  a  constant  of  integration. 
But  -77  is  zero  when  x-c,  and  therefore  ci-y-c*. 


Hence 


(|)'.._<^.^,.,.. 


This  is  precisely  the  same  equation  as  was  obtained  in  the  first 
case,  and  between  the  Hmits  0  and  x 


<s|^  =  Bin->^, 


sii  t'^at  the  motion  is  isochronous,  and'the  time  of  a  complete  oscil- 
lation is  _ 


When  X-  —  i, 


aiul  lience 


or 


-2(ca-P)-2gffc, 
c»-/»  +  2iA. 


If  /i-O,  i.e.,  if  the  weight  is  merely  placed  upon  the  rod  at 
thf  •nd  A,  C'±l,  and  the  amplitude  of  the  oflcUlation  is  twice  the 
statical  elongation  due  to  P. 

I  lie  ro<l  may  be  safely  stretchetl  until  ita  length  is  L  +  ?,  while 
A  further  elongation  c  fnight  prove  most  injurious  to  its  elasticity, 
will li  shows  the  detrimental  efifect  of  vibratory  motion.    If  «  small 


228 


THEORY  OF  STRICTURES. 


downward  force  Q  is  aiipliod  to  P  when  it  has  reached  the  end  of 
its  vil)ration.  it  will  produce  a  corn's])()ndiiig  descent,  and  the 
weight  P  will  then  ascend  an  ("(jual  distance  above  its  neutral  j)osi- 
tion.  At  the  end  of  the  interval  corresponding  to  P's  natural  jierioil 
of  vibration.  a|)p!y  the  force  again,  and  /'  will  descend  still  further, 
liiis  process  may  be  continiii'd  indelinitely,  until  at  last  rupture 
takes  place,  however  small  /'  and  <J  may  \h\  If  Q  is  applied  at 
irregular  intervals,  the  an.plitude  of  the  oscillations  will  still  be 
increased,  but  the  increase  will  be  followed  by  a  decrease,  and  so 
on  contimially.  In  practice  the  problem  becomes  nuich  more  com- 
plex on  .-iccount  of  local  conditions,  but  experience  shows  that  a 
fhirtmition  of  stress  is  always  more  injurious  to  a  structure  than 
the  stre.-is  due  to  the  maxiimnu  load,  and  that  the  injury  is  aggra- 
vated as  the  iM'Hods  of  fluctuation  and  of  vibration  of  the  structure 
become  more  nearly  synchronous. 

An  example  of  a  fluctuating  load  is  a  proce.s.sion  marching  in 
time  acro.-^s  a  susiwnsion  bridge,  which  may  strain  it  fa;  i  lore 
.s(>verely  than  a  nnicli  greater  dead  load,  and  may  set  up  a  syn- 
chronous vibration  which  may  prove  :d)solutely  dangerous.  In 
fact,  a  bridge  has  lu-en  known  to  fail  from  this  cause. 

The  coellicient  of  elasticity  of  the  rod  may  be  appro.xiniately 
founil  by  means  of  the  formula 


r-I. 


T  U'ing  the  time  of  a  compl(>te  oscillation.     For  supjwse  that  the 
rod  emits  a  musical  note  of  n  vibrations  per  second,  then 


is   the   time   of   travel   from   C  to  (\; 
therefoie 


,        '/  ,,  ,.     PLi-'n^ 

I      ,    ,  o.     an<l  tieiice     /•.,  =  —:    . 

I--"-  .1      (/ 


Si!pp<is<"  that  (lie  weight  i'^  p-crferfly  fre-e  tf)  sliiie  along  the  rod. 
When  it  returns  to  .1.  it  will  leave  the  end  of  the  hmI  and  rise  with 
a  certain  initial  \elocity.    This   velwity  is  evidently  n  2(//i,  and 


TIIEOliY  Of'   VIHKATIOSS. 


229 


the  weight  accordingly  a.'icemls  to  1),  then  falls  again,  repeats  the 
foiiiK  I-  oiKTation,  and  ho  on.  The  equations  of  motion  are  in  this 
f;bf  uiiiy  true  for  values  of  x  between  x=  +c  and  x=  —I. 

4.  On  the  Oscillatory  Motion  of  a  Weight  at  the  End  of  a  Ver- 
tical Elastic  Rod  of  Appreciable  Mass.— Suppose  the  mass  of  the 
roil  10  1)''  taken  into  account,  and  assume: 

((M  That  all  the  jiarticles  of  the  rod  move  in  directions  parallel 
III  tlir  axis  of  the  rod. 

/o  That  all  the  particles  which  at  any  instant  .are  in  a  plane 
l^riundicular  to  the  axis  remain  in  that  plane  at  all  times. 

As  hcfore,  the  rod  0\  of  natural  length  L  and  sectional  area  A 
i-  ii\(il  at  0  and  carries  a  weight  Pi  at  .1. 

Take  0  as  the  origin,  and  let  OX  \yo  the  axis  of  the  rod. 

l.ct  f,  ^  +  (jI?,  and  x,  x  +  dx,  Ix*  re.sjiectively  the  actual  and  natural 
ili«taii(i's  from  O  of  the  two  consecutive  sections  MM, 

M'M'.  O 

l.(t  f>,  Ih'  the  natural  density  of  the  rod,  and  p  the 
i!tii>iiy  of  the  section  MM,  distant  c  from  0. 

Tlic   forces   which   act   \iiw)n   the   rod  are: 

1(1    The  upward  and  constant  force  /*o  at  0. 

i/ii  The  weight  /',  at  .4. 

ii  I   The  weight  of  the  rod. 

ih  \  force  -V  |M'r  unit  of  mass  through  the  slice 
iHiuiiil.d  hy  the  planes  MM,  M'M',  distant  f  and  ?  +  d$, 
n-lMi  lively,  from  O. 

Sii|i|)ose  the  roil,  after  ecjuilibrium  has  In'en  estab- 
li>lii  i,  to  be  cut  at  the  |)lane  M'M'.     In  order  to  main- 
tiiiii  till'  etiuilibrium  of  the   portion  OM'M'   it   will   be 
wi(— aiy  U)  apply  to  the  surface  of  this  plane  a  certain  force  P, 
ml  'lie  ('(juation  of  e(|uilibrium  iK'Comes 


M 


M 


a 


Fm.  309. 


-  /\,  +  /**  pAd^X  +  P  ^p^njAx  -  0. 

Jo 


Kit  if  the  thickness  dt  of  the  slice  .U.U'  is  indefinitely  dimini.«'ied, 
/'  i-  I  vidently  the  ela.stic  reaction,  and  its  value  b 


>^^i^-^^i^ .). . 


230 
Hence 


THEORY  OF  STRUCTURES. 


Differentiating  with  respect  to  x, 


But  pd^=podx, 


therefore 


or 


PqAX+EA-^+P^A  -0, 


Also,  poAXdx  is  the  resistance  to  acceleration  arising  from  the 
inertia  of  the  slice,  and  is  therefore  equal  to 


so  that 
Hence 


-poAdx-^, 


X-- 


d^    Ed^ 


.  .  (1) 


To  solve  this  equation.~ln  the  state  of  equilibrium, 


^■<^-^) 


is  tlip  tension  in  the  soption  of  which  the  distance  from  O  is  x,  and 
••ounlcrbalancos  the  weight  P,  and  the  weight  poA{l-x)g  of  thtt 
IM)rtioii  AMN  of  the  rod. 

Therefore  E^{^-l) -Pi-^PoAgil-x), 

or  ^-ix^a.'W/;     -\ 


SPECIFIC  WEIGHT. 


231 


Integrating,  '  =  ^  +  £j^+^('^~|")-  •    •    ^    ^    •    •    (2) 

There  is  no  constant  of  integration,  as  x  and  f  vanish  together. 
This  value  of  f  is  a  particular  solution  of  (1),  and  is  independent 


of<. 


Put 


e-x.^x.'-^ 


fl-^-D^'. 


z  being  a  new  function  of  x  and  t.    Then 

<?$         Po   J^  A     ^     ^ 

Hence,  from  eq.  (1), 

cPz    E(Pz       .<Pz       ,  .    E 

-m-—  7-5" ^'i^Zs<    where    vi'  - — . 

The  integral  of  this  equation  is  of  the  form 

Z'-F{x  +  rit)+j{x-vit), 

t'ii=\l—)  being   the   velocity   of  propagation  of   the  vibrations. 
\     ^Po/ 
The  full  solution  of  (1)  is  therefore  of  the  form 


f  =  i  + 


|^T+^|'(/x-f )  +F{x  +  v,t)+nx-v,t). 


5.  Specific  Weight;  Coefficient  of  Elasticity;  Limit  of  Elasticity; 
Breaking  Stress. — lioforo  the  strongth  of  a  Iwdy  can  be  fully  known 
<crtaiii  physiral  constants  whose  values  depend  upon  the  material 
must  lie  determined. 

((t)  Specific  Weight. — The  speeific  weight  is  the  weight  of  a 
(iiiit  of  volume.  The  six>cific  weights  of  most  of  the  materials  of 
cniistruption  have  l)een  carefully  found  and  tabulated.  If  the 
s|K'cific  weight  of  any  new  iiiatcriul  is  re<juired,  a  convenient  approxi- 
mate method  is  to  prepare  from  it  a  numl)er  of  regular  solids  of 
ileterminate  volume  and  weigh  them  in  an  ordinarj'  pair  of  scales. 
lilt'  ratio  of  the  total  w«-ight  of  these  solids  to  their  total  volume 
is  t.ic  specific  weight.    It  must  be  remembered  that  the  weight 


232 


THEORY  OF  STRUCTURES. 


may  vary  consulorahly  with  tin>o,  etc.;  tlu.s  a  sample  of  groenheart 
-■.Phed  (0.75  lbs.  per  cubic  fc.t  when  first  cut  „ul  of  xl^Z 
only  .u  lbs.  ju.r  .ubic  foot  at  the  en.l  of  si.x  n.onths.  When  the 
strength  of  a  tnnber  is  being  ,let,.rn.ine.l.  it  is  in.portant  to  not 
l-a.nc,unt  of  water  present  in  the  test-piece,  since  this  appears 
to   have  a  great   mlhience  upon  the   results  ' 

^^^_The  straining  of  a  structure  is  generally  largely  due  to  its  own 

The  total  had  upon  a  structure  includes  „//  the  external  forces 
applied  to  ,t,  and  in  practie.>  is  designated  dem  U^rmanent)  or  lire 
(--..V)  accordM.g  as  the  forces  an-  gradually  applied  an.l  steadv 
or  sud.lenly  apphed  and  accon.panied  with  vibrations.  For  exan  le' 
•.<•  -■..  t  of  a  bridge  is  a  dea.l  load,  while  a  train  p.ssingC  i 
.s  a  hve  load;  the  we.ght  of  a  roof,  together  with  the  weight  of  anv 
snow  winch  n.ay  have  accumulated  up<,n  it.  is  a  dead  load;  ^ 
cav.ses  at  tnne.s  excessive  vibrations  in  the  n.en.lKTs  of  a  structure 

^:t::r;rt;i:  "''""•' ^^^'''-'''^^^^ 

'n.e  dead  loads  of  n.any  structures  (as  masonry  walls,  etc)  are 
o  grea      hat  extra  or  accidental  loads  n.ay  be  safelv  di  rega  dod 
oI.l  chmates,  great  n.a.sses  of  snow  an.l  the  ..netrating  efTe  t 
of    I...   frost   necessitate  ^,.ry  deep  foundations,   which  pro^rtion 
ately  increase  the  dead  weight.  "poruon 

of  Z  exN^r  I'f  "^    ^'"^''•''^•-^"--"y   «l-aking.    a   knowledge 
of    he  <.x  en  1  forces  aCmg  upon  a  structure  .liscloses  the  n.anner 

n  a   on  of  the...  .....nbers  ca..  only  !«>  e.stin.ate,l  bv  means  of  the 

0    (hcen     of  elastuMty,    which   exprcs...   the   relation    l.tween   a 
Mress  and  the  correspon<ling  strain 

In  pn.cti,.e  it  is  usually  sudicient  to  asstune  that  a  material  is 

.   n  >s  ,na>    be  found  ,f  the  coefJieients  of  elasticitv,  of  forn.,  an.l 
ot  \niui,ie  are  known.  '  ' 

In  a  ho„,og,.neous  solid  there  may  be  twentv-one  distinct  coeffi- 
|-n.s  ..f  ..lasfcty,  whieh  are  usually  ..|assified 'under  the  fdl^::;^ 

ami  normtr<;;,r''''-i;'^  """  't''"^'  '"''^^■^""  longitudinal  strains 
ami  normal  sfre.s.M.s  u,  th,.  same  direction. 


II:;;' 


COEFFICIENT  OF  ELASTICITY. 


233 


(2)  Transrerne,  expressing  the  relation  between  tangential  stresses 
ami  strains  in  the  same  direction. 

i'.i)  Lateral,  expressing  the  relation  between  longitudinal  strains 
liiit!  noriTial  stresses  at  right  angles  to  the  strains,  i.e.,  a  lateral 
ivsistance  to  deformation. 

I)  Oblique,   expressing  other  relations  of  stress  and   strain. 

If  a  body  is  isotropic,  i.e.,  equally  elastic  in  all  directions,  the 
■  (iilij-one  coellicients  reduce  to  two, 
\]/...  the  coefHcients  of  direct  elasticity 
;ii!(l  (if  lateral  ela.sticity.  Huch  bodies, 
•iiiwcver,  are  almost  wholly  ideal.  In 
a  perfectly  elastic  body  E  would  he 
ihc  same  both  for  tension  ami  com- 
|iirssion.  In  the  ordinary  materials 
(if  construction  it  is  slightly  less  for 
(■(iin|)ression  than  for  ten.sion;  but  if 
the  stresses  do  not  exceed  a  certain 
limit,  the  difference  is  so  slight  that  it 
i.iay  be  disregarded.  ^'«-  ^lO. 

The  (Miuation  f  =  Ee  may  be  illustrated  graphically  by  the  straight 
line  POP,  the  ordinate  at  any  point  being  the  stress  required  to 
jiKnUice  the  strain  represented  by  the  corresjx)nding  abscissa. 

The  work  ilone  jx-r  unit  of  volume  in  producing  the  extension 
(ir  comim'ssion)  OX 

f 
2E- 

} 


-  area  of  triangle  POX  =  ^-OX  ■  PX  - 1  - 


Also,  tan/'aV=-  =  /i. 
e 

CocHicients   of   ela.sticity    nuist    be    determined    ex|)erimentally. 

Tile  coelficients  of  direct  elasticity  for  the  different  metals  and 
liiiilicrs  are  sometimes  obtained  by  subjecting  bars  ol  the  material 
tn  forces  of  extension  or  compres-sioii,  or  by  observing  the  deflections 
i<i  beams  loaded  transvers(>ly.  The  coefHcients  for  blocks  of  .stone 
ind  masonry  might  also  1m'  found  by  transverse  loading;  they  are 
'if  little,  if  any,  jiractical  use,  as,  on  account  of  the  inherent  stifT- 
;  ■■-s  (if  masonry  structure.'^,  their  deformations,  or  :'et(linfjs,  are  due 
lailicr  to  defective  workmanship  than  to  the  natural  play  of  elastic 

'■'ices. 


234 


THEORY  OF  STRUCTURES. 


The  torsional  coefficient  of  elasticity,  i.e.,  the  coefficient  of  elastic 
resistance  to  torsion,  has  boon  shown  by  experiment  to  var>'  from 
two  fifths  to  three  eigliths  of  the  coefficient  of  direct  el"i-*icity. 

When,  e.g.,  it  is  retjuired  to  find  the  PJ  for  a  material  a  '-.<t-piece 
is  prepared  and  is  held  in  the  grips  (or  holders)  of  a  testing-machine. 
An  intermediate  portion  of  a  definite  length  L  (usually  8  or  10  ins.) 
and  far  enough  from  the  ends  of  the  grijis  to  be  unaffected  by  their 


-f- 

1 
1 

•i 

f  — 

t 

I 


G 


Fig.  311.      Fio.  .'Jl^.    Fio.  .313.     Fro.  314. 

action  is  then  marked  off.  liefore  the  loading  of  the  specimen 
commences  the  telescope  is  set  at  zero.  Under  a  load  P  the  tele- 
scoi)e  gives  a  reading  R,  indicating  that  the  length  L  has  been  in- 
creased bv  R. 


Therefore 


p-eaS-, 


A  being  the  sectional  area  c.  the  sfiecimen. 

If  the  load  is  now  increasod  by  an  amount  JP,  there  will  be  a 
corresponding  increase  of  JR  in  the  ext<>nsion. 


Therefore 


Hence 


P  +  J/'  =  AM 


li+JR 

L     ' 


JR   A' 


from  which  the  value  of  E  can  be  calculated. 

F.v.  1.3.  \  ,>nhl'>.-irr!  sprnmrn  of  1  /,,•.  ,li,irr,rir-  ;„-  phicnl  in  the  Icaiw,/- 
mar/nne  ami  a  ihslance  of  10  in/,.  (/.)  6r/»w/(  thr  mrnsuring-poinls  is  marked 
eff;  determine  the  average  E  from  the  following  Inbulntcil  obaervntions: 


235 


COEFFICIENT  OF  ELASTICITY. 

Load  in  Tons  Increment  Extension  Increment  of 

of  2240  Lbs.,  of  Load,  in  Inches,  Extension, 

P.  JP.  R.  JR. 

0  0  0  0 

4  4  .0042  .0042 

8  4  .0081  .0039 

12  4  .0121  .0040 

Thus,  for  an  increment  in  the  load  of  4  tons,  the  corresponding  average 
"increment  in  the  extension 


and  therefore 


.0042 +  .0039 +  .0040    .0121 
3~' 


E- 


3 
10 


:0121221_ 
3      7  4^*^ 


=  12,629  tons. 


The  ends  of  the  test-piece  are  enlarged  and  are  connected  with 
the  main  body  of  the  specimen  by  giving  the  shoulder  a  suitable 
curve  (Fig.  311).  Abrupt  changes  of  section  (Figs.  312  and  313) 
mast  be  avoided,  as  at  such  points  great  stresses,  of  which  the  dis- 
tribution is  unknown,  are  induced.  The  fracture  of  a  hard  speci- 
men almost  invariably  takes  place  at  an  abrupt  change  of  section, 
while  with  a  ductile  material  there  will  be  a  floiv  (p.  254)  toward 
the  narrower  portion,  which  will  prevent  its  full  contraction  before 
fracture  takes  place.  The  apparent  strength  of  the  material  is 
therefore  increased.  When  test-pieces  are  sheared  out  of  boiler- 
plate, about  a  quarter  of  an  inch  of  the  metal  should  be  removed 
on  each  side  between  the  measuring-points  (Fig.  314),  in  order  to 
eliminate  the  upsetting  effect  of  the  shearing  action.  All  speci- 
mens should  he  long  enough  to  allow  the  portion  between  the  meas- 
uring-points to  contract  freely,  as  otherwise  they  seem  to  be  under 
<()nstraint  and  fracture  takes  place  before  the  contraction  is  com- 
jili'te.  Thus  the  apparent  strength  is  greater  than  the  actual  strength, 
wliich  is  the  breaking  load  per  square  inch  of  the  original  sectional 


area. 


In  ordinary  practice,  however,  the  engineer  does  not  care  to 
know  this  actual  ultimate  stress,  but  demands  the  stress  which  the 
material  would  bear  if  the  sectional  area  of  the  test-piece  jemained 
\i!iohanged  up  to  the  point  of  fracture.  In  other  words,  he  requires 
'.ho  breaking  load  jier  square  inch  of  the  original  sectional  area,  and 
lie  can  then  decide  what  fraction  of  this  load  may  be  safely  applied 
to  any  structural  member. 


lu 


ToQ 


THEORY  OF  STRICTURES. 


Again,  tho  .Ustanco  In-tween  tlio  '•inoasuriiig-points  "  of  a  test- 
l"m-  is  .sul„l.vi,|.-.l  i„;n  a  .lumber  (S  or  ]()}  of  l-i„ch  lengths,  and 

Actual  Extensions 


Kqulvuluut  l^teusiuua 


Fiii.  3  J  5. 

after  fracture  has  taken  place  the  extension  of  each  division  is  care- 
fully (leternnued.     If  tiic  fracture  is  near  the  centre,  the  .H^.ventage 

of  elongation   is  -^.     If  the  fracture,  however,   is  at  a  distance 

from  the  centre,  as  in  Fig.  315,  the  equivalent  elongation  is  obtained 
in  the  following  manner: 

The  elongation  of  each  division  is  shown  i.i  the  figure  On  the 
right  of  the  divLsion  containing  the  fracture  and  in  which  the  local 
elongation  i,  aeces.sarily  the  greatest,  i,m,jinary  divisions,  shown  by 
dotted  hues,  are  a<i.le,l,  and  th,>se  divisions  have  the  same  extensions 
as  the  corresponding  divisions  b  and  c  on  the  left  of  0  The  total 
eqmralent  elongation  is  then  AB-8",  the  length  AB  including  8  divi- 
sions.    In  the  specimen  in  question  /IB  =  10". 7. 

Therefore  the  erjuivalent  elongation  =  2". 7   and   the   equivalent 
percentage  =  —  X  2.7  =  33^,  very  nearly. 

Also,  if  a  is  the  frarlnred  area,  the  percentage  of  reduction  in  area 


m{i-^). 


These  two  percentages  are  measurements  of  the  ductiUtv  of  the 
material.  " 

In   the  above  ^pvcmwn   tl.<.   initial  diameter  =  l.'u59,  and  the 
final  diameter  =  0".G.S1 . 


Therefore  "-J^^Y^,,,-^ 


and  tho  reduction  of  area=  lOOfl -.413.-))  =r),S.6r)  per  cent 

(;■)/''■'«'■'  0/  Ela,tmty.~\yhon  the  fon-es  which  strain  a  bodv 
fa     below  a  certain  limit,  the  bo<ly,  on  the  removal  of  the  forces 
will  resume  its  original  form  and  dimensions  without  sensible  change 


LIMIT  OF  ELASTICITY.  '*' 

(ilisrojjanling  any  offpcts  due  to  the  development  of  heat)  and  may 
1(1'  treated  as  jierfectly  elastic.  But  if  the  forces  exceed  this  Hmit, 
the  body  will  receive  a  jx>nnanent  deformation,  or,  as  it  is  termed, 
a  .yd. 

Such  a  limit  is  called  a  limit  oj  eloKtinty,  and  is  the  greatest  stress 
tiiat  can  be  applied  to  a  body  without  producing  in  it  an  appreciable 
and  permanent  deformation. 

This  is  an  unsatisfactory  definition,  as  a  body  passes  from  the 
ela.stic  to  the  non-elastic  state  by  such  imperceptible  degrees  that 
it  is  impossible  to  fix  any  exact  line  of  demarcation  between  the 
two  states. 

Bauschinger's  experiments  also  indicate  that  the  application  to 
a  body  of  any  stress,  however  small,  produces  a  plastic  or  perma- 
nent deformation.  This,  perhaps,  is  sometimes  due  to  a  want  of 
imiformity  in  the  material,  or  to  the  bar  being  not  quite  straight 
initially.  In  any  ca.se,  the  deformations  under  loads  vt^hich  are 
less  than  a  loail  known  as  the  elastic  limit  are  so  slight  as  to  be  of 
no  practical  account  and  may  be  safely  disregarded,  and  for  such 
loads  Hooke's  Law  may  Ije  regarded  as  substantially  correct. 

So,  too,  the  liysteresiK  effect,  i.e.,  the  lagging  of  the  relation  of 
stress  to  strain,  is  so  slight  within  the  elastic  limit  as  to  be  safely 
disregarded.  lieyond  the  ela-^^tic  limit  the  creeping  becomes  very 
marked  for  the  first  few  minutes  and  may  continue  much  longer, 
but  at  a  diminished  rate.  At  last  a  point  is  reached  at  which  the 
material  draws  out  and  breaks  without  any  further  increase  of  load. 

Fairbairn  defines  the  limit  of  elasticity  more  correctly  as  the 
stress  below  which  the  deformation  is  approximateh'  proportional 
to  the  load  which  produces  it,  and  lK>yond  which  the  ileformatior 
increa.ses  much  more  rapidly  than  the  load.  In  fact,  both  the  elastic 
and  ultimate!  .strengths  of  a  material  dejx^nd  ujion  the  nature  of 
the  stres.st's  to  which  they  are  .subjected  and  upon  the  frequency  of 
their  application.  For  example,  in  exix-rimenting  upon  bars  of 
iron  having  an  ultimate  tenacity  of  46,704  lbs.  jx^r  square  inch  and 
a  tluctility  of  20  jkt  cent,  Wohler  found  that  with  repeated  stresses 
of  ecjual  intensity,  but  alternately  tensile  and  compressive,  a  bar 
failed  after  56,430  repetitions  when  the  intensity  was  ;>3  0(rf>  lbs. 
jx^r  square  inch;  a  second  bar  failed  only  after  19,187,000  rejjetitions 
when  the  intensity  was  18,700  lbs.  {)er  square  in;    while  a  third 


238 


THEORY  OF  STRUCTURES. 


bar  rpinainod  intact  af^-  :  .  re  than  i;?2.000,000  repetitions  when 
the  intensity  wjvs  1(.  *  ..,o.  yxr  s(|uure  inch.  These  experiments 
therefore  indicated  that  the  limit  oj  elasticity  for  the  iron  in  (juestion, 
under  rejx'ated  stn-sses  of  e(|ual  intensity.  l)iit  alternately  tensile 
and  conipressve.  lay  iK-tween  Ki.tKK)  and  IT.tMK)  il).s.  per  s<iiuire  inch, 
which  is  much  less  than  the  limit  under  a  steadily  applied  stress. 
Similar  results  ha\-e  Ix-en  shown  to  follow  when  the  stresses  fluctuate 
from  a  maxinmni  stress  to  a  mininmm  stress  of  the  same  kind. 

Cienerally  sjM>aking,  then,  the  limit  of  elasticity  of  a  material 
subjected  to  rejx'ated  stress(>s  is  a  certain  maxinmm  stress  below 
which  the  condition  of  the  body  remains  unimpaired. 

Hauschinj^er  defines  tlie  ela.stic  limit  as  the  jMtint  at  which  the 
stress  ceases  to  Im-  sj-nsibly  pro|M)rtional  to  the  .strain,  the  latter 
h-ing  mejusured  by  a  mirror  apjiaratus  reading  to  about  .00001  inch. 
If  the  yield  iHiint'is  exceedeil  it  will  ri^e  so  long  as  th'>  bar  remains 
loaded.  If  the  load  exceeds  the  clastic  limit  this  limit  will  rise 
until  it  approaches  the  yield  ix.int,  when  it  rapidly  falls.  If  the 
load  is  then  increa.s«'d  U'vond  the  yield  jM.int  the  limit  will  ri.se  again 
until  it  miches  a  much  liigiier  |M»int  than  its  former  value. 

The  main  object,  then,  of  the  theory  of  the  strength  of  materials 
is  to  determine  whether  the  stresses  develo|K-d  in  any  particular 
memlKT  of  a  structure  exctvd  the  limit  of  elasticity.  As  .soon  as 
they  do  so.  that  memlMT  is  |M'rmanently  deformed,  its  stnMigth  is 
im|)aired.  it  JH-comes  i)re<lis|>osed  to  rupture,  and  the  safety  of  the 
whoK'  structure  is  threatene.l.  Still,  it  must  1k'  lK)rne  in  mind 
that  it  is  not  abs(»lutely  true  that  a  material  is  always  weakened 
by  iM'ing  subject. -d  to  forces  .suiM'rior  to  this  limit.  In  the  manu- 
facture of  iron  bars,  for  in.«tance,  eadi  (»f  the  pr(K'ess<'s  through 
which  the  metal  i>a.s.s<>s  changes  its  elasticity  and  increases  itssln'tigtli. 
Such  a  material  is  to  lie  Invited  as  In-ing  in  a  new  state  and  as  |Mtsse.<w- 
ing  new  pro|H'rties. 

Again,  when  a  bar  has  U-en  overstrained  so  that  its  mole<'u- 
lar  conditH.n  is  «-|ianged,  the  ap|>iication  of  a  small  load  produces 
an  immediate  exten.sion  foll.nve.l  by  a  vrcepinii  which  only  shtvly 
disap|K-ars  when  the  bar  is  relieved  of  load,  if  the  bar  i«"allow<Ml 
to  rest  it  gradually  recovers  its  elastic  pr(»|»«'rties  and  the  o'covery 
Ijeiomr-s  niiiie  and  mon-  cfimplete  as  the  time  of  rest  inrn>aw>t. 
Muirs  ex|»erinients  indicate  that  the  bar  may  regain  its  elMticity 


BREAKING  STRESS. 


23» 


by  being  immersed  for  a  few  minutes  in  boiling  water.  It  is  also 
loiiii.l  that  this  process  gives  the  metal  a  higher  elastic  limit.  The 
(ivcrstniining  dcvelopes  a  resistance  to  plastic  deformation  or  pro- 
duces a  hardening  effect  which  may  be  eliminated  by  heating  the 
imtiil  to  redness  and  then  allowing  it  to  cool  slowly,  i.e.,  by  annealing. 
The  strength  of  a  material  is  governed  by  its  tenacity  and  rigidity, 
iiiid  the  essential  requirement  of  practice  is  a  tough  material  with 
a  liigh  elastic  limit. 

This  is  especially  neceesarj-  for  bridges  and  all  structures  liable 
to  constantly  repeated  loads,  for  it  is  found  tha  these  repetitions 
lower  the  elastic  limit  and  dimmi.sh  the  strength. 

In  the  majority  of  ca«os  exix'rience  naa  fixed  a  practical  limit 
for  the  stresses  much  below  the  limit  of  elasticity.  This  insures 
greater  safety  and  provitles  against  unforfseen  and  accidental  loads 
which  may  exceed  the  prattical  limit,  but  which  do  no  harm  imless 
thcv  pass  beyond  the  elastic  limit. 

Certain  ojierations  have  the  effect  of  raising  the  limit  of  elasticity; 
a  wrought-iron  bar  steadily  strained  almost  to  the  {K)int  of  its  ulti- 
mate strength  and  then  rcleasjnl  from  strain  anil  allowed  to  rest, 
exiHTienees  an  elevation  Iwth  of  tenacity  and  of  the  elastic  limit. 
If  the  bar  is  stn'tched  until  it  breaks,  the  tensile  strength  of 
the  br''  <'n  pieces  is  greater  than  that  of  the  bar.  A  similar  result 
follows  in  the  various  processes  employed  in  the  manufacture  of 
iron  and  stcn-l  bars  and  wires:  the  wire  has  a  greater  ultimate  strength 
than  the  bar  from  which  it  was  drawn. 

Again,  iron  and  steel  bars  subjected  to  long-continued  compreB- 
>ion  or  extension  have  their  resistance  incn-ased  mainly  becausie 
time  is  allowed  for  the  molecules  of  the  metal  to  assume  such  posi- 
tions as  will  enable  them  to  offer  the  maximum  resistance;  the 
increase  is  not  attended  by  any  appreciable  change  of  density. 

liider  an  increa-sing  stress  a  brittle  material  will  Ik;  fractured 
without  any  great  deformation,  while  a  tough  material  will  become 
[iliistic  antl  undergo  a  large  deformation. 

(ft)  lireaking  .<»/rf.i.«.— When  the  loa«l  upon  a  material  increases 
indefinitely  the  mat«'rial  may  merely  suffer  an  increasing  defonna- 
tion,  but  gPwaJly  a  l>">>t  »«  reached  at  wl'ih  fracture  suddenly 
takes  place. 

('a»t  iron  is  perhaps  the  most  doubtful  of  all  materials,  and  the 


1 


I 


240 


THEORY  OF  STRUCTURES. 


greatest  care  should  l)e  obser\e(l  in  its  employment.  It  posse&ses 
httle  tenacity  or  elasticity,  is  very  hard  and  brittle,  and  may  fail 
suddenly  un.ler  a  shock  or  an  extreme  variation  of  tem|)eraturc. 
rne(|ual  cooling  may  predispose  the  metal  to  rupture,  ami  its  strength 
jnay  Ix'  still  furlJKT  diminished  by  the  presence  of  air-holes. 

Cast  iron  and  similar  mat.Mials  receive  a  sensible  set  even  under 
a  small  loa.l.  an.l  tlie  set  increases  with  the  load.  Thus  at  no  jwint 
will  the  stress-strain  curve  Ix-  absolutely  straight,  ami  the  jjoint 
oi  fracture  will  1m>  r(>ached  without  any  ureot  change  In  the  sloijeof 
the  curve  and  without  the  d.'velopnient  of  nmch  plasticity. 

ll>,»/,<//,/  iron  and  slei-l  are  far  more  uniform  in  th<<ir  liehaviour, 
and  olx-y  with  tolerable  regularity  certain  theoretical  laws.  They 
an.  tenacious,  ductile,  have  great  compressive  strength,  and  are 
most  reliable  f„r  structural  puriK,ses.  Their  .stn'ngth  ami  elasticity 
may  Ih"  considerably  reduc-.l  by  high  temiKTatures  or  severe  cold. 
When  a  bar  ..f  such  material  is  tested,  the  «/rc.x.v-.s<rai/i  curA-e 
<I=±Ee).  sxs  has  already  Inrn   pointed  out,   is  almost  absolutely 

straight   within   the  ela.stic  limit,  e.g., 
from  O  to  .t   in  tension  and  from  0 
to    li    in    compre.s.sion  (Fig.  31()).    As 
the  load   increa.^es  U'voml   the  elastic 
limit,    the   increasing  deformation    be- 
comes phistic  and  |)ermanent,  and  the 
stress-strain   diagram   takes  an  appro- 
ciable    curvature    Ix'twcH'n    the    limits 
A    and    />  and    the   point.s   B  and   E 
corres|M (tiding  to  the  ma.xinmm  loads. 
In    tension,   as  .noon   as    the   jwint   D 
is  reached,   the  bar  rapidly  elongates 
and  is  no  lunger  able  to  sustuin  the 
maximum  load,  its  actional  area  rapidly 
diminishes,     and     fracture     ultimately 
takes   place   under   a   load    much   1«« 
than   the   maxinuim   lo-ul.     Tlie   |K>int 
of  fractun-  is  repres<'nt(H|  in  the  figure 
by  the  iH.int  F,  the  ordinate  ..f  F  being  the  actual  uUimate  vitewHty 

,    ,  fiiiiil  lotui  on  the  ipor 

o/ .v/rcsv  -  -         — : , _ 

urea  of  Iraclured  Ketiton' 


Flo.  3ia. 


BHEAKISU  STRESS. 


241 


Tho  exact  form  of  tho  stross-strain  curve  between  D  and  F  is 
iiiikiiown,  as  no  dofiiiito  relation  has  Ix^n  found  to  exist  between 
the  stress  and  strain  durittg  the  elongation  from  1)  to  F. 

It  is  also  iiiiiKirtaiit  to  note  that,  as  the  defonnation  gradually 
incri'ases  under  the  increasing  load,  the  molecules  of  the  materia! 
r('(|Uin'  greater  or  less  time  to  adjust  themselves  to  the  new  condition. 

During  the  tensile  test  of  a  ductile  material  there  is,  at  some 
|Hiint  hevond  the  ela'stic  limit,  an  abrupt  break  (iH  in  the  continuity 
(if  tlic  .stres.s-strain  curve,  the  curve  again 
iHcnmiiig  continuous  from  //  to  /),  Fig.  317. 
Tln<  jthenomenon  is  always  very  marked  in 
mill  I  steel  and  other  ductile  materials,  and  the 
(icfiirtnation  after  passing  GH  is  almost  wholly 
pJHsti''  or  {XTmanent. 

The  point  (/  is  called  the  j/i'cu.  i)oint  and 
seems  to  be  always  higher  than  A,  the  true 
elastic  limit. 

In  (om fjreifition  there  is  no  local  stretch 
a-  ill  tension,  and  there  is  con.sequently  no 
(•oii>iiierable  change  in  the  curvature  of  the 
(■oiiiinession  stress-.strain  curve  up  to  the  point  of  fracture. 

Tiiiibtr  is  asually  tested  by  being  subjected  to  the  action  of 
l(ii>ile,  compressive,  or  transverse  loads.  Other  characteristics^ 
Imwivcr,  must  be  known  l»efo«?  a  full  conception  of  the  strength 
(if  the  wcxkI  can  U*  obtained.  Thus  the  sjM-cific  weight  must  Ije 
fnmi'l;  the  amount  of  water  present,  the  lass  in  drying,  and  the 
(■.itr.s|i«)t  ding  sliinkage  should  be  determined;  the  structural  difTer- 
ciiccs  of  the  several  si)eciinen8,  the  rate  of  growth,  etc.,  should  be 
(iliM  Tved. 

Tlie  chief  object  of  ex|H>riment8  u|)on  ntOKonnj  and  brickwork 
i-  III  discover  their  n-sistance  to  compn>s.Hion,  i.e.,  their  cru.shing 
^iri  !:;;th.  In  fact,  their  stiffness  is  so  great  that  they  may  be  com- 
['■■1  -cd  up  to  the  iK)int  of  fracture  without  s<'tisible  change  of  form, 
•iti.i  It  is  then'fore  very  ditiicult,  if  not  iin|K)Hsible,  to  obnTVi.  the 
liiiiil  (if  ela-xticity. 

riie  rement  or  mnrtar  uniting  the  stones  and  bricks  is  most  irregu- 
lar in  (|uaiity.  In  every  im|M>rtant  work  it  should  ix'  an  invariable 
rule  tu  prei)are  speciniens  for  testing.    The  crushing  strength  of 


Fig.  317. 


242 


THEORY  OF  STRUCTURES. 


cement  and  of  mortar  is  much  greater  than  the  tensile  strength,  the 
latter  being  often  exceedingly  small.  Hence  it  is  advisable  to  avoid 
tensile  stresses  within  a  msiss  of  masonry,  as  they  tend  to  open  the 
joints  and  separate  the  stones  from  one  another.  Attempts  are 
freijuently  made  to  strengthen  ma.sonry  an«l  brickwork  walls  by 
ii>.><erting  in  the  joints  tarred  and  sanded  strips  of  hoo|)-iron.  Their 
utility  is  doubtful,  for,  unless  well  protected  from  the  atmosphere, 
they  oxidize,  to  the  tletriment  of  the  surrounding  material,  and 
besides  this  they  prevent  an  cfjuable  distribution  of  pressure.  They 
are,  however,  far  preferable  to  lx)nd-timl)ers. 

The  working  stress  is  the  greatest  stress  which  ^  material  is  to 
bear  in  ordinary  practice,  and  the  ratio 

ultimate  strength 
working  stress 

is  called  the  factor  of  safety. 

The  value  of  this  factor  is  governed  by  many  important  consider- 
ations, as,  for  example,  the  nature  of  the  nmterial,  the  character  of 
th<  stresses,  (he  effect  of  bad  workmanship,  etc.  The  material  shwly 
deteriorates  with  time,  and  its  life  is  also  influenced  in  an  important 
degree  by  variations  of  tem|x>rature.  A  marked  change  in  the 
magnetic  properties  of  iron  and  st(H'l  when  ex|)o.<»ed  for  a  considerable 
length  of  time  to  a  tem|KTature  as  low  as  1()0°  to  150°  F.  shows  that 
tluw  metals  undergo  a  gradual  molecular  change,  but  they  may  be 
restored  to  their  original  condition  by  re  annealing.  Again,  the 
iM'havior  of  a  material  largely  (le|»ends  ui)on  the  existence  of  initial 
internal  stres-tf^s  w.iich  may  \m>  due  to  xct  cau.sed  either  by  temperature 
variations  or  by  the  action  of  previou-sly  applied  forces.  This  is 
w.-jl  illustrated  in  a  cylindrical  casting  in  which  an  outer  shell  cools 
first  and  contracts,  anil  is  then  compn>s.sed  tangentially  by  the  cooling 
and  c.>ntraction  of  an  adjoining  inner  shell,  while  the  metal  of  this 
inner  s]uA\  is  puilcl  out  radially.  It  has  U-en  |M>inted  out  by  J. 
Thomson  that  the  <l.-fect  of  elasticity  un<ler  small  loads  observed  by 
IliHlfrkiiist.n  in  ea,st  iron  is  probably  due  to  the  existence  of  interna' 
.sttv-xs.  There  must  necessarily  Ix-  an  ei|uilil)rium  In'tween  tlicse 
initial  stnsMs  .n  any  si-ction  <il  llir  material,  as  there  is  no  external 


ioiid. 


iiiid  ii  I  ill-  iiiaicriat  is  whoiiv  fri'c  imm  such  st 


to  \k'  in  u  "late  oj  c«,se  i  Pearson).     \um\\\ 


ress<'s  it  is  said 
ing  will  give  a  condition  of 


wUhler'S  experiments.  243 

aliimst  complete  ease  to  a  plastic  metal,  but  even  then  test-pieces, 
win  II  first  loaded,  show  elastic  defects  which  may  arise  in  part  from 
iniiial  internal  stress  and  which  are  reduced  or  may  disappear  under 
ir|Kat('d  applications  of  the  loads. 

A  factor  of  safety  is  rarely  less  than  3,  and  the  following  factors 
art"  in  accordance  with  the  best  practice: 

For  timber 3  for  dead  loads 

"       "     6    "  live 

"  metals 3    "  dead     " 

"     6    "  live       " 

"    masonry 6    "  dead     " 

"         "         10    "  live       " 

Kxperiments  also  indicate  that,  under  a  steady  (or  static)  load, 
tinilMT  may  be  strained  almost  to  the  point  of  fracture  without  appar- 
ent injury  to  the  material. 

The  factor  of  safety  is  largely  influenced  by  the  variability  of  the 
stresses  to  which  a  material  may  be  subjected.  The  extension  of  a 
liar  under  a  gradually  applied  load  is  much  less  than  when  the  load 
i-  jilacod  on  the  bar  and  suddenly  released,  while  it  is  still  greater  if 
the  l)ody  is  in  motion  and  commences  its  action  on  the  bar  by  a 
shixk  or  a  blow.  This  is  followed  by  a  vibration  (Art.  2)  of  the  weight 
alidiit  a  neutral  position  and  the  vibration  continues,  but  with  a 
liiininishiiig  amplitude,  until  it  finally  disappears  imder  the  action 
lit  the  molecular  frictional  resistance  of  the  material.  The  effect, 
then,  of  a  suddenly  applied  load  or  a  shock,  even  if  it  strains  the 
iii.iiiTiul  only  slightly  l)eyond  the  limit  of  ela.sticity,  is  detrimental, 
while  the  same  load  may  be  gradually  applied  without  doing  any 
liaiiii.  The  kinetic  energy  due  to  a  shock  or  blow  is  exjMMided  in 
ilniiii;  the  work  of  straining,  and  if  the  resulting  strain  is  so  great  that 
lii.  limit  of  elaslicit>  .s  ('xc<'eded.  a  local  hardening  is  |)roduced  which 
i(  iileis  the  material  less  al>le  to  take  up  the  work  as  an  ela.stie  strain, 
.Hii  lis  rapacity  for  doing  s<»  may  Im*  rapidl>  cxhau-sted  by  re|M'tition8 

I'!  -Ui'll  slio<'ks 

6.  Wtthier'h   Experiments.— Fatigue.— It    is  known  that  variable 

Inn.  s.  eiiiistaiidy  re|K'ated  loads,  aial  cuntimied  vibrations  diminish 
ilii'  -trength  of  a  material,  whether  they  produi-e  .stresses  approximat- 
uiL'  Id  the  elastic  limit,  or  exceedingly  .«inall  .«tres.s,'s  occurring  with 


244 


THEORY  OF  STRICTURES. 


groat  rapidity.  Indood  many  structiiros  aro  dosignod  so  that  tho 
several  monilM>rs  arc  always  subjected  to  the  same  kind  of  stress 
thus  avoiding  the  detrimental  elTect  of  alternating  tensile  and  com- 
jir(>ssive  stresses.  Although  the  fact  of  a  \ariahlo  ultimate  strength 
had  long  been  tacitly  acknowledged  and  often  allowed  for,  Wolilor 
was  the  first  to  give  formal  expression  to  it,  and,  as  a  result  of  ob.^cr- 
vation  and  exiM-rinieiit  extending  over  a  period  of  twelve  years, 
enuMciat<'(l  the  following  law: 

"That  if  a  stress  /.due  to  a  -tatic  load,  cause  the  fracture  of  a 
bar,  the  l)ar  may  also  Ix'  fractured  by  a  series  of  often-repeated 
stresses,  e.ach  of  which  is  less  than  /;  and  that,  as  the  differences 
of  stress  increase,  the  cohesion  of  the  materials  is  affected  in  such  a 
manner  that  tho  mhiimum  stress  required  to  i)roduce  fracture  is 
di.niuished." 

This  law  is  manifestly  incomplete.  In  Wiihler's  exjH'riments 
the  applications  of  the  load  followed  each  other  with  great  rapidity, 
yet  .1  certain  length  of  time  was  ro(iuire<l  for  the  resulting  stresises 
to  attain  their  full  inten.sity;  the  influonee  due  to  the  rapidity  of 
application,  to  the  rate  of  increa.se  of  tho  .stress,  and  to  the  duration 
of  individual  strains  still  remains  a  subject  for  investigation. 

The  ex|)erimonts.  howexcr.  show  that  the  rate  of  increase  of 
reiH'tilions  of  stress  re<|uired  to  produce  fracture  is  nmch  more 
rapi.l  than  the  rate  of  decrease  of  the  .stre.s.ses  themselves,  and 
de,Kiids  Intth  upon  the  maiimiim  .stress  and  uiK>n  the  difference  or 
lliirfiirttioii  ((/  .sYrc-i-v. 

The  effect  of  re|K'ate«l  stres.ses  of  enuul  intensity,  jjut  alternatt-ly 
tensile  ami  compressive,   has  In-en  already  jKiinte*!  out  in  Art.  4. 

iViis  of  the  same  material  re|H'at«M||y  JM'nt  in  one  din'ction  Ijore 
HI.I:L'  IIh.  |K'r.s(|uare  inch  when  the  load  W!ls  wholly  removed  In'tween 
ea»!i  ben; ling,  ;uid  l.VT.'U  lbs.  [mt  s(|uare  inch  when  the  stn\ss  fluctu- 
ated JM-tween  4.").7;{;{  lbs.  and  24,1(41  lbs. 

The  tai)lo  on  pagi-  215  gives  the  results  of  similar  exiHriments 
on  .steel. 

riie  .ixle  sf(H'l  W!is  foiuid  to  Ix'ar  22,h;M)  lbs,  jjor  .-jquare  inch 
when  subjected  to  reiK'atetl  shears  of  equal  inten.sity  but  npjmsite 
in  k'uul,  ,iiid  21»,  J4()  lbs  [kt  .square  inch  when  the  shears  were  of 
the  >viiw  knid.  It  would  therefore  ap|>«>ar  that  the  shearing  strengths 
of  the  metal  iu  the  two  cas<'s  are  alKuit  |  of  the  stnnigtlw  of  the 


FA  TIGUE. 


245 


saino  metal  under  alternate  bending  and  under  bending  in  one  direc- 
tiiiii  resjx»ctively. 


Character  <pf  Fluctuation. 


Alt.Tiiatiiin  .stro(w«'a  of  (filial  intensity 

1  ..iiipU'tc  rflief  from  ^t^■ss  iu-twiiii  each  benil- 

l';irtml  n'iicit  from  stress  betwwn  eacli  bending 


Maximum  RwiKtance  to  Kepeatad 
StresKCs  in  Lbii.  (ler  Square  Inch. 


Axle  StwI. 


29,(K10,-29,(K;0 
lO.SOO.  0 

.s;i,i!0,    3r.,3S0 


Spring  Steel 
( uiihardenetl ). 


.52,000,         0 
'XiSM).   62,240 


Irom  torsion  exiKrinients  with  various  qualities  of  >*teel  the 
i'lportant  result  was  deduced  that  the  maxiniuin  resistance  of 
ihf  steel  to  alternate  twisting  was  J  of  the  inaxinium  resistance 
III  tlie  same  ste<'l  to  alternate  bending. 

For  shearing  stresses  in  opi>osit«  directions  Wohler  fountl,  in  the 
ca^.'  of  Krupp  cast  steel  (untemperetl).  that  h -39.500  lbs./s«i.in. 
a;wl  .»- 23,000  Ibs./sq.  in.,  or  about  |  of  the  corresponding  values  for 
stresses  which  are  alternately  tensile  and  compressive,  antl  it  may 
Ik  jr,.iicrally  jvssumed  that  the  value  of  the  working  stress  for  shear- 
iiiii  stres.s('s  is  \  of  its  value  for  stresses  which  are  alternately  tensile 
ami  compressive,  and  for  which  the  ratio  of  the  maximum  tensile 
t(i  the  maximum  compressive  stress  is  the  same. 

Wohler  pro|)osed  2  as  a  factor  of  safety,  and  considered  that 
the  iiuixinmm  jK^-missible  working  stre.ss<>s  should  be  in  the  ratios 
nf  1:2:3,  according  as  members  are  subjected  to  alternate  tensions 
a:;,l  comjm'ssions  (alternate  lK>nding),  to  tensions  alternating  with 
c  Ml  ire  relief,  or  to  a  steady  loatl. 

Fatigue.— It  is  a  fact  of  practical  and  scientific  im|X)rtance  that 
ir .  1  and  .steel,  and  probably  all  materials,  are  weakened  by  re;M'ated 
ViiiiMtions  of  stn-ss,  the  weakening  elTect  iH'ing  called  jnHijite.  A 
r  ri;iiii  variation  may  Ix'  iH-rmissible  provided  tiiat  the  stn\s«'s  ai-e 
w.  1]  within  the  elastic  limits,  but  it  must  not  Im'  forgotten  that  when 
!!,-.•  liiuits  are  exc(>«>dcd  even  the  toughest  bar  may  !>«>  fractured 
u\  ;\  very  few  iM-ndings.  It  is  .strange  that  the  material  in  old  rails, 
titvv  etc..  which  have  Ihhmj  long  in  u.s<'  ami  have  Imh'ii  almost 
-  .';:tu^tt-tl  {;y  fafijr;!!'  'Am\M  havf  Ihcis-i!!-  ;i  si'i'niingly  totally  difTcr- 
iiit  materiul.  but  yet.  under  the  ordinary  tests  lor  phisticily  and 
^ir.ngth,  should  give  results  which  do  not  tlifTer  in  any  marke«l 


246 


THEORY  OF  STRUCTURES. 


dpgroo  from  tliosp  ohtainod  for  the  now  inatorial.     A  period  of  rest 
tends  to  restoro  tlie  elasticity  and  |)ossil)Iy  the  strength  of  a  fatigued 
piece,  and  the  effect  of  fatigue  may  be  entirely  "ehminated  by  heat- 
ing tile  piece  to  redness  and  then  allowing  it  to  cool  slowly,  i.e. 
by  annealing. 

The  phenomena  of  fatigue  have  been  fully  explained  by  Prof.  J. 
Thomson's  deductions,  verified  by  Baaschinger's  ex|K'riments.    There 
are  in  reality  two  limits  of  elasticity.     Bauschinger  first  loaded  a 
number  of  wrought-iron  bars  until  a  jMiint  was  reached  at  which 
the  tensile  stress  was  no  longer  s(>nsibly  proportiotial  to  the  strain. 
He  then  reversed  the  process  and  loaded  the  bars  in  compression  until 
the  compressive  stress  was  no  longer  sensibly  proportional  to  the 
strain.     He  found  that  these  oi)erations  had  "the  effect  of  lowering 
the  elastic  limit  in  tension  and  raising  that  in  compression  until 
finally  the  two  limits  were  equidi.stant  from  the  line  of  no  load.    These 
he  called  the  natural  elastic  limits  of  the  wrought  iron,  the  correspond- 
ing stress  being  ^  tons  p<'r  square  inch,  which  is  practically  the 
same  as  that  obtained  by  Wcihler  in  bars  of  the  same  material  tested 
alternately  in  tension  and  compression.   liauschinger's  e.\periments 
also  .showed  that  although  either  oue  of  the  limits  might  be  changed, 
even  to  the  ordinary  breaking  stress,  the  range  of  elasticity  remains 
about  the  same.     This  accords  with  Thomson's  theoretical  deduc- 
tions in  which  he  defines  the  limits  as  a  .vM/xrjor  and  an  inlerior 
limit,  for  the  reason  that  they  are  not  nece.-warilv  ecjual  and  equi- 
(hstant  from  the  line  of  no  load,  and,  if  the  overstraining  is  suffi- 
cient, may  both  lie  on   the   .same   si<le   of   this   line.     The   tensile 
elastic  limit  of  bars  fresh  from  the  rolls  is  raised  bv  processes  of 
manufacture  and  treatment  to  a  higher  jMiint  than'Bauschinger's 
natirnl  hmit,  ami  is  at  once  lowered  when  subjected  to  stresses 
which  are  alternately  tensile  and  compressive.     At  the  same  time 
the  limit  in  compression  is  raised  and  this  change  continues  until 
both  hniits  are  the  .same.     It  is  not  the  straining  l)evond  one  elastic 
hmit  which  is  in.irious  to  a  material,  but  the  repeaUnl  straining 
beyond  the  Itco  elastic  limits. 

On  the  basis  of  Wohlers  ex|H'riments.  empirical  formula"  have 
l>eon  deduced  whic},,  jt  i«  rlaitne,!.  are  more  in  aocordanee  with 
the  results  of  exjH'riment,  give  smaller  errors,  and  insure  greater 
safety  than  the  incorrect  assumption  of  a  constant  ultimate  strength. 


ill 


LAUSHARDT'S  FORMULA. 


247 


The  formulae  necessarily  depend  upon  certain  experimental 
rf  suits,  but  in  applying  them  to  any  particular  case,  it  must  be 
roinembered  that  only  such  results  should  be  employed  as  have 
hcon  obtained  for  a  material  of  the  same  kind  and  under  the  same 
(oii.litions  as  the  material  under  consideration.  The  effects  due 
t'.  faulty  material,  rust,  etc.,  are  altogether  indeterminate,  .so  that 
IK)  formula  can  be  perfectly  universal  in  its  application.  Hence 
till"  necessity  for  factors  of  safety,  with  values  depending  upon  the 
character  of  the  stresses  as  well  as  upon  the  nature  of  the  structure 

still  exists. 

In  the  formula?  the  following  assumptions  are  used: 

t  is  the  ultimate  strength  of  the  material  under  a  static  or  under 

M  very  gradually  applied  load.      • 

H  is  the  strength  when  the  material  is  subjected  to  a  number 

oi  repeated  stresses,  the  stress  in  each  repetition  remaining  unchangeil 

in  kind;  that  is,  being  wholly  tensile  or  wholly  compressive  or  wholly 

shearing. 

s  is  the  ultimate  strength  of  the  material  when  subjected  to 
niM'ated  stresses  which  are  of  equal  intensity  and  are  alternately 
tensile  and  compressive. 

/  is  tiio  working  .stress  per  unit  of  sectional  area. 

.1  is  the  effectives  sectional  area. 

jA  is  the  numerically  absolute  maximum  load  which  the  material 
h;is  to  carry. 

I.nitnhnrdt's  Formula— \M  Oi  Ix'  the  ultimate  strength  of  a  bar 
,.i  a  sectional  area  .4  when  the  bar  is  subjected  to  stresses  which 
\arv  Ix'tween  ai*and  a  mininmm  stress  a-i  of  the  same  kind. 

Let  «i  -a2,  the  fluctuation  of  strejix,  =d. 

.\s  the  result  of  experiment, 

aiocd  =  Fd, 

r  ix'ing  a  numerical  coefficient  whose  value  must  be  determined 
i\[K'riment. 

Ii  ^2    0,  i.e.,  if  the  fluctuation  is  between  the  maximum  load 
1  complete  rest,  then 

ai=d-H    and    F-1. 


24S 


THEORY  OF  HTHUCTL  lit:^. 


If  (/  =  0,  i.e.,  if  there  is  no  fluctuation  of  stress  so  that  the  bar 
is  under  a  static  load,  then 

^1=02  =  '    and    F=»x. 
Launhardt's  assumption  that 

satisfies  these  extreme  conditions  and  also  gives  intermediate  values 
of  fl,  which  approximately  agre*'  with  the  results  of  the  most  reliable 
experiments.     Hence 

<-»,-(.-.-)(.-2-;), 

and  .heretore       a,  =«(■  +^' Jj)  =  „(,  +i^^), 


ai 


where 

Taking  3  as  a  factor  of  safety, 

the  working  load^jA  =.4^  =— Yl  +-1^^ 


(L) 


Weifrnuch'^  Fonn„h.-Tho  bar  is  now  subjected  to  stresses  which 
vary  from  a  nnnwrimlly  maximum  stress  a,  to  a  minimum  stress 
02  of  an  upimsite  kind.     Thus 

the  nctunl  ilurtuation  of  stress=ai  +a2-rf, 
and  again  as  the  result  of  exjx'riment 

«i=F(/. 

If  «i.-0.  i.e.,  if  the  fluctuation  is  between  the  maximum  load  and 

(*oni[)letc  rofft, 

ai=d  =  u    and    F  =  l. 


weyrauch'S  formula. 


249 


If  ai=a2=s,  i.e.,  if  the  fluctuation  is  between  maximum  stresses 
of  equal  intensit/  but  of  opposite  kinds, 

ai=a2  =  -j-  =  s    and    **  =^» 


Weyrauch's  assumption  that 


F  = 


u—s 
2u  —  s  —  ai 


satisfies  these  extreme  conditions  and  gives  intermediate  values  of  oi 
which  appr  viniately  agree  with  the  most  reliable  results  of  the 
few  cxjierinuMits  yet  recordetl.  It  is  also  in  accordance  with  Wohler's 
deductions  tlnit  oi  incn»ases  as  d  diminishes  and  vice  versa. 


Hence 


Ol 


==Fd 


u—s 


2h— s-oi 


(01+02) 


or 


where 


2u— a— oi  =  (u— s)(  I H — -) 

=  (H-S)(1+^), 

^    ai 


(S  ^  H     \ 

iiiul  taking  3  as  a  factor  of  safety, 

the  working  load=fA  =A-^ 

Axil       s— M,\ 

(L)  and  (W)  may  be  written  in  the  form 

the  working  load 
the  workmg  stress—  —  ■■   -. 


(W) 


-i(-'?*> 


250  THEORY  OF  STRUCTURES. 

and  in  ordinary  practicp  it  may  Ix*  assumed  that 

M  is  30,000  lbs.  for  wrouglit  iron  and  48,000  lbs.  for  steel. 
Also,  if  the  varying  stresses  are  of  the  same  kind, 

/  — «  .    1  9 

—^  IS  ^  for  wrought  iron  and  ry  for  mild  steel, 

while  if  they  are  opposite  in  kind, 

/-" •      1  5 

—^  IS- 2  for  wrought  iron  and  -jj  for  mild  steel. 

Ex.  14.  Find  the  proper  sectional  area  of  a  bar  of  axle  iron  which  has 
to  carry  loads  varying  from  a  maximum  pull  of  110,000  lbs.  to  a  minimum 
pull  of  44,000  lbs. 


Therefore 


The  working  stress  =  10000 ( I  +^  {^^)  -12000  lbs. 


the  required  sectional  area  = =9*  so.  in 

12000        •     ' 


Ex.  1.-).  Find  the  working  stress  of  a  mild-steel  girder  which  has  to  carry 
a  dead  load  of  I)  lbs.  and  a  niaxinmni  live  loud  of  L  lbs.,  and  apply  your  result 
to  the  case  of  a  lattice  girder  when  the  live  is  .i  times  the  dead  load. 


The  working  stress  =  l(iOOOM  +—  --) 

V       11   /,/ 


=10000( 

=  2a3G4  lbs. 

Ex.  16.  A  mild-steel  bar  has  to  carry  loads  which  vary  between  a  max- 
imum tension  of  .W.CKK)  lbs.  and  a  maximum  compression  of  42,000  lbs.  Find 
its  sectional  area  (disn-jiarding  buckling). 

The  working  stress  in  Iba./sq.  in. 

-leooofi--  ■*'^^^) 

\       1 1  5<)000  / 


-16000X 


29 
44' 


Therefore 


the  sectional  area  — 


.yiooo 

ItJOOOX} 


T  -5.31  sq.  inches. 


UNWINS  FORMULA. 


251 


I'mvin's  Formula. — Unwin  has  proposed  to  include  all  cases  of 
lluctuating  stress  in  the  formula 


ai  =  :^  +  Vt{t-nd), 


u  being  a  coefficient  to  be  determined  by  experiment,  while  the  other 
svinbols  are  the  same  as  before. 


Then 


rf  =  fli±a2. 


If  d=0,  the  load  is  steady  and  aj  =<. 

If  d  =  ai,  02  =  0,  and  the  load  alternates  with  entire  relief. 


In  this  case 


fli 


ai  =  -^  +  Vfi-nait 


and  therefore  oi  =.2t(Vl  +  n^-n). 

If  rf  =  ai,  oi  =  — 02,  and  the  stresses  are  alternately  tensile  and 
coinpressive,  but  of  equal  intensity. 


Tliorefore 


£ai 


01  =  ",^  +\^t{t-2ain) 


:tii( 


2n 


In  these  extreme  cases  if  n  is  put  equal  to  1.42  for  wrought  iron 
and  to  1.06  for  steel,  results  are  obtained  almost  identical  with  those 
^Mvon  by  Launhardt  and  Weyrauch.  The  formula  may  therefore  be 
assumed  to  ho  approximately  correct  for  intermediate  cases. 

A  mean  \alue  of  n  for  iron  and  steel  seems  to  be  about  },  so  that 
the  formula  mav  be  written 


..-|.J.(.-I4 


F.x  \7.  A  diagonal  of  a  bowstring  trtutg  haa  a  sectional  area  of  3  sq.  ins. 
iiiiil  carries  ItHids  which  vary  between  a  maximum  tension  of  14  tons  and  a  nuuei' 
mum  n^mpression  of  6  tons.     Find  the  statical  strength  (t)  of  th«  material, 

14        J     J     14+6    20 
o.--    and    d'-^'j. 


■i 


:  It' 


252 

Therefore 


THEORY  OF  STUlcnRES. 
niul  /-  I0.I7.">  Utm. 


6.  Remarks  upon  the  Values  of  t,  u,  s,  and  f.— As  ypt  the  value 
of  (/  ill  (•(ii.ij,rcs...i„M  lias  not  Ihhh  satistj.ciorily  (U'tcniiiiiocl,  and  for 
tiK'  prc'viK  its  value  may  Ik.  assuiiir.1  to  ih-  tlii'  sami'  lK)th  in  tension 
ami  coMipn'ssinn. 

If.  as  Wohjcr  states,  "  n-|H.at<'.l  strcs.ses "  an'  dotriinental  to 
the  strength  ..f  a  n.at.Tial.  then  tii.-  values  of  „  and  n  diminish  ii.s 
(lie  re|K'tili..ns  inerease  in  numlK-r.  an!  are  minima  in  struoturra 
(lesjr|„.,|  lor  .,  practically  unlimited  life. 

Only  a  very  lew  of  Woliler's  ex|H-rinien(s  pivc  the  values  of  t, 
K.  s.  and  a.  so  that  Launhardts  and  Weyraucl.s  !i.s.sumptions  for 
the  value  of  /  nmst  Im'  r.'narded  as  tentative  only  and  require  to  Ix- 
verilied  l.y  further  exiH-rinients.  The  cltsc  aKnvment  of  Wohler's 
results  from  tests  upon  uiitem|H-n-d  cast  steel  (Krupp).  with  those 
jfiv.  II  l.y  l.auiihai'its  formula  may  l»e  s..en  from  the  following: 

K(.r  /  I  KM)  centners*  |Mr  square  zoll  WV.hler  found  that  u-500 
centners  |»er  s(|uare  zoll.     Thus  (i»i  InTomes 


\        .>   r»i/ 


and  therefore  „,-'  -  .VM)<(,  -  iWNWr.,    0. 

"'•"••"•''"•  "-•    l>.       .'.-^».     4<H»,     m\.     1100, 

l.aunliardt'<  fornmla  j:i\es 

"I     ■"•(((1.     7I(»,     SIM).     <MM),     1|()0, 
w         W  ii|||rr'-i  e\iH'riiiieiii>  gave 

"I     .VM),     7(M»,     ,S(M).     <MM»,     \m). 

.\>raiii.   will,    j'lMiiix   iron.   f.,r  /     .'MM)  c-ntners    {mt  w,.   «oll.  u 
was  found  to  l„.  ;mm)  centners  [Mr  square  zoll.  and  then-f.jr.' 

"I    ••"»<>( I  •  !.'"■) 

•  A  cniiMT     1 10  .'.I  ,».u.,.U      A  -.|.mr.-  mil  -  I  IWIM  «,,iiin-  inchrt!  ' 


or 


liEMARKS  Ul'OS   THE  VALfES  OF  t,  W,  s,  AND  }. 
a  1=2 -300fl, -25002=0. 


253 


If  aj  =  240,   01=436.8,  which    tilmoist   exactly  agrees  with   the 
ult  giv«>n  by  the  tension  exiK'rinients. 

In  general,   the  iV(hnis.sible  stri'ss  |)or    square  unit  of  sectional 
;ir  ;i  niay  be  expressed  in  the  form 

/  =  r(l±m0), 

r  ;in(l  w  being  certain  co(»lfieients  which  deijond  upon  the  nature 
i>l  the  material  and  aliso  U|H)n  the  manner  of  the  loading.  Cnn- 
-iiliT  thnH'  eases,  the  material  in  each  ea.><e  Ix'ing  wrought  iron; 

{(i)  Let  the  stn*»«'H  vary  U'tween  a  maximum  tension  and  an 
i<|Ual  ma.xinmm  compression;    then 


;iiiil  therj'fore 


Vi  =  l, 
/  «=  700(  1  -  J )  =  3 ')()*  |)er  cer.  ». 


f/»)  l.'.'t  the  material  Im'  subjected  to  stresses  which  are  either 
iMi-ile  or  compn*s}<ive,  and  let  it  always  ^>tur^  to  the  original 
iiii-l rained  condition;  then 


ilicrefore 


ai-0,    or    Oi^-^O,     and     •^-O; 
/  =  7O0(l±O)-7(K)*  iMTcent^. 


((■)  I  yet  the  material  l>c  continually  subjectetl  to  the  same  dead 
Io.hI;  then 

oi  -^at, 

'I  therefore/ -700(1  4  J)-10:i0*  pj-r  cent»-14,n34  lbs.  |ier  square 
■ ' 'i.  which  is  aliout  one  third  of  the  ultimate  breaking  strength. 
Thus  in  the.si'  thn^e  nis«>s  the  a<lmissible  «tn's«e«  an*  in  the  ratios 
I  2:3,  rati(w  which  have  lieen  already  adopted  in  machine  con- 
•1  let  ion  as  the  result  of  ex|x*rience. 

"iViihlf-r,  from  hisj-xiK-riments  uiMtn  untenijieri'il  ea.xt  steel  (Kru|)p), 
(  'K'hided  'hat  for  alterations  lietwwn  an  unloaded  condition  and 
•  '1   r  a  tension  or  n  compp'ssion,  /-llOO*  fXT  cent*,  and  for  alter- 


254 


THEORY  OF  STRUCTURtii. 


nations  U'tween  equal  compressive  and  tensile  stresses,  /=580*  per 
cent-. 

It  has  not  boon  unusual  to  take 

.        01+02 

for  stresses  alto.natoly  tensile  and  compressive,  it  being  assumed 
(hat  if  the  stn'ss«'s  arc  tensile  only,  their  admissible  values  niav 
vary  from  0*  to  700*  per  cent*. 


cf         .     "2  700.4  .  rti 

Smce^  =  -     o,  =  T— -r     and     /-x=T 
«i  1+0  '.4     1 


700 


+  <!> 


(A> 


Again,  taking  m=-21(X)  k./cm.', 
safety,  Woyrauch's  formula  Ijocomos 


M— .S 


■  J,  and  3  as  a  factor  of 


/=£.'-7Ooa-i0) (B) 

Comparing  the  results  of  (A)  anti  (B) 

for  .A-  0,        1.        i,         I        1, 

rA)  gives/    700.     rm.    4f.7.     100,  MiO, 

an<l  (li)  gives/ ,700,    012,    525,     4.17.  .{.W. 

7.  Flow  of  SoHd«.-\Mien  a  duetilo  Inwly  in  strained  beyon.« 
the  .-lastie  limit.  i(  :ipprnaches  a  purely  plastic  ornlihon  in  which 
■A  siillicicntly  great  force  will  .i.inrni  t!ic  i-ody  indcfinih-ly  Indcr 
Mich  a  lurcc  the  elasticity  disapinars  and  the  material  "is  said  ro 
Ih'  in  a  Jtui,l  state.  U'liaviim  precisi-ly  like  a  lliiid.  For  example. 
II  llnw<  ihr.iiigh  nrilic.-.:  and  shows  a  contraci.-.l  section.  The  stres- 
d.v.-ln|«M|  ii,  the  material  is  called  the  //,/,//  i,res.sure  or  vm'ilu-ind 
ii'i  ifimlih/. 

The  general  principle  ..f  the  (low  of  solids,  ileduci'd  by  Trcsea, 
may  Im>  enunciat'd  as  follows 


FLOW  OF  SOLIDS. 


255 


A  pressure  upon  a  solid  body  creates  a  tendenqi  to  the  relcUive 
viotion  of  the  particles  in  the  direction  of  least  resistance. 

This  gives  an  explanation  of  the  various  effects  produced  in 
inatrrials  by  the  operations  of  wire-drawing,  punching,  shearing, 
roiling,  etc.,  and  in  the  manufacture  of  lead  pipes.  I'robably  it 
ulso  explains  the  anomalous  behavior  of  soliils  under  certain  extreme 
coiuiitions. 

Hiiils  which  have  been  in  use  for  some  time  are  found  to  have 
acquireil  an  elongated  lip  at  the  edge.  This  is  doubtless  due  to 
the  flow  of  the  metal  under  the  great  pressures  to  which  the  rails 
lire  continually  Sabjected.  Other  examples  of  the  flow  of  solids 
lire  to  l)e  obwTved  in  the  contraction  of  stretched  bars  and  in  the 
swelling  of  lilocks  under  compression.  The  period  of  fluidity  is 
trri'uter  for  the  more  ductile  materials,  and  may  disappear  altogether 
tor  certain  vitreous  and  brittle  substances. 

In  punching  a  piece  of  wrought  iron  or  steel,  the  metal  is  at 
tirst  compressed  and  flows  inwardj<,  while  the  xhfiarintj  only  cuin- 
lucncos  when  the  opposite  surface  begins  to  open.  A  case  brought 
iiiiilcr  the  notice  ')f  the  author  may  l)e  mentioned  in  illustration  of 
tliis.  The  thicknes.s  of  a  cold-punched  nut  was  1.75  in.'^..  the  riut- 
holc  was  .3125  inch  in  diameter,  and  the  length  of  the  piece  punched 
iHii  was  only  .75  inch.  Thus  the  flow  must  have  tak»'ii  place  through 
ii  (li'pth  of  1  in.,  and  the  .^hearing  through  a  <lepth  of  75  inch  Hence 
tile  surface  really  shorn  was  s  X  .HrJ5x  .75  =  .7;W»  square  inch  in 
;iic,i.  and  a  meo!<ii,-e  of  the  shearing  action  is  the  pro«luc(  of  this 
urlacc  an'a  and  the  fluid  ifre.s.ture.  Tlie  nature  of  the  flow  may  be 
n|,~irvc(|  by  .splitting  a  cold-punched  nut  in  half  and  treating  the 
iradured  .surfaces  with  acid  after  having  plane<|  them  and  given 
tlnin  a  bright  (Kilish.  The  metal  lM)rdering  the  con-  will  Im*  found 
liitvtMl  downwards,  the  curvature  increiu'ing  frtMn  the  bottom  to 
till  top,  ami  \vell-d«iined  curvj's  will  mark  the  si'pnrating  planes 
"t  tlic  plates  which  were  originally  uwl   in  piling  and  rolling  the 

In  r\|)erinientitig  ujkhi  lead,  Tn'sca  placed  a  numlier  of  plate«, 
I  111  alMive  the  other,  in  a  strong  cylinder  (Fig.  HIH)  with  a 
I  "Ic  in  the  iMitloiii  I'lMin  applying  pressun'  the  lead  wjw  always 
!"iind  to  flow  vlun  tin-  cueiprienl  uj  fluidity  was  alnnit  2iS4l  llw. 
!'r  st|u»re  inch,  the  differcfuc  of  xtresn  In-ing  double  this  amount. 


256 


TIIKDHY  OF  STRl  CTURES 


Tlir  s»>pnratiii>i  planes  assuniod  ciirvod  forms  analogous  to  tho  corre- 
filHMidinji  surfaces  <if  flow  when  water  is  substituted  in  the  cylinder 
for  tile  lead. 

The  lliiw  (tf  ductile  metals,  e.p..  cop|M>r.  lead,  wrought  iron,  and 
soft  steel,  commences  as  soon  as  the  ela-tic  limit  is  «'xce<'ded,  and 
in  order  thai  llie  How  may  Iw  continuous  the  distorting  stress  must 
C'iini;nitly  increase.  On  the  other  hanil,  in  the  case  of  truly  plastir 
Ixidies.  (low  conunences  and  continues  under  the  same  constant 
stress.  It  evidently  de|H'n<ls  upon  the  hardness  of  the  material, 
and  has  Ikhti  called  the  vocjlirieut  of  hitrduess.  Hio  lomier  the  stress 
acts  the  greater  is  the  deformation  which  gradually  increasi's  indefi- 
nitely or  at  a  diminishing  rat<'. 

lA|H'riment  shows  that  there  is  very  little  alteration  in  the 
density  of  a  ductile  Ixnly  during  its  phtstie  deformation,  and  Trf>sca's 
analvticj'.l  invi'stigatiotis  are  I)as4>d  on  the  a>sumption  that  the  ixxly 
is  deloriiied  without  sensible  change  of  volume. 

("misider  a  prismatic  bar  undergoing  plastic  deformation. 

Let  /,  Ih'  the  length  and  .1  the  section  of  the  bar  at  commence- 
ment of  deformation. 

l.it  /,  ij-  Im'  the  length  and  a  the  section  of  the  bar  at  a  sul>- 
s«M|uciit   |H'riod. 

I.il  i>  Ih'  the  intensity  of  tht>  fluid  pressure. 

Siiire  the  volume  remains  unchanged, 


LA  =(/-±xVi, 


(1) 


the  iNoitive  or  negative  sign  In-ing  taken  according  as  the  bar  is 
in  li'iision  or  compression. 

I^'t  /'i  Ik-  initial  force  on  b.nr. 

Let  /'  Im-  force  on  bar  when  its  length  is  L±x.     Then 


and  hence 
Hence 


/'i     .1     I.  XX-       • 
P(A±J-)     I'll,    a  constant. 


(2) 
(3) 


tSm. 


FLOir  OF  SOLIDS. 


257 


\ 


\ 


Via.  318. 


;uiil  tliP  form  diminishes  as  ihc  har  .-Irctches  and  increases  us  the 

l,;i:  contracts  uiulcr  prcssuro.     If  ('(|iia- 

tii.ii    {'.it    \k'     rcfcrrcil     to     rectangular 

;i\i<.   the  onliiialcs  represent inji  differ-     j 

i!i!    values  of   /'  and   the  abscissa'  the     i 

(•i.in  <|HHidinR    values   of   j,    the    stress-     j^ 

Mr.iiii    diagrams,    //    in    tension    and    vc  xt- 

III  ( uiiipression,   are   hy])»^rlM)lic   curves,     j 

lia\iiiH    Jis    asymptotes    the    axis    of   x, 

\(i.\.   and   a   line   parallel   to   the  axis 

III    /  at  a  distance  troin  it  ('(jual  to   the 

liiiriii  A  of  tlw  l)ar. 

,N7 /)>«/.— {'onsiiler    a    metallic    mass 
z    had)    resting    n\xm    the    end    CI) 

ot  ;i  (vliuder  of  radius  II  and  (iiling  up  a  space  of  depth  /). 
A  hole  of  radius  r  is  made  at  the  centre  of 
the  face  CI),  through  which  the  mass  flows 
under  tiie  jircssine  of  jluidili/  exerted  l>y  a 
piston.  Wlien  the  ma.ss  has  JM-eii  compres.s<'d 
to  the  thickness  l)<>  x.  let  ij  Ik'  the  corre- 
s|)oiiding  leiiglh  /vA'  of  the  "jet." 

/•'(/>/,  a.'^sume  that  the  s|H'cific  w.vrht  of 
the  nia.s-*  remains  c<iiistant. 

If  lis  lie  the  diminution  in  the  lliickncss  l)f> 
corres|)on<ling  to  an  incresuse  dy  in  the  length 
of  the  jet,  then 

r/W/  +  rr»rft/=0.      .     .     .     ^4) 

hiii«;r:i'ing,  and  nMneinl¥«ring  that  >)-{)  when  j^-/), 

!{■{!>    X)     r-'v    0 (">) 

.'<>iotid,  !^^*^umo  that   the  cylindrical   |K)rtion   EFHII  is  ijrnduidhj 
ii:iii  inrmi'd  int.i  .\MI'LK(J.\  ,  of  which  the  part   I'MSQ  is  cylin- 
vhil<'  the  dianider  of  the  part  I'l.ki)  ijriuliudhj  inrre!US4>s  from 
ai-c  of  tl  «•  cylinder  to  /\7.(     KF)  at   the  end  of  the  jet.     Thi-n 
/ ' )(//    amount  of  metal  which  (lows  into  llw  n-ntral  cvlimler 


■h  !■ 


^'iTvrdrx,       .... 
ill;;  the  depth  to  which  the  metal   i)enetrat<»s. 


W 


2oS 


THEORY  OF  STRICTURES. 


Third,  assume  that  tho  diminution  of  the  diameter  of  the  cylin- 
drical j)ortion  l'M.\Q  is  directly  proi)orti()nal  to  the  said  diameter. 
Then,  if  2  be  the  radius  of  the  cylinder  PQXM, 


drjh 
r      z' 


By  oqs.  (6)  and  (7^  and  therefore 


Integrating, 


(/?»-r2)  log.x  =  2r2  1og,«+c. 


c  lx>ing  constant  of  integration. 
When  x  =  D,  z  =  r, 


and  therefor? 


(/P-r2)log.^=2r»log.p 


or 


Hi) 


*Lzr! 

2r> 


By  eqs.  (5)  and  (8), 


If-r* 

2rt 


which  is  the  eqijation  to  the  iirofile  I'l.  or  QK. 
If /i'2  :ir2.  e(|.  (Sj  represents  a  straight  line. 
If  f>-     2r-.  "      •'  "  "  piir.MlM.la. 

8.  Thin  Hollow  Cylinders ;  Boilers ;  Pipes. 

l/'t  r  Im-  flic  niilius  of  thf  cylinder. 

lA't  I  In-  tin-  thickness  of  the  metal. 

Ik*   the   fluid   i>resKure 


(7) 


(8) 


upon  each  unit 


or  cotnpresHive  unit  stress, 
nial  or  an  external  pres- 


tieoiw  and  free  from  initia 


THIN  HOLLOW  CYLIXDERS;  BOILERS;  PIPES. 


259 


(2)  that  t  is  small  as  compared  with  r; 

(3)  that  the  pn-ssurcs  arc  uniformly  distributed  over  the 

ititcnial  and  external  nurfaees; 

(4)  that  the  ends  are  kept  ixTfectly  flat  and  rigid; 

(5)  that  the  stress  in  the  metal  is  uniformly  distributed 

over  the  tliickness. 

The  last  a^umption  is  e(|uivalent  to  sup(x)sing  that  it  is  the  mean 
rirciiiiiferential  stn^ss  which  is  governed  by  the  strength  of  the  met^, 
wliili-  in  reality  it  is  the  internal  or  maximum  circumferential  stress 
which  is  so  governed. 

The  figure  represents  a  cross-section  of  the  cylinder  of  thickness 
unity. 

A  section  made  by  any  diametral  plane,  as  AB,  must  develope 
:i  total  resistance  of  21},  and  this  must  ho  e<jual  and  opposite  to  the 
n-siiltant  of  the  fluid  pressure  upon  each  half,  i.e.,  to  2pr.     Hence 


2</  =  2pr,     or    tf  =  pr. 


(1) 


This  formula  may  be  employed  to  determine  the  bursting,  prool, 
(ir  trorking  pressure  in  a  cylindrical  or  approximately  cylindrical 
iHiiliT,  provided  that  /,  instead  of  Ix'ing  the  tensile  or  compn»ssive 
unit  stress,  is  .some  suitable  coefficient  which  has  l)een  determined 
by  I  \|K>riment.     If  t)  is  the  etticiency  of  a  riveted  joint,  the  formula 

ijtf-pr 

iu:iy  Ik'  employed  to  determine  the  working  pressure  in  a  cylindrical 
or  Mppntxitnately  cylindrical  IniihT. 

Ill  (inlinary  practice  the  vahu^s  of  ij  an«i  /  are  given  by  the  follow- 
ing! tiible: 


.<t...  1 


MMTial. 

Jainl. 

» 

1  ill  eoumlii  iwr 
Hquwc  Inch. 

mtit  inm    

."^innle-rix  rtiMi 

I)iHll>li'-rnH(M| 

Trolilc-ns  fti-)! 

Sitiirlc-rAffitl 

i)<>iili|(<-riv<>te<l 

Tn-lilf-riveftMl 

H  to    K', 

N.tttH)  to  tt.lNNI 

N.INNI  "  <I<NII( 

. 

HUM!  "  IMNNI 

l2.INN)tii  i:<.IMNI 
12.INK)  '■   la.lKM) 
ia.(l«XI  "   ia,(KKt 

liir  cast-iron  cylinders  th«'  working  vtdue  of  /  may  \w  taken  at 
!il'"ui  JdOOHis.  i^wr  square  irjch. 


260  TIlEOIiY  OF  STRVCTUUES 

The  total  pressure  upon  each  of  the  flat  ends  of  the  cylinder 

=  rr2p. 
The  longitudinal  tension  in  a  thin  hollow  cylinder 


"  -Ir.rt     -It ' 


f2> 


and  is  one  half  of  the  circumferential  stress  /. 

Let  the  cylinder  hr  siil>jcct<'(l  to  an  external  pressure  p',  as  well 
as  to  an  internal  pressure  /).     Then 

ft-pr-p'r' (3, 

r'  iM'iiiK  the  radius  of  the  outside  surface  of  the  cylinder.    /  is  a 

tension  or  a  pressure  acc«irdinji  as  pr^pY. 

(ienerally  the  ditlerence  iM'twet-n  r  ami  K  is  very  small,  and 
ef],  (,'})  may  l»e  written 

It    rip- p') 

g.  Spherical  Shells.  I..t  the  data  Im>  the  same  as  before. 
'I  he  -ictiiin  mailf  liy  any  diametral  plane  must  develop  a  total 
resistance  of  Jrr//.     Then 


or 


2r.rlj     r.f^p, 
atf    pr.    . 


n) 


Ifeiic(»  a  spherical  shell  is  tiriw  as  strong  as  a  cylindrical  shell 
«)f  the  sam.'  diam.'t.T  and  lliickiiess  of  m«'tal,  so  that  the  strongest 
parts  of  t,j(f-,t„U'il  lM)iliTs  are  the  ends. 

Let  the  slicll  Im'  suhjccted  to  an  external  pressure  p' ,  as  wdl  ■ 
to  an  internal  pressure  p.    Then 


and 


lir'  *r)l    r-'/'     r'-p'. 


(2) 


/  is  a  tension  or  a  firessure  according  as  r-p^r^'^p' 


PRACTICAL  Rt:MARKS.  261 

(lonerally  r'-r'ia  very  siimll,  und  tho  relation  (2)  may  be  written 


it  =  ry{p-p'). 


For  a  thick  hollow  sphere  Ilankine  obtained 
P  °  *'  ^^T^  approximately. 


(3) 


(4) 


10.  Practical  Remarks.  —  A  conmion  rule  requires  that  the 
wdikiiig  i)ressure  in  fresh-water  Injilers  should  not  exceed  one  sixth 
(it  the  bursting  pressure,  and  in  the  case  of  marine  boilers  that  it 
>lii>iii(l  not  excit'd  one  seventh. 

An  lOnglish  lioard  of  Trade  rule  is  that  the  tensile  working  stress 
in  I  lie  lM)iler-plate  is  not  to  exeeetl  (»()00  lbs.  i)er  .square  inch  of  gross 
Mction,  and  French  law  fixes  this  limit  at  4250  lbs.  jier  scjuare  inch. 

riie  thickness  to  ho  given  to  the  wrought-iron  plates  of  a  cylin- 
iliiral  l)oiler  is.  according  to  French  law, 

<  =  .0036nr  +  .lin.; 

.icinnling  to  Prussian  law, 

I  ^  ^g.oo3»  _  J );.  ^  J  in  =  .003nr  4- . i  in.,  approximately, 

r  1«  iiiR  the  radius  in  inches,  and  n  the  excess  of  the  internal  above  the 
iM'  riial  pn'.s8ure  in  atmo-spheres. 

The  thickness  given  to  ciist-iron  cylindrical  lK)iler-tulx»s  is,  accord- 
I! ;:  t..  Fn'ru'h  law,  five  times  the  thickness  of  equivalent  wrought- 
!r.!!  Tain's;    according  to  Prus-sian  Ian' 

(  =  (,.oi"_  l)r  + J  in.=.OlHr  +  i  in.,  approximately. 

Stram-lH)ilers  IWon*  In-ing  uwmI  .should  l)e  subjected  to  a  hydro 
i     ■■  test  varyujit  from   IJ  to  3  times  the  |)res.sure  at  which  (hey 

'<!  Ik-  worki'd. 

I  liiliaini    cimducted   an   extensive  series  of   experiments    upon 

■  nll;i|)siiig    slriMiptli    of    riveted    plate-iron    flues,  by    enclosing 

'!i    llius  ill  larRer  cylinders  iiiid  subjecting  them  to  hydraulic  pn's- 

-  I  Fniin   these;  exix'rinieMt.-  he  deiiund  the  following  formula 

fir  ,1  urouijlit-iron  cylindrical  flue  or  tul)c; 


262 


THEORY  UF  STUICTURES 


Collapsing  prcssiirc  i  /2.i8 

(Is  (H>r  .sciuarc  iiicli  of  surfaco  j  =/'  =  '*03150  -^, 


in  poun 

/  iM'ing  the  tiiickness,  r  the  radius  in  inclifs,  and  /  the  length  in  feet. 

This  formula  cannot  be  relied  u]Hm  in  extreme  cases  and  when 
the  thickness  of  the  tul)e  is  less  than  i|  in. 

//(  practice  t-  tuny  lie  ifenemlly  uxcd  inxUml  of  /*  •«.  The  exjieri- 
ments  also  showed  that  the  strength  of  an  elliptical  tulx'  is  almost 
the  same  :us  that  of  a  circular  tuln-  of  which  the  radius  is  the  radius 
of  cun'ature  at  the  ends  of  the  minor  axis.  Hence,  if  a  and  b  are 
the  major  and  minor  a.xes  of  the  ellip.se,  the  above  formula  becomes 


/)  =  403i:)0 


«-'  Ir- 


By  riveting  angle  or  T  irons  around  a  tulie  its  length  is  vir- 
tually dimini.slied  and  its  strength  is  then-fore  increased,  as  it  varies 
inversely  as  the  length. 

The  thickness  of  tulx's  subjected  to  external  pressun*  is,  accord- 
ing to  French  law.  twice  the  thickness  of  tuln-s  subjecteil  to  interior 
pressure  h\n  uiuler  otherwi.se  .similar  conditions;  according  to 
l'ru.^«ian  law  the  thickness  of  heating-piix-s  is 

t  =  .OOiud^' n  +.05  in.,  if  of  sheet  iron, 


ami 


t  =  .Olrfi^i  +  .07  in.,  if  of  brass. 


According  to  Keuleaux,  the  thickness  (/)  of  a  round  flat  plate 
of  radius  r,  subjected  to  a  normal  pressure,  uniforndy  distributed 
am*  of  intensity  p,  is  gi\-en  by  the  fornmla 


or 


/        )2  /. 


iKc.rding  as  the  plate  is  merely  supported  around  the  rim  or  is 
rigidly  fixed  around  the  rim.  as,  e.g.,  the  en<l  plates  of  a  cylindrical 
boiler:  /,  as  iM'fore,  is  the  coefficient  of  strength.  The  correspond- 
ing deflections  of  the  plate  are 


im  »"-  |(r)*l- 


TABLES.  263 

A  practical  rule  for  the  thickness  t  ins.  of  a  loam-mouldc<l  cast- 
iron  water-main  under  a  head  of  /i  ft.  of  water,  and  of  diameter 

d  ins.,  is 

Formerly  guns  and  cylinders  were  cast  round  chills  for  the  pur- 
pi)-('  of  equalizing  the  stress  over  a  cn)s.s-.section.  The  insid(>  hot 
iiiital  was  at  once  cooled  and  was  subjected  to  compression  by  the 
(Miitraction  of  the  more  slowly  cooling  outside  metal.  With  the 
>:iiu(>  object  in  view  it  is  now  a  common  practice  either  to  shrink 
ring  upon  ring  or  to  wnd  wire  around  an  internal  tube. 

TABLES. 
The  Strbnoths,  Elasticities,  and  Weights  or  Varioi-s  Ati-ors,  etc. 


Mkterud 


Weinhi 
{in  HiiuikIh 


Tenaiie 
Sirpimth 
in  riinu- 

l'iiun<l« 


II-: 


Aiiiiuinium  .  

•nnekled 

r»«t         

rolteti.  .  

bninie  (90%  Cu.  10%  Al) 
rft.«t.  . 
<irilin»ry    yrllow.   rB»t    (66% 

.14%  Znl 

nnlinary  yellow,  rollol. 

»llM>t 

wiTV 

I   'ppiT.  ra.'-t 

rollwl  or  wn»ucht 

with  .2  ti.  .4%  P 

wire,  aDiiralrf) 

h»nl  ilrawn 

*'.*Tmun-(*ilver  wire 

I. "I  I.  .I'liwri.     

i: . ,  mptsi  (90%  Cu.  10%  Sn) 

l-'i.i 

<*IUi( 

wire 

P'  iMniim  wire 

I'lM^lthor  hninM 

cmif.  .  

wir»(h«nl) 

"     (UIIMSM).  .  .  . 

Si-  n-   'Irawn 

s  -1  -iii.ier 


Cu.  Ft.        I-""""'"   1    p„uml. 


I'u, 


/in.' 


.■•»>it 

r..ll«l 

Mirni.  r.i|ie.     .  .  . 
luiOtrt  heltlns. 


163 
lliS 
III 
170 

I    490 

>     1" 

530 

490  to  530 
53:1 

550 


555 

laoo 

540 
710 
700 


1.140 


ASA 

4«A 
436 
450 


C«fn- 

pretiHive 

Streiiicili 

ill  Thou- 

I   MUxtn  of 

Pountlx 


2H  lo  33 

13  5 

9  to  13 

13  lo  23 

90 

30 

30  to  27 
33  to  54 

:iO 
45  111  56 
IM  |o27 
29  to  ,36 
45  lo  49 
40  to  47 
58  to  68 

tl7 
3A  to4l 
27  to  38 

19 
3 

3  1 

no 

36  to  40 

55 

100  to  150 

50  to  60 

43 

6  7 

2  3  tn5  ( 

2  3toS  ' 

30 

9  toll 

4  5 


MfKluItu 

of  Uan- 

tirity  in 

Milliona 

o(  PouiiiIk 

l*r 

Sq.  In. 

(E). 


MimIuIus 

of  Itifid- 

ity  in 

MilUonii 

of  Poiinils 

l>er 


9tul0      3.4to48 


10. S 


14.5 

9  to 
13.5 


5.6 

1,4  5  to 
I   5.6 
5  5 


5  8 


9  to  13.5 
14.3   I 
1  II  to  14  I  4.3  to  5 
'  12  to7  <   4.7  to6.5 


7.1 


17   I 
I 
13 

10  to  11.3 
.72 

1 
24 
13.5 
14 


II 
« 


4  7 

4  3 

'^.6to«  7 


5.35 


MICROCOTY    RESOIUTION    TEST   CHART 

iANSI  and  ISO  TEST  CHART  No    2l 


1.0 


I.I 


i^  12.8 

m 

..  y2 

m 

I:i^ 

t   Ui 

zo 

IL25  ■  1.4 


.8 


1.6 


A 


^PPLIEn  IM/1GE  J 

■•■'1^  Km  ua-    .i.i.,t 

"ycm.f,   Nv«   tj.i,        ueOT       ISA 

' ' '  s)  «»;'     ojoo     p^on» 


264  THEORY  OF  STRUCTURES 

The  Strexoths.  FxAsiiriTiEs,  and  Weicjhts  of  Iron  and  Steel. 


Com- 

Modulus 

Modulus 

Weinht 
in  Pounds 

StretiKth 

pn   »  \e 
Str-    .th 

Shearinc 
otrength 

of  Elas- 
ticity in 

of  Rigid- 
ity in 

Material. 

i:.     ■.   u- 

in  ihou- 

Millions 

Millionit 

,-  '"^r 

saii,.s  of 

sands  of 

of  P<iunds 

of  pounds 

Cu.  ft. 

per  .siq.  In. 

I'oiinds 
perSq.  In. 

Pounds 
per  8q.  In. 

per 
8q.  In. 

per 
8q.  In 

(E). 

(G). 

Cast  iron.                   

430  to  470 

11  t..33 

5ii  to  145 

13  to  20 

10  to  24 

3.8  io7.« 
6.3 

<  ast  iron.  averaRi' 

450 

l.S 

00 

20 

10 

Hary.  Lowmnor  ami  Yorksiiire. 

480 

53  to(i5 

averaKe 

480 

50 

36  to  45 

45 

28  to  30 

'0  5 

>triicttires. 
IMatrs,  finest  Lownioor,  Vork- 

480 

40 

50 

40 

28  tc  30 

10.112 

>liire.  or  SiaffordHtiire  twitii 

lilirei 

'       480 

58  to  65 

I'hiic*-*.  finest  Lowmoiir.  \  ork- 

-iiirc.  or  Slatforfinhire  (acroM} 

fihrc.     .  . 

480 

53 

riatc-.  averase  (with  filire)  .  .  . 

'       480 
480 

52 
45 
45 

78  t<i  OO 
ti7 

18  to  27 
36  to  45 

25 
27 

e.5 

(arros.s  fibre).  . 

Soft  iron.     .  . 

10.8  to  11.3 

Wire,  charcoal  (haril  <lrawii) 

(annealed) 

Harx,  "ociial.                    

ill 

100  to  1,30 

eu  to  150 

120 

20  to  42 
30 

(  !i»t  .H(vl.  iiiiiemi>erecl 

12 

"      drawn 

teinixred 

400 

33  to  1(X) 

36 

14 

('astinK-* 

annealeil 

50  to  78 

180 

8.1 

54  to  80 

Chrome  steel 

.Maiittanese  steel,  cant 

i 

.Mild  '  verv)  steel 

steel  liars  and  iilalen  (  2*^ 

j 

(1. 

400       1 

03  to  72 

28  to  3) 

13  to  13  5 

steel     liars,     plates,    and 

shapes  for  structures. 

400 

52tn68 
75 

00 

200 
f'lO  to  KM) 

120 

47  to  56 

32 

30 
30 

Nickel  sttvl.  iinhardenedi 

I'l'"  (')  annealed  . 

'■     (12'-;  ci     •• 

Soft  steel,  tinhanlencil 

400 
400 

hardetied           

Si.tI.   hiirh  carlion   (hardi-ned 

l>\'  sn((tiier''ion  1.  . 

266  l»  400 

Steel  rails  i ,  1' ,  (   I 

7S  to  lOll 

!  ittlK-Ien  steel. 

ItK) 

\\  ire.  ordinal  v 

tetniiereti 

224  to  .330 
270  to  336 

28 
26 

pianoforte 

Sthenoth  or  Wire  Ropeh. 

Ilrrakinit  strength  in  p.iuiids  (for  working  strength  factor  of  s»fefy-5). 


I»iltneter 

llM-he-, 


* 

I 

1 

it 
II 

a 

It 
2i 


Swe.|i«h  Iron. 

lleniji  ( "eiitre. 

''  Striiiid-, 

U' Wires  lo 

Straii.l 

.'lOno 

S  >.  1  I 

I!l  KKI 
.((.IKMI 
.',()  (KHl 
7.'  I  lot) 
'»»".  O'Hl 

I:m  iioii 

I'K.IKKl 
tf;-:  jiiiii 
2-.'>t,{Hio 


Cast  Steel. 

Hetnp  Cenife; 

•  1  Strniids, 

in  Wire-  lo 

St  mild 

Kl.tNIO 
1 7  r,i  111 

.•«•  HIK) 
(IS.IKIO 
IlKI IKHI 
I  It  IMUI 
IIIL'IHUI 
L'4><  110(1 
312  000 

--n  arm 

4.'MI<IOt) 


Crieiiile  steel. 

Ileiiiii  Centre; 

Ci  Siraiuls, 

ill  W  ire.  to 

Strand. 

1 1  ..'.(M) 
20  21  HI 
4  t  IMMI 
7«0OO 

IKIlKKI 

WW  IHIO 

?'.' '  (Kill 

2S8,IMI(I 

.'Kit. 1 100 

'i  i  1  T>-'ni 

Mi  (ton 


Plough-steel. 

Hemp  Centre; 

Criieihle  steel 

6  Strands, 

7  wires  lo 

1(1  W  ires  to 

Stiaid 

Strand 

13.100 

ii.ino 

'.'■■  .HI'll 

10400 

.'ilKHIO 

42  (KHl 

8fl(H)0 

74  000 

niodo 

1I2(KK) 

in2iKH) 

158,000 

2"(i  (100 

.■130  (too 

41(1  (KKI 

.vw  (ion 

«■  10  (KK) 

TABLES. 


265 


Table  giving  Strength  and  Ductilitv  Tests  of  Hahd-drawn  Copper  Wire 
FOR  Telegraphic  Purposes, 


Diameter,  Mils, 

Weight  per  Mile. 

Breaking 

Weight. 

Twiitsi 

B&S. 

<i;i'iKe 

.N(j. 

Average     Mini- 

Maxi- 

Average'   Mini- 

Maxi 

Average 

Mini- 

Average 

Mini- 

Keqijireil   mum. 

mum. 

Kequireil    mum 

mum. 

Uequired 

mum. 

Kequinxl 

mum, 

12 

80 

79  3 

81.2 

102.6       100  8 

105.7 

334 

327 

40 

36 

1(1 

101  9 

10U.9 

102. 9 

105. U    '    162  7 

169.2 

53,-. 

516 

34 

29 

V 

1U4. 

103. 

lo.-.. 

172.8    1    169.5 

176.2 

555 

536 

33 

28 

114.4       113  4 

11,'i  4 

200.1       20.-)  .-j 

212.8 

670 

640 

.30 

25 

8 

128.5        127  5 

129.5 

263  9       2.')9.8 

268 

840 

811 

27 

22 

?» 

137.          1,36 

13.^ 

3(X)           295.6 

304  3 

950 

917 

^.5 

21 

144.3    1    143  3 

145.3 

332.8    i  328  2 

337.4 

1013 

1013 

24 

20 

The  ductility  test  is  made  by  pn.perly  gripping  the  cample  wire  at  the  ends  by  two  vises 
whcise  jaws  are  6  inches  apart  and  causing  one  vise  to  revulve  uniformly  at  right  angles  to  the 
« irp  at  a  uniform  speed  of  about  one  revolution  per  second.  1  he  tests  are  reckoned  by  the 
inimlxT  of  complete  revolutions  made  by  the  revolving  vise  at  the  moment  the  wire  breaks. 
I'he  wire  should  be  capable  of  being  wound  in  six  close  turns  around  wire  of  its  own  diameter 
:iti<l  unwound,  without  breaking. 

Iron  Telegraph-wire. 

The  wire  should  be  soft  and  pliable  and  capable  of  elongating  15  per  cent  without  breal|ins 
after  being  galvanised 

The  breaking  strength  should  not  be  less  than  2i  times  the  weight  of  the  wire  in  pound* 
1  (T  mile. 

The  ductility  test  should  be  matle  as  follows;  The  wire  is  gripped  by  two  vises,  whose  jaw» 
:irp  6  inchc-i  apart,  anil  one  vise  is  causeil  to  revolve  uniformly  at  right  angles  to  the  wire  at  a 
uniform  s[)eed  of  one  revolution  iier  second.  The  number  of  twists  in  6  inches  should  not  bu 
U—'  than  15. 


Strbngth  or 

Manila  Rope. 

(  irnimference, 
liiihes. 

Breaking       , 
Strength,       | 
Pounds. 

Circumference, 
Inches 

Breaking 
Strength, 
f'ounils. 

t'ircumference, 
Inches. 

Breaking 
Strength, 
Pounds. 

» 

1 

u 

2 

2» 

3 

4 

560 

784 

2,700 
4,300 
6.100 
8„'>00 
ll„'i00 

6 
7» 

14,800 
18,400 
21,900 
25..')00 
29,100 
32,700 
36,300 

8 
9 
10 
11 
12 
13 
14 
1 

39.900 
47,(K)0 
54.200 
61.4IH) 

75,700 
82,900 

Tim:  Hmkakino  Weiohts  and  ("oEFFiriEMs  or  Mending  Strength  in  Tons  (op 
•_»2I(>  i.iiH,>  OF  Vaiiiois  Uectangilau   Hkams,  the  Weights  having  iieen 

I'nIFUHMI.V    DlSTHIIlI-TKD. 


Material. 


>•  II  .'V  iii'ie  (Q'b.lw),  II  joj.ts 

2  limtma 

'  '•  '  ililii''!.    3  beams 

II  (olsis 

i>«wcli»h>.  2  loists        

.     <•  I  H.iiiii'i,  2  Iwams.         . 
;.Mii.  r,..|vv.Kh|  deal  iWvherg),  ajolsls 
-ri-f  .I»!il.  I, St.  Joiin),  3  |.»it"  with 
I'li'lirltiK  ple.'es 


I'lear 

Mean 

Span 

Mreaillh 

Depth 

Brrnkiiis 

CiwHiclent 

lielween 

111 

in 

W.ighl  of 

..r  llen.ling 

8upiM)rt  8 
In  Inches. 

Inches. 

Inches. 

ea<h  .loist 

Strength. 

or  Henni. 

142 

3 

9 

3  IV. 

■J    '8 

142 

34 

11 

7  89 

1  OH 

126 

14 

15 

no  07 

1  m 

126 

14 

14 

46  6 

10 

142 

3» 

11 

8  20 

2  t)« 

142 

3 

9 

5  7 

2  40 

126 

13) 

13t 

58  43 

3  34 

142 

3 

B 

5  7.'i 

3  53 

142 

3 

e 

6  HI 

>.«• 

266 


'THEORY  OF  STRUCTURES. 


The  Stkenoths,  Klasticities,  axd  Weights  of  Timders. 

This  table  contain"  the  results  nf  the  most  recent  and  most  reliable  experiments,  but,  (jen- 
erally  siiealtinn,  only  small  specimens  of  the  material  have  lieen  tested.  It  is  found  that  tha 
strength,  elasticity,  and  weight  of  a  timber  are  affected  by  the  soil,  age,  seasoning,  per  cent  of 
moisture,  position  in  the  log.  etc..  and  hence  it  is  not  surprising  that  specimens  even  when  cut 
out  of  the  .same  log  show  results  which  often  differ  very  widely  from  the  mean.  Additional 
exiwriments  cm  large  limlwrs  are  needed,  and  in  each  case  should  be  accompanied  by  a  com 
plete  history  of  the  specimen  from  the  time  of  felling 


Description 

of 

Timber. 


Ten.sile 

Stretigth 

in  Tons 

pcr.Sq.  In 


Acacia 

Alder 

Apple 

Ash,  Canadian  . 
Ash.  English. . . 

BefB'i 

Birch 

Box 

Blue  gum 

Cedar 

Chestnut 

Kbony. 

Elm.  Canadian. 
Elm,  English  . .  . 
Oreetiheart. 

Hawthorn 

Haiel 

Hctnibeam 

Iro!d>ark 

Ironwooil.    .  .  . 
Jat-nih 

I-atHTWlMUl         .   . 

La;,  h      

Lignum  vita*.  . . 

I.iicn-t 

.Mahogany, Span 
ilond. 

.Maple 

Mora 

•  Onk.  Vni 

( >ak.  Atn    reil.    . 

white.    .  .  . 

■•      l:rig. 
Pine.  Ihuitiic. 

"      Memel.   . 

pilch 

»"       red 

"      red. 
•' '       yellow  , 

"       vcllow  . 
•"       white. 

"       white.  .   , 

Plane 

Poplar 

•  .'^iTHce. 


4.5  to  6  ;i 

8.8 

2.45 

5.35  to  l.^iS 

4.9  to9.N 

6  ti9 

9 

2.7 

2.23  to  4.(1 


Com- 
pressive 
.Strength 
in  Tons 
pcrSq.  In. 
along 
Eibres. 

7.1 
3.1 


Shearing 
Strength 

in  Tons 

I>er  Sq.  In. 

along 

Fibres. 


4  3 
5  89       I 
2.7  to4.ll4 
4,li8 
8.48 
9  1 
7   12 

4  31 
1    31 

3  1)  tofi  : 

3.92to4..V 

5  211 

4  .')  tofi  : 
17  to7  3 
13  to3  ij 
4  7  to7 

4   1 


2.5 
4 

3  fi7 
.47  to  2.2; 

4.fi 

3.u78 

2.66 

2.9 

4  0 
46  to  6.5 


.2  to  .312 
25  to  !364 


2  fi 

4  ,54 

5  21 
3.2 


42  to  2.45 
4  46 

1  ;J3 
3.3 

2  23 
4  4 


.308 


Young's 
Modulus 

E 
(in  tons). 


620 
723 
lit  17 
734 
803 
509 
217 
509 

1100 


759 


.5.33 
3  3 


446 


Coeffi- 
cient of 

of 
Ri«i<lity, 

O 


Coefficient 

of  Bending 

Strength 

in  Tons 

perSq.  In. 


Synirnore. 
Teak 

Willow.  .  . 


4  46 

8  8 

r,  4 

3  5 

4  5 

4  II 

1  7  toO  t.7! 

,2;;t.ir.  s7 

I    13  lo5.1 
I       5.4 
2.94 

2  99  to  5  97 
i  3 

5  8 

4  7toO  7 

3  ,'■. 

.  4.6  to6.25j 


,      830 

. . '         !      .^77 

89  to  2.6  ..324  to  .446 


2.84 
4  4 

.33.'*  to  .131 

664 
1025 

3  5 

♦ 

j    670 

2  4  to  3 

962 

2^4  to3  6 
2  24 

779 
.  227       900 

..  484 

.119  to  .1641 

604 

1  8 

'    340 

594 

.r  10 

5  .3.^ 
2  7 

4.38  to  737 
.113  to  .167    700 

464 

1000 

1  S 

629 

5  86 


7.18 

560  In  1,3.39 
712  to  879 


8.13 
4.13 


Weight 

in  pounds 

per 

Cu.  ft. 


2.712 
4  56 


2  93 

3  71 

3  255 

4  51 
2   146 
3.03 


2  18 
1.63  to  2.86 


50 

50 

47 
43  to  53 
43  to  53 
45  to  49 

64 

35  to  47 
35  to  41 

47 
34  to  37 
58  to  72 


*7i 


42  to  63 
32  to  38 
41  to  83 

53 

35 

49 

57  to  68 

01 

61 
49  to  58 

36 

34 
41  in  58 

34 

34 

32 

30 
40 

23  to  26 


29  to  32 

:'.r,  I,,  43 

41  to  52 
38  to  57 
24  l<i  :)5 


•The  results  for  these  timbers  are  dediired  from  er.periments  carried  out  by  B>utchinf«r, 
Lania,  and  others  un  cumparatlvely  larga  siieclmens. 


TABLES. 


267 


The  nRE.\KiNO  Weiohts  and  Coefficients  of  Bending  Strength  in  Tons  (or 
2240  i.B.s.)  OF  Various  Rectangui-ar  Bf.ams  Loaded  at  the  Centre. 


Material. 

Clear  Span 
between 
Supports 
>n  Inches. 

Breadth 
in  Inches. 

Depth 
in  Inches. 

Breaking 
Weight 
in  Tons. 

Coefficient 

of  Bending 

Strength. 

Remarks. 

129 
129 
45 
45 
45 
45 
45 
45 
129 
129 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
45 
41 
45 
45 
45 
45 
45 
45 
45 
1.19 
147 
147 

14 
14 

5 

5 

5 

5 

2i 

2i 
14 
14 

5 

b 

5 

5 

l\ 

5 

5 

2i 

2* 

5 

5 

P 

5 
I* 

6 
6 

15 
15 

7 

7 

5 

5 

J' 

15 

7 
7 
7 

7 

it 

7 

ii 

i' 

7 
3 

|l 

12 
12 

38.15 
34 

5  9 

5  7 

3   1 

3.05 
.925 

1.075 
59  25 
60.25 

7.8 

9.75 
10.65 
11 

16 

1.35 

7 

8.5 

1  125 
1.2 

14.0 
15  A 

2  65 
2.6 

14 
11.45 

3  85 

4  00 
3  55 

24.5 
7.5 
8.45 

2  34 
2.09 
1  62 
1.57 
1  67 

1  64 

2  04 
2.37 

3  64 
3  7 
2  14 
2  68 

2  92 

3  03 

3  52 
2.97 
1.91 
2.34 
2.48 
2  6< 

4  1 
4.29 
5.84 

5  73 
7  5« 

6  31 
9  H25 
8.81 
7.82 
8.87 

1  01 
2.  IS 

>•                  t  « 

•  1                  It 

Old  timber 

•  •                     4* 

'  '                 '  * 

•  t                     «  t 

•  •                     It 

Old  timber 

■  1                     It 

*'        " 

.,          •• 

•t          t , 

.\merican  elm 

Oldtimbw 

««               •• 

tt           >« 

t  I               <* 

,,            II 

■  1               *l 

II 

•  1 

II 

• . 

N.B.— The  results  contained  in  the  last  two  tables  are  mainly  deduced  from  experimenti 
carrif.l  out  under  the  supervision  of  W.  Le  Mesurjer,  M.  Inst.  C.  E.,  Dock  Yard,  Liverpool. 

Average  of  the  Results  obtained  bt  the  Author  with  Beams  of  Large 
Scantling  in  the  Testing  Laboratort,  McGill  University.* 


Canadian  Timbers. 


Coefficient 
of  Bending 
Strength  in 
Pounds  tier 
8q.  In. 


Coefficient 

!  of  KlB.<<li<'iiy 

in  I'ounds. 


Weight  per 
Cubic  Foot 
in  i'oun<ls. 


DoiiKlas  fir— sperinlly  selected,  free  from  knots,  and 

nut  of  log  at  a  distance  from  the  heart.  .  .  ■ 

I>  MiglaK  tir— ordinary  first  quality 

IN'  I  pine.  .    


•  ■■..'  tielaits  of  the«?  tests  the  reaiier  is  referred  to  the  Transactions  of  the  Canailian  Society 
f  (  ivil  Kngineers,  Vol.  IX,  1895,  and  Vol.  Xlt,  18B8. 

f'unTHEii  Results  obtained  by  the  Author  with  Beams  of  Large 

Scantling, 


Timber. 

Coefficients  o(  Strength  for 

Shear. 

8p.  Wt. 

Bending. 

Tension. 

10.000 

tt,M)>t 

8.000 

9,000 

Compression. 

3.800 
A.9M 
3.200 
3.200 

Ui.ite  Dine.  . .  ....,, 

4.800 
5.400 
8,000 
8,000 

340 

.WO 
380 
360 

2fl 
3.1 

Mirniock 

33 

30 

Ji; 


26S 


THEORY  OF  STRUCTURES. 


Coefficients  (a)  of  Linear  Expansion  per  Unit  of  Length  from  32°  F 

212°  F  ^^     r 


Materiala. 


Brass,  .  .  . 
Bronae.  .  . 
Cast  iron, 
('"pper.  .  . 

Fir 

GlaiK 

Gold 


.001K68 

.U0I82 

.OU1075 

.001718 

.00352 

.00801 

.001 406 


Materials. 


Gun-metal 

Iron  wire 

Lead 

,  Oak 

Platinum 

Silver 

Steel,  unhardened. 


.00181 

.00144 

.0002848 

.000746 

.000884 

.001009 

.001079 


Materials. 


Steel,  hardened |  .001247 

Tin I  (X)2173 

Wrought  iron  (bar).  .;|  !ooi235 

Wrought      Iron!  .001182 

(smith's) I  .002941 

§!"<•■'■»»* , !  .003108 

^inc,  hammered 


The  Weights  and  Crushing  Weights  of  Rocks,  etc. 


Material. 


Asphalt 

Budalt,  Scotch.  .  . 

Green.-^tone. 

Welsh 


Weight  per  Cubic 
Foot  m  Lbs. 


B.'lon 
Brick, 


common 

stock  (Kng.). 
Sydney,  N.  S. 


yellow-faced  (Kng.). 
□lue. . . 


StafTordshire 

**       lire 

' '       prcsseil  (best) 

Bnckivork 

Cement,  Portland..  . 
Uoman.  .  .  . 

Clay 

Concrete,  ordinary.  . 

„    "         in  cement. 

Earth 

Firestone 

Freestone 

Glass,  Hhit 

"       crown 


common  green. . . , 

"       plate 

Granite,  Aberdeen  gray, 
red... 

"         Cornish 

"         Sorrel 

Irish    

"         I".  S.  (Quiiicy). 

■•         Argyll ',.. 

Gneiss 

Limestone    

Lime,  (jiiick 

Mnriur  

favcrjige) 

Masonry,  t*(,rnnion  brick  ' 
in  cement 
rnbble 

Marble,  statu. iry 

niiscellanenus 

Ocilile,  Portliiiid  stone 

Htith  stone 

Sand.  c^uiiriK 

"       n\  cr 

"       pit 

"       fine 

Sandstone,  re^l  'Kna. ) 

Pcrltv-  itrit 

PHvinR  (Kng.) 

.■^...trll 

IS 

Rcori.'\  paving  lilor-ks  (nn  ed"") 
(on  flat),  . 

S'ltf'irles 

Slate,  .Vtiglesi^A 

"   Welsh 

Trap.  .  . 


1S6 
184 
18. 
172 


100  to  13fi 


150 

112 
86  to  94 

100 

119 

119 

137 
77  to  128 

112 


192 
157 
158 
172 
163 
165 
166 
167 


96  to  175 
154  to  •6'? 

.53 

8(j  to  119 

106 

116  to  144 

170 
168  to  170 

151 

123 

177 

117 

100 
95 

l.^l 

150 
I.WtoI.57 
153  to  155 


88 
179 
157 
180 
170 


Crushing  Weight 
in  Lbs.  p^r  Sq.  In. 


8,300 

17,200 

16,800 

800  to  1,400 
650  to  800 
2,250 
2,200 
1,440 
7,200 
1,700 
10,200 

1,700  to  e.oco 


460  to  775 


19,600 
3,000  to  3,500 
27,500 
31,000 
31.000 

10,800 

14,000 
12,800 
10.450 
15,000 
10,900 
19.600 
7,500  to  9,000 

120  to  1:40 

500  to  800 

760 
jS  of  cut  stone 

3.200 
8,000  to  9,700 

4,100 


6,700 

3.100 

8.700  to  6.000 

5,300  to  7.8C'1 

5.300 

lO.IHK) 

11,250 

io,nnn 

24,000 


TABLES. 


269 


Crushino  Strength  of  Granitbs,  Limestones,  and  Sandstones. 


LocBlity. 


Aberiieen 

<'<iruish 

Dublin 

Newry 

I'atapsoo 

Bav  of  I'undy 

City  Point 

Dix  l:<lanil 

Da'.utli 

Fo  c  Island 

^ifpawiclt 

Ha  bor  Quarry 

Hurricane  Island 

New  Haven 

I'lirt  Deposit 

Quincy 

KDckport 

VinalnaVfcD 

Westerly 

Hunn  Island 

Kcene 

Sit.  Raymond 

Monsoon 

Npw  London 

iUohmond 

Siiiiiv  ("reek 

Mt.  Johnson  (dark  gray) 

Nei>eaii  (bed  vertioal) 

(bed  horiiontal) 

Stanstead 

(black  and  white).  .  .  • 

ynmy.  Mann,  (bed  vertical).    ■ 
(l)ed  horiiontal). 

Si    INiilip,  Q.  (reililish) 

(bed  vertical) 

(bed  horiiontal).  .  .  . 

Sr.  Pliiiip 

Arbruath    

\iiui;i  Creek 

^''ikiliire 

Albion 

Hellfville 

It.TC'l 

("It'vcland.. 

I'raialcinh 

liiir.'hcMter.    

I-  .ril  du  I^ac 

H;tv»'rstraw 

Kri'.ota 

l.ittio  Falls 

M;ifi|tiette 

M.i-illim 

M.-.liua 

Mi.Mleton 

.Nirih  .\mherst 


Material. 


Granite 


Authority. 


HasweU 


Gillmore 


Laurenttan 


Sandstone 


\  Mrrnillon.  .  .  . 
\V;irrriisbun|. 
AliHinount.. . 

IMi'iburch.  .. 


HerriU 


Bovey 


Ha.swell 
Gillmore 


IliiMirnelstown 

■I'Tdnn        

I.oiitt  Moadow 

Mri'iiiiiu 

Mi.'liiEiin 

New  (iunnison 

()<wi-eii 

"'uwlins 

Taylrir's  Falls 

k^uebao  (heil  wrtlcal).     . 
"       (be<l  horiiontal). 


Merrill 


Bovey 


Cniahins 

Strength 

in  Pounds 

per  Sq.  In. 

10,768 
6,339 
1(1,430 
12,850 
5,340 
12.020 
15,093 
15.1I.)0 
19,000 
15.U62 
11,700 
10,837 
14,937 
9,750 
19  755 
17,750 
19,750 
lii.9.j() 
17,750 
20,650 
10,375 
5.970 
15„390 
12,.'i00 
19.104 
16,750 
23.500 
l(i,746 
19,0.^3 
13. .500 
13.526 
23.522 
26.800 
23  ,.'500 
24,520 

so.o.w 

26,000 

7,850 

5,340 

5,710 

13. .500 

11,700 

10,250 

7.910 

12.000 

9,412 

6.2,'M) 

4. ,350 

1 1 .675 

9,sr>o 

7,4.50 
8.7.50 

17.725 
6,9.50 
6.6.50 

10. .500 
8,8,50 
5.000 
1,149 

12.000 

12,752 

16,610 
3,750 
8,812 

13,046 
6.323 
9.903 
6.220 

lO.KW 
5.500 
7.344 
B.130 


2''0  THEORY  OF  STRICTURES. 

Crushino  Strength  of   Limestones,  Marbles,   etc. 


Locality. 


Material. 


English  Magne»ian 

Baril!<town 

Joliet 

Marquette 

Marhleheail 

North  IJivfr 

Williamsville.  . .  . 

Beilfdrd 

Quebec  (be<l  vprlican.  . 
i\ietl  horizontal) 

Stockbridfte 

Ilorset 

?uincy.  III 
uckahoc 

Italian 

Vermont 

Scotland 

Connecticut 

Fairhaven 


Limestone 


Marble 


Whiiistone 

Freestone 

Slate 


Cnuhini 

Strength 
in  Pounds 

Authority. 

per  8q.  In. 

Ha-xwell 

3.130 

Gillmure 

16.250 

" 

16,600 

8.050 

' ' 

12.600 

* ' 

13,425 

' ' 

12.375 

Merrill 

10.125 

Bovey 

12.597 

' ' 

12,330 

HaaweU 

10.382 

Gillmore 

8.670 

' ' 

9.878 

' ' 

13.594 

Merrill 

12,156 

*  ' 

13.400 

Haswell 

8.300 

'* 

3.3ig 

Merrill 

12.870 

Average  Coefficients  of  Shearing  Strength  for  \'ariou8  Woods  (Bovet). 


White  oak 690 

I  loiigla.-  fir ,3H0 

Bi-fh SV) 


Yellow  pine 400 

Maple 700 

Kim 590 


White  uh 660 

Hemlock 400 

Cottonwood 260 


..i-^K  ■"''t'"??*  «»ren(fths  are  for  planes  parallel  to  the  axis  of  the  scantling       The  tensil* 
^iT^Vr^        '"*'"'*  '"•'  "'"""lerable  variation.      The  direct  tensile  strength  is  1^  uch  girater  thaS 
.♦!?!  l^TfrL""!'""*.-""  7*"«>h.  and  the  failure  of  a  beam  loa,le<l  transverseK-    bv  *criMing^ 
likely  to  take  place  under  a  load  much  less  than  the  uiatenal  could  beai 


the  compression 
iti  tpiision 
Kiln 
sheari] 
although 


siile. 


viln  drying  largely  increa.se8  the  direct  compressive  strength,   but  greatly  diminishes  ,h^ 
ng   strength,    whde    the   direct    tensile   strength   d.*s   not    appear    to    bfT  m,^h    !«Sl.*j 
igh  in  the  majority  of  ca.ses  it  is  diminishcl.  aSd  sometimes  con^derablv  '*"*••' 


Average  REst-LTs  obtained  in  Cement-testing  Laboratort, 
MrGiLL  T'xivehsity. 

Neat  Portland  cement  after  1  to  4  week-,  has  a  crushing  strength  of  from  3000  to  6000  lb* 
per  square  mnh  and  a  tensile  strength  of  from  400  to  fiOO  lbs.  per  square  inch 

Neat  Natural  cement,  after  1  .0  4  *eeks,  has  a  crushing  strength  of  from  600  to  1200  lb.  per 
square  mch  an,l  a  ten.ile  ^.rc^f,  of  from  101  to  200  lbs.  per  square  inch. 


Material. 

Common    brick     piers    (cemeni 

mortiit) 
PrcJserl     bri<  k     piers     demeTii 

mortar'* 


'rn-*hing  Load 
l.bs.  |ier.Si|.  In. 


Material. 


uon  .     „u-u,    li  !\"''^  '''"'  "'""■  mortar) 
KOOtoKHV)    '1   Co.urefe  (I    2  .ll   (145  to   IM 
Ih-j.  per  e-u    ft   ) 
llnOtoI2Hrt    -    Cnr-^te'l    2   4.  3  to  (1  mnnth- 
old  I.. 


Crushing  Load, 
Lbs  perSq.  In. 

2!W  to  400 

400  to  800 

2350 

.After  2S  <i.i\--'  imriier'^iMii  the  a\eratre  <'fim'tre---i\e  ^trf>>ifftl,M  ,,f  ....,.,.»»«    4  :  1.  j 

ofcoo.1  I'ortla-i.I.einent  a.ul  '  < omrete  4  m.  cubes,  made 

(a)  sharp  angular  «.i.i.|»t(me  gravel  in  the  ratio  of  1  to  I.  1  to  2    1  to  .1    1  »„  4    .«<.~.  »«,.-^ 
1.,  !,..  if;,>;    t  i.>t>  o-v-t        *  ,■  V,  ■)  ■     1  '  '    "  ^.  were  four.a 

1,4     ™      .  .  I  ■  ■'.'  '       "  '''-.I""   '•!>'»■,■  i.i.h  re-oecflvelv 

6)  smiM.tli  «,iier_wr,rn   Laureotian  gravel  r,  the  ratio  ..f  I  to  1     1  to''    1  t.i  1    I  .0  *    .... 
f-.und  to  b-.  Vm.  9.VI.  800.  an.l  5m)  lb.    ,H,r  square  inch    resVi.'iively 


EXAMPLES. 


271 


Table  of  Average  Valceb  of  E,  G   K,  in  Millions  of  Lbs./Sq.  In.  and  of  a. 


Bras* 

Cast  iiim 

I  a<t  stpel,  tempered 

Cast  steel,  untempered 

Copper 

Delta  metal,  rolled 

(ilass 

(iun-metal 

Inm,  boilpr-plate    

Load 

Suft  iron 

Soft  steel.  unhardene<l 

Soft  steel,  hardene<l 

Steel   platea  (i  to  1  per  cent  C).  .  . 

Steel,  boiler-plate 

Tin   

Wood. 

Wrought -iron  bars 

Wrought  iron  plates 

Water     

India-rubber 

Zinc 

Slate 

Ciratiite 

Granites. 

•  Stanstead 

•  Peterhead 

•Lily  Lake 

•  We.sterly 

•  Quincy 

•  DoweDs 

\Urble. 

Marbles 

•  Carrara 

•  Tennessee 

•  Ksserite 


E. 


K. 


II 
3.14 


5  to  6.05 
7.9  to  8.07 

8.1 
7.2  to7.6 
6.75  to  7.9 
6.65  to6.95 


8.4 

9  to  9.5 

9.67  to  10 


5  to5.S 

5  to  7.6 

14 

12 

5.6to6.7 

5.25 
3.3  to3.9 
3.7 
14 
.27 
lO.Sto  11.3 
11 
11 
13 

13.5 
22 
.1  to  .17 
10.5 
9.5 
3.14 


15.3 


17.1  to  24. 
14.4 


5.8 

21.1 

5.3 


5.1  to  5  4 
3.2 
1.8 

2.22to2.46 

3.2  to3.3 
3.4 

2.98  to3.I6 

2.8to3.26 

2.64  to  2.92 

1.7 

2.8 

3.6  to3.8 

3.72  to  4.1 


21.6  to  26.7 


21.1 
3.2 


3.68to4.1 
4.56  to  4.96 
6.2 
4  to  4.75 
3.66  to4.5 
4.55  to4. 9 


5.21 
5.8  to6.32 
6.5to7 


3.1  to3.3 
3  to4.7 


2.9  to3 
3.9 


3.6  to  4. ft 


3.6 
2 


4 
4.3 
5.1 
4.3 
5.0 
3.9 


3 

4 
3.9 


64 


*  These  ref  ults  have  been  obtained  by  Dr.  Adams,  F.R.S.C..  of  McGitl  University,  in  hi* 
experiments  on  the  flow  of  rocks. 


EXAMPLES. 

1.  How  many  square  inches  are  there  in  the  cross-section  of  an  iron  rail 
wpighinp  30  lbs.  per  lineal  yard?  How  many  in  a  yellow-pine  beam  of  the 
s.iinc  lineal  weipht?  Ans.  3  sq.  in.;  45  sq.  in. 

2.  A  vertical  wTought-iron  bar  60  ft.  Lnft  and  1  in.  in  diameter  is  fixed 
at  the  upper  end  and  carries  a  weight  of  2000  lbs.  at  the  lower  end.  Find  the 
liutors  of  safety  for  both  ends,  the  ultimate  strength  of  the  iron  being  50,000 
lbs.  per  square  inch.  .fins.  19i\;   ISyVi. 

3.  .\  vertical  rod  fixed  at  both  ends  is  weighted  with  a  load  w  at  an  inter- 
iiiiili.ite  point.  How  is  the  load  distributed  in  the  tension  of  the  upper  and 
(•iiniprc^sion  of  the  lower  |X)rtion  of  the  rod?       Anx.  Inversely  as  the  lenpths. 

4.  Find  the  length  of  u  steel  bar  of  sp.  gr.  7.8  which,  when  suspended 
vcrtiiaily,  would  break  by  its  own  weight,  the  ultimate  strength  of  the  metal 
1.1  iiijr  CO.OOO  lbs.  per  .squ.'irc  iii-h.  .In.*.  17,72.1  ft. 

5.  The  iron  composing  the  links  of  a  chain  is  ^  in.  in  diameter;  the  chain 
is  liniken  under  a  pull  of  10,000  lbs.  What  is  the  corresponding  tenacity 
pir  .sfjuare  inch?  Ani^.  ,57,272A  11*. 

0.  A  vertical  iron  suspension-rod  90  ft.  long  carries  a  load  of  20,000  lbs, 
Ht  it!«  lower  end;  the  rod  is  made  up  of  three  equal  lengths  square  in  section. 


*IH 


272 


rilEOHY  OF  STRUCTURES. 


m 


S"fi<!*ihr^'°""'  *"■'*•  °^,^^^\'«"gth,  the  ultimate  tenacity  of  the  iron  being 
50,000  lbs.  per  square  inch,  and  5  a  factor  of  safety. 


A         -'00 

9y  "*''  '"■ 


L'OOOO 


in.; 


2000000 


99" 


, —  sq. in. 


7.  If  the  rod  in  the  previous  question  is  of  a  conical  form,  what  should 

at;«or;roi\t"r r  "'^ '-' ''-  '"""^*^'-  «^  ^'^  ^-'-  -  ^« 

8    Tl,     ,    i ''•''•  rTV'?-. '"•'   ^^""-^^ '*'■'••  »»»»--''b«-I^'-«q''a>-e  inch. 

8.  The  dead   oad  of  a  bridge  is  .1  tons  and  the  live  load  10  tons  per  panel 
he  c«rrcs,,ondm.;  factors  of  safety  being  3  and  6.     If  the  two  loads  are  taken 

together,  making  9  tons  per  panel,  what  factor  of  safety  would  you  use? 

Ino  r/'^'.ri"^  ■''  ^'T  ^^  '"•  *''°'»'^  '^'*^  «"  "  ^-a"  "f  masonrv.'^'inf"be 
loaded  w,th  10  tons,  what  length  of  bearing  surface  is  necessan",  the  safe 
crushing  stres.s  for  stone  being  1.50  lbs.  per  square  inch?  Ans    min 

"0  to"ns  th!^ «  f  "^^--^""K  surface  at  the  base  of  a  column  loaded  with 

.0  tons,  the  same  stress  being  allowed  as  in  the  preceding  question. 

11.  What  should  be  the  diameter  of  the  stays  of  a  boiler Tn  whicrthe 
pressure  is  .iO  lbs.  per  s,,uare  inch,  allowing  one  stay  to  each  U  sq.  ft.  of  sur- 
face and  a  stress  of  ;{.-.00  lbs.  per  square  inch  of  .section  of  iron?    .L    Urn 

12.  \  short  ,|ast-,ron  ,K,st  is  to  sustain  a  thrust  of  64,000  lbs.,  the  ultimate 

t;:'      S  ;  '";■  '^•""-"^•^«-«^  ^^-  P^  square  inch  and  10  a  factor 

ot  .saf,t>.     Inid  the  <l.mcnsio„s  of  the  post,  which  is  rectangular  in  .section 

with  ,he  sules  in  the  ratio  of  2  to  1.  ^„J  ^  -^^  'JI^J'"" 

13    How  many  ;-,.,.  rivets  must  lx>  used  to  join  two  wrought-iron  plates, 

nv  le^l  ;.h,:    ,7f  S"-  'T:  ^''''"^  ^'^'^  "'■'"'  ""'^y  ^  -strong  as   ; 

mtiri'io  to'.) '  •     '  '"'^  '^"■""^*''  "'  wrought-iron  beinj  in  the 

,^     .  .        ■  ■      ,  ,  -4"s.  19  rivets  (18  3) 

14.  .\  horizontal  cast-ron  bar  1  ft.  long  exactly  fits  between  two  vert  cal 

plat,.s  of  ,n.„.     How  min-h  should  its  temperature  be  raised  so  that  it  m  gh 

renK.,,.  supported  l.twccn  ,he  plates  by  the  friction,  the  coefficient  of  Sn 

.5.  A  bri,.k  wall  2  ft.  ,hi,.k,  IL'  ft.  high,  and  weighing  112  Ibs.'"p^f  cubic 
foo    ,s  supported  upon    solid    pi,ch-pine   columns   9   in.  ,n  diameteriO  f^ 

If  the  crushing  stress  .,f  pitch-pine  is  r,300  lbs.  per  square  inch  and  the 
factor  of  safety  10,  find  the  height  to  which  the  wall  may  be  buiU. 

,ft    ,w        •         u      ,.  •*"■''••  •■>0"03  1b.s.;  ,510.5  lbs.;  12.46  ft. 

iH>  ubs  t  rrr  '.k  ^;"T*'""  "/  ""-  -■•"""''t-i-n  .olumns  nhich  might 
lH>  su  .stituted  for  the  timber  columns  in  the  preceding  example  allouL 
a  work,,,,  s,r,.ss  ,n  the  metal  of  7.500  lbs.  per  squ.re  in.h.  .1,7  "  36  iT 

two'n;rallH"H.!"'"''  I'/.t'"'  ''\\'''l^P^'^'^'<^  "^  «-'°  F-  Just  fits 'in'^bet ween 
find    h     J     '      .  K  ""■'  ^  ^'"'''^  "f^  ^»  •■'  "^Pcrature  of  1620  F . 

find  the  stress  induced  in  the  steel,  the  walls  being  assumed  unyieW.ng.     YoJ 


EXA  MPLES. 


273 


may  assume  that  the  rooffirient  of  oxpaasion  of  stool  nor  dojn'oo  F.  is  tincnnt 
an.lthati'=  12,000  tons  jxt  square   nch.  .1/i.v.  SJ  tons/sq.  inch. 

if  the  coefficient  of  friction  between  the  liar  and  plate  is  i,  what  vertical 
i.i.'ui  will  the  bar  just  carry?       Aiis.  2]  tons  for  each  square  inch  of  section. 

i8.  'I'here  is  a  thrust  of  10"),000  lbs.  along  a  strut  (prevented  from  Ixjnding) 
made  up  of  four  4"x4"xi"  angle-irons  and  40  ft.  long.  Find  the  load 
\»";  s()uare  inch,  and  the  amount  by  which  the  strut  is  shortened,  E  being 
J^,ll(H),(KX)  lbs.  per  square  inch.  Ans.  7000  lbs.;   .12  in. 

iQ.  A  force  of  10  lbs.  stretches  a  spiral  spring  2  in.  Find  the  work  done 
i:i  >•  retching  it  successively  1  in.,  2  in.,  3  in.,  up  to  6  in. 

Ans.  I,  V-,  V,  V,  ^IS  ^1"  in.-lbs. 

20.  A  roof  tie-rod  142  ft.  in  length  and  4  sq.  in.  in  sectional  area  is  subjected 
to  a  stress  of  80,000  lbs.  If  E  =30,000,000  lbs.,  find  the  elongation  of  the  rod 
ami  the  corresponding  work.  Ans.  1.130  in.;    3786J  ft. -lbs. 

2  1.  .\n  iron  wire  |  in.  in  diameter  and  250  ft.  in  length  is  subjected  to  a 
tpnsi'in  of  COO  lbs.,  the  consequent  strain  being  .jls  Find  E,  and  show  by  a 
diasrram  the  amount  of  work  done  in  stretching  the  wire  within  the  hmits 
of   elasticity.  Ans.  14,C61,818,*t  lbs. 

22.  A  timber  pillar  30  ft.  in  Idngth  has  to  support  a  beam  at  a  point  30  ft. 
from  the  ground.  If  the  greatest  safe  strain  of  the  timber  is  »',(),  what  thick- 
m  --  of  wedge  should  be  driven  between  the  head  of  the  pillar  and  the  beam? 

An.t.  ,'„  ft. 

23.  During  the  test  of  a  given  piece  of  J-in.  hard  round  steel  bar  in  com- 
pn^sion,  a  length  of  20  ins.  was  shortened  by  .007  in.  under  a  load  of  2  long 
tnns.     Find  the  corresponding  E.  Ans.  2S,9()1,G10  Ibs./sq.  in. 

: 4.  An  iron  bar  of  uniform  section  and  10  ft.  in  length  stretches  .12  in. 
under  a  unit  stress  of  25,0(X)  lbs.  Find  E,  the  bar  being  1  sq.  in.  in  section. 
If  L'".0()()  Ibs./sq.  in.  is  the  proof  stros'i,  find  the  modulus  of  resilience.  De- 
t  Tiniiie  the  work  stored  up  in  the  bar  in  foot-pounds,  and  compare  it  with 
till  work  which  would  be  stored  up  if  for  half  its  length  the  rod  has  its  section 
iinTeised  to  4  sq.  ins. 

Ans.  2.1,000,000  lbs. ;  25  in  in.-lb.  units;  125ft.-lbs.;  78 J  ft. -lbs. 

25.  A  ship  at  the  end  of  a  600-ft.  cable  and  one  at  the  end  of  a  500-ft. 
rnble  stretch  the  "ables  3  in.  and  2J  in.  respectively.  What  are  the  corre- 
ppoiidiiig  strains?  Ans.  y^Vir. 

26.  \  rectangular  timber  tie  is  12  in.  deep  and  40  ft.  long.  If  £'=  1,200,- 
0(i()  Ills.,  find  the  proper  thickness  of  the  tie  so  that  its  elongation  under  a  pull 
of  270,000  lbs.  may  not  exceed  12  in.  Ans.  7i  in. 

27.  A  wrought-iron  bar  GO  ft.  long  is  stretched  5  in.  by  a  pull  of  50,000 
li  ?.     liiid  its  diameter,  E  being  25.000,(X)0  lbs.  Ans.  .fi  in. 

28.  A  wrought-iron  rod  084  ft.  long  alternately  exerts  a  thrust  and  a  pull 
of  .".2,'.H0  lbs.;  its  cross-section  is  9.3  sq.  in.  Find  the  loss  of  stroke,  E  being 
-".MMio.OOO  lbs.  Ans.  4.G32  in. 

2q.  .\  wrought-iron  bar  2  sq.  in.  in  sectional  area  has  its  ends  fixed  between 
tu  I  immovable  blocks  when  the  temperature  is  at  32°  F.  If  fi -29,000,000 
lbs.,  what  prei:>sure  will  be  exerted  upon  the  blocks  when  the  temperature  is 
100°  F.?  a-.0C125.  .Ans.  27388J  lbs. 


274 


THEORY  OF  STRUCTURES. 


30.  A  wTOught-iron  rod  25  ft.  in  length  and  1  sq.  in,  in  sectional  area  is 
subjected  to  a  steady  stress  of  5000  lbs.  What  amount  of  live  load  will 
instantaneously  elongate  the  rod  by  J  in.,  E  being  30,000,000  lbs. 7 

Ans.  6250  lbs. 

31.  The  following  examples  a  to  v  give  the  observations  made  in  the  actual 
tests  of  sjx!cimeiis  of  different  materials.  In  each  case,  where  possible,  draw 
the  stress-strain  diagram  and  calculate  the  stress  at  the  elastic  limit,  at  the  yield- 
point,  and  at  fracture.  Also,  determine  Young's  modulus,  the  reduction  of 
area,  and  the  equivalent  elongation. 


(a)  Tensile  test.     Distance  between  gauge-points -10  ina. 
Section  =1  in. X. 735  in. 


Load  io  Lbs.    Ext.  in  Ins.  Load  in  Lbs.    Ext.  in  Ina. 

100      0  800     .00438 

400    .00219       1200    .00652 

Ans.  B.\V./sq.  in. -15,455  lbs.;    E- 


Load  in  Lb*.   Ezt.inlna. 
1600  .00871 

11360  B.W. 

'2,500,000  lb8./8q.  in. 


(6)  TentUe  test.    Distance  between  gauge-points  ■ 
Section  =.98  in.  X.74  in. 


■10  ins. 


Load  in  Lbs.    Ext.  in  Ins. 


100 

0 

200 

.00092 

300 

.00178 

400 

.00263 

500 

.00.354 

600 

.00440 

700 

.00531 

Load  in  Lbs.    Ext.  in  Ins. 


800 

.00622 

900 

.00713 

1000 

.00799 

1100 

.00892 

1200 

.0098 

1300 

.01069 

1400 

.01145 

Load  in  Lb*. 

Ext.  in  Ina. 

1500 

.01237 

1600 

.01318 

1700 

.01404 

1800 

.01485 

1900 

.01567 

2000 

.01648 

5730 

B.W. 

Ans.  B.W./sq.  in.  =7900  lbs.;  £:- 1,595,000  Ibs./sq.  in. 


(c)  Tensile  test.     Distance  between  gauge-points  - 10  ins 
Sectional  area  =.7425  sq.  in. 

Load  in  Lbs.    Ext.  in  Ins.  Load  in  Lbs.    Ext.  in  Ins. 

IfO     0  800    .00647 

400    .00281        1000    .0084 
600    .004.59 

.4ns.  B.W./sq.  in.  =8350  lbs.;    £-1,435,000  Ibs./sq.  in 


Load  in  Lba.   Ext.  in  Ini. 
1200  .01032 

6200  B.W. 


((0     Tensile  test.     Distance  between  gauge-points  =  18  ins. 
Section  =1.97  in.  X  .5  in. 


Load  in  Lbn. 

Ext.  in  Ins 

Load  in  Lbs. 

Ext.  in  Ins. 

t«ad  in  Lbs. 

Ext.  in  In*. 

1000 

0 

5000 

.0.50 

9000 

.09 

2000 

Ol.j 

6000 

.000 

10,003 

.105 

3000 

.025 

7000 

.07 

19,830 

B.W. 

4000 

.037 

8000 

.OS 

Ans.  B.W./sq.  in.  =20,132  lbs.;  £-1,624,000  Ibs./sq.  in. 


EXAMPLES. 


275 


{()  Tensile  test.     Distance  between  gauge-points -10  ins. 
Section -.98  in.  X.73in. 


I,i>a.l  in  Lb*    Ext.  in  Iiu. 


100 
200 
400 

600 

800 

1000 


0 
.00076 
.00219 
.00358 
.00492 
.00609 


Load  in  Lbs.    Ext.  in  loa. 

1100         .00762 


LowlinLb*.    Ext.inlsi. 


1200 

1400 
1600 
1800 
2000 


.00907 
.00907 
.01051 
.01194 
.01332 


2200 
2400 
2600 
6800 


.01446 

.01572 

.01718 

B.W. 


Ana.  B.W./sq.  in.  -9500  lbs. ;    E  -2,043,000  lb8./sq.  i" 


(/)  Tensile  test.     Distance  between  gauge-points 
Section  =.98  in.  X.73  in. 


Lnaii  in  Lbs. 

100 
200 
400 
000 
800 


Ext.  in  Ina. 
0 

.00077 
.00217 
.00360 
.00513 


Load  in  Lbs. 

1000 
1200 
1400 
1600 
1800 


Ext.  in  Ina. 

.00672 

.00831 

.00990 

.0115 

.01307 


-Sins. 

Load  in  Lba. 
2000 

2200 
2400 
7600 


Ext.  in  Ini. 

.01441 

.01591 

.01748 

B.W. 


Ana.  B.W./sq.  in.  =  10,620  lba:  £?- 1,500,000  Iba^sq.  in. 


(g)  Tensile  test.     Distance  between  gauge-points -16  ins. 
Initial  diameter  =  .29  in. 
Final         "        =.213  in. 
Elquivalent  extension— .14  in. 


Load  in  Lba.    Ezt.  in  Ina. 


Load  in  Lba.    Ext.  in  Ini. 


200 

.002 

400 

.009 

COO 

.cn 

SOO 

.Ola 

•000 

018 

IJ(H) 

.022 

14(X) 

.026 

1600 

.030 

1800 

.033 

2000 

.028 

2200 

.040 

2400 

.044 

2600 

.049 

2800 

.053 

Load  in  Lb*. 

Ezt.  in  Ina 

3000 

.060 

3200 

.066 

3400 

.075 

3600 

.086 

3796 

Broke 

lbs 


.l/,v.  B.W./sq.  in.  =57,432  lbs.;    £  =  14,348,800  Ibs./sq.  in.;  E.L. -33,294 
Ml  in.;  Y. P.  =48,424  Ibs./sq.  in.:  reduction  of  area -46%;  equiv.  elong. 


(h)  Tensile  test.    Distance  between  gauge-points -36  ios. 
I.;'tial  diameter -.136  in. 
Final  "        -.085  in. 


I.'it'Hn  Lbu.  Ext.inlna. 

!.■)()  .01 

:i(H)  .02 

400  .03 

.■.no  .04 


Load  in  Lbs. 

600 
700 
800 
900 


Ext.  in  Ins. 
.05 
.06 

.075 
.095 


.\nx.  Load  at  fracture -75,692  ibs./sq.  in.; 
reduction  of  area  -61%. 


Load  in  Lba. 

1000 
1110 


Ext.  in  Inn 
.150 

B.W. 


.B -24,772,000  Ibs./sq.  in.; 


IfF^ 


276 


(0    Tensuc  tent. 


THEORY  OF  STRUCTURES. 

Distaiiop  l.vtwoon  gaiig(>-i)oints  =30  ins. 
Sectional  area  =0.0")  in.  X.26  in. 
Wt./lin.  fi.-|.14:{  ll)x. 


Ijiiacl  ill  Llw.      Kxt.  in  Ins. 


0 

iO(M) 
150(1 
L'(KH» 
2'><K) 
3000 


0 

.67 
.01 
.L'3 
52 
.7.S 
04 


Loa<l  in  Lhs. 

3.")(K) 
40<X» 
4r)(X) 

m)o 

5.500 
0000 


Kxl 


in  InA 

.2S 


Loail  in  Lbs.    Ext.  in  Ini. 


L'..)L' 

2  73 
2.04 

3  14 
3  32 


OoOO 
7000 
7500 
SOOO 
S.t00 
8000 


3  .W 
3.71 
3.80 
4.02 
4.22 
B.W. 


Ans.  Load  at  fraoture  =.3320  ll«./sq.  i„. ;  E  =263,000  Ibs./sq.  in. 


(j)   Tciixilc  teM.     Distanco  l)ot\voon  paupe-|>oints=S  ins, 
Initial  dianiptor  =.005  in. 
Final  "        =.0<)1  in. 

Equivalent  ('lonirati()n=.405  in. 

L<>ii  I  in  LliM.    Kxl.  in  In-s. 


■JiHM) 

4(KM 

iilHX) 

10,(Nri) 

14.(Mm) 

;s.(KKI 


0 
.00075 
.(N)t45 
002H2 
(K)4I2 
(X)542 


Lim.l  iti  .,!..<.    Est.  in  Iii». 


20,000 
24,000 
2S,(KK) 
30.(HI0 
.32,0<Mi 
3(i,()0!) 


.00615 
.00735 
.00025 
.0OOS7 
.010."i7 
01217 


.1/i.N.  Load  in  il)s./,si|.  in.  at  K.L.  =  4,s,,S.")0;  at  V 
-110,240;  led.  of  area  (i.7  (ht  cent; 
cent ;  E  -  32,200^000  lbs. /sq.  in. 


IxHul  in  Lh!i. 

38,000 
42,000 
46,00() 
4S,(XX) 
50,(KK) 
92,6,'-,0 


Ext.  in  Im. 
.01287 

.01445 

01622 

01712 

Yield  I't. 

B.W. 


r  -64,280;    at  fracture 
I'quiv.  elong.  -5.06  per 


(k)   Tctisilc  test.     Di.stancc  l)ef\veen  Kauj!r-iK)int.s - 
Section -2.01  in.  x.25in.  initial. 

'  \.m  ill.  -  .U  m.  final 
iMfuivalcnt  rloii>.'ation     2.22  in.s. 


ms. 


t:l'l  in  I.l».» 

.".00 

E\{.  in  Uin 

I.'.lfl  i'l  l,l>. 

lAl    ill  In«. 

Lnail  in  Lb.. 

r.xt  in  Ini 

0 

KMNN) 

(K).-)44 

20,(XX) 

.OIOi»8 

20(N) 

(10086 

12.(MN) 

(KMi.V) 

21. (KX) 

Oll.Vt 

KNH) 

()0I07 

14.(M)() 

(Mt766 

22  (XX) 

01200 

(ilNM) 

m-iio 

MKNN) 

(10877 

23,(XKI 

.oi:<(X) 

.S(KX> 

(H)t3l 

IS.OtX) 

(X)0.S8 

20,34(» 

H.W. 

Ann 


Load  at  K.L.  ^30,.S(X)  ll«./«q.  i„,;  „f  Y.P  =45,8(X)  11.. /s.).  in.;  at 
inHture-.',s..ji«)  il,s./.s,,.  ,„.;  ,.(•.!.  of  urtu  -55.4.t;  equiv. 
elong. -27.75':; ;   A -2S,32O,0(J0  ll«./.q.  in. 


rl| 


EXAMPLES. 


277 


(/)  Tensile  test.     Distance  between  gauge-points  = 
Section  =2  ins.  X 1  in.  initial. 

"      =1.37in.  X.GL'iii.  liual. 
Equivalent  elongation  =2.78  ins. 


8  ins. 


u;l.|  ill  Lbs. 

Kxt.  in  Ins. 

Lowl  in  Lbs. 

Kxt,  in  In». 

Loa>l  in  Lba. 

Ext.  ir.  Im. 

:;ii.iHHl 

0 

.■)2,000 

.00613 

66,000 

.00822 

lil.OIKI 

.00463 

56,000 

.(K)()(j2 

68,000 

.00860 

II.IKH) 

.00513 

(i0,000 

.00717 

70,000 

.00910 

4>.(MH) 

.00")f)i 

64,000 

.(X)786 

119,900 

B.W. 

.1/-..  Load  in  Ib.s./sq.  in.  at  E.L.  =31,000;   at  Y.P.  =35,000;   at  fracture 


=  59,9.j0;    red.   of  area- 
31,500,000  Ibs./sq.  in. 


■57.5%;    equiv.    elong. -34.75%;    £- 


(«))  Tensile  test.     Di.stanre  between  gauge-points— 6  ins. 
Initial  sectional  area  =.4359  sq.  in. 
Final         "         "     =.1452    " 
Equiv.  elong.  in  S  ins.  =2.05  ins. 


L«>;i>l  in  IJw. 

Ext.  in  Ins. 

Ijoail  in  Lbs. 

Ext.  in  Ins. 

Iins<l  in  Lb«. 

Ext.inlna. 

HHMI 

0 

11.000 

.(X)461 

16,000 

.00691 

7(I(MI 

(M)2S1 

12,000 

.00.506 

17,000 

MMK» 

(M)32fi 

13,000 

(10551 

23,900 

B.W 

'MKKI 

()0371 

14,000 

00.599 

KI.IKK) 

(H)416 

15,000 

.(KX)44 

.1'-.  Load  in  lbs./,s(|.  in.  at  E.L.  =29,820;  at  Y.P. -;}0,000;  at  fracture 
=  54,S;«);  red.  of  area  =66.7';; ;  equiv.  (long,  in  8  ins.-  25.63% 
f;  =30,590,000  Ibs./sq.  in. 


d')  TniKilrtrst.     Distance  Ix'tween  gauge-point.s —8  ins. 
Initial  diameter -.757  in.x. 
I''inid  "  .617  ins. 

E(|uivalfnt  clongal  ion  =  1.84  ins. 


i.t,  1,1,., 

lAl,in  Ins. 

Liittil  in  Lb« 

Kvl.  in  ln« 

I.'xkI  in  Lhs. 

Ext.inliw 

".(HI 

0 

!K)UO 

(KMi67 

I4,0(X) 

01086 

IHH) 

tKH(39 

10,(100 

(K)747 

15,(K)0 

01175 

'.1  |i  it  1 

00203 

II. (KM) 

(K),S29 

I6,(HK) 

01280 

M  M  M  ( 

(M(.r>H 

12,(NN) 

00912 

24,060 

H.W. 

<¥)5I  I 


I3.()(HI 


(K>997 


.1'  .  Load  in  llw./sq.  in.  at   K.L. -24,445;   at   Y.P. -35,.').'>6;  at  fracture 


.'")3,4(>7 


(i.     of 


22.6<K),(X)0  lb.s./sq.  in. 


area  ■=33.5' ,  ;     equiv.    elong. -23',' :     A?- 


278 


THEORY  OF  STRUCTURES. 


(o)  Tensile  test.     r)istanrolx'twccngaugp-point9=8  ins. 
Initial  diameter  =  .87")  in. 
J'inal  "       =.472  ill. 

iMjuivalent   elongation  =2.33  ins. 


^oad  in  Lbs. 

Ext.  in  Ins. 

Loail  in  Lbs. 

K.xi.  in  In.f. 

L<>a<t  in  Lb«. 

Ext.  in  Ids. 

1000 

0 

11,000 

.00439 

17,.)00 

00757 

2000 
4000 

.00049 
.00141 

12,(XX) 
13,000 

.00488 
.00.")36 

18,000 

19,000 

.008:'2 
.00929 

WXX) 

.00222 

14,000 

.00.">89 

31,500 

B.W. 

StKK) 

.(X)309 

15,000 

.()0(i44 

10,000 

.00392 

10,000 

.00()99 

Ans.  Load  in  lbs./  .s,,.  in.  at  K.  L.=l«,«i20;   at  Y.P.  =31,590;  at  fracture 
-■)2,.{()0;     red.    of    area  =70.9';; ;     equiv.    elong.  =29.19J.-    E- 
31,790,000  Ibs./sq.  in.  '^^' 

(p)   Tensile  test.     Di.stance  l)otween  gauge-points  =  8  ins. 
Diameter  =.409  in. 


Load  in  Lhs. 

^  ^  100 
S  -  200 
C  i  300 


Ext.  ill  In'*. 

.001 

.00138 

.00178 


Load  in  Lbs. 

400 
500 


Ext.  in  Ins. 
.00218 

.00258 


Lowl  in  Lbs. 

600 
700 


Ext.  in  Ins. 
.00299 

.(K)340 


Ans.  ff  =  1.5,220,000  Ibs./sq.  in. 


('/)   Tensile  test.     Distance  Ix-tween  gauge-jwints  =8  ins. 
Diameter  =  . 31)9  in. 

L'lail  in  Llis, 
■r.    i    100 


-      2(K) 
-  (300 


Ext. in  Ins. 
.001 

.00143 
.00182 


Lua<i  in  Lbs. 

4(K) 
500 


Ext.  in  In». 

.00219 
.002.J8 


Luarl  in  Lb^.. 

(HX) 
700 


Ext.  in  Ins. 

.00299 
.00341 


Ans.  £'-18,017,000  Ibs./sq.  in. 

(r)  Tensile  lest.     Distance  between  gauge-points  =-10  ins. 

Initial  diameter:  external  -1.28  in.,  internal -1.06.-)  in. 
I"?""'  "  "        -1.205  in.,      "      =1.03  in. 

K<|iiivalent  elongation  ^ .S<)  in.  for  8  ins. 
I'ilK-  under  internal  pressure  of  1000  Ibs./sq.  in. 


Loll' I  in  L))n. 

0 

100 
KNN) 
2(KK) 

:um) 

4000 


5000 


Ext.  in  III'. 
0 

.0<K)30 
.(M)I07 
00193 
(K»2K4 
(Ki;{71 
.(H)I.V.) 


LoB't  ill   UlM. 

Ext   in  Ins. 

Liiaii  in  Lh.'<. 

Ext.  in  In 

5.".0<) 

(lO.VJl 

!XXX) 

.00801 

(•(XK) 

(X)544 

9.500 

.00845 

O.J(X) 

(X).")87 

1(),(XK) 

.00897 

7(XX} 

IK)(i29 

10,:)00 

.009r.O 

7.5(K) 

(XMi70 

1 1 ,000 

.01170 

StXXI 

(<<)7I2 

18,.'>60 

B.W. 

H.-)<XJ 


(Xt 


Load 


,.).t 


Ml 


lb 


sq,  in.  ut   K.L.  =.•... v.w)0;  .,t  Y.P. -.33,200;  at  fracture 


52,.'«X):     red.    of    area -23*^, 


29,(kX),(XX)  Uw./sq    in. 


%;     equiv.    elong. -11.1%;     E- 


i 


EXAMPLES. 


279 


usi  Compression  test.    Distance  Ijetween  gauge-points -1.25  in.; 
1  division  =.1/250,000  in. 
Sectional  in. X.5  in. 


Loail  in  Lbs.  Comp.  in  In«. 

1000  .00362 
2000  .00402 
4000    .00485 


Luad  in  Lbs.  Comp.  in  Ins. 

6000    .00570 
8000    .00665 


Load  in  Lbs.  Comp.  in  Ins. 

10,000        .00743 
53,100  B.W. 


.t  ms.  Load  in  Ibs./sq.  in.  at  fracture  - 106,200  lbs.  ;£- 14,764,000  lbs. /sq.  in. 
(t)    Compression  test. 


Load  in  Lbs.  Comp.  in  Ins. 
1000  .00288 

2000         .00307 
4000         .00351 


Distance  between  gauge-points  =  1 .25  ins. 
1  uivision  =  1/250,000  in. 
Section  =  1  in.  Xi  in. 

Load  in  Lbs.  Comp.  in  Ins. 
0,000  .00398 

8,000         .00439 
10,000         .00485 


Load  in  Lbs.  Comp.  in  Ins. 

12,000  .00528 

14,000  .00569 

43,400  B.W. 


Ans.  Load  in  Ibs./sq.  in.  at  fracture -86,400;  £-28,646,000  Ibs./sq.  ia. 

(u)  Compression  test.     Distance  between  gauge-points -1.25  ins. 
1  division  =  1/2.50,000  in. 
Section "1  in. Xi  in. 


l,.ia.l  in  Lbs.  Comp.  In  Inf. 
KHK)     .00425 
2000    .00447 
4000    .00490 


Luail  in  Lbs.  Comp.  in  Ins. 

6,000         .005.T2 

8,000         .00573 

10,000         .JH)613 


Load  in  Lba.  Comp.  in  Int. 

12,000  .00055 

14,000  .00698 

68,000  B.W. 


A,,..  Load  in  lb9./sq.  in.  at  fracture  =  136,000;  E -29,761 ,000  Ibs./sq.  in. 

(i)  Compression  test.     Distance  lietween  gauge-points  =  72  ins. 
Section  -3.03  in.  X 3.02  ins. 


Liia.li:i  Lb*. 

1000 
2000 
3000 
4000 
5000 
6000 


Cnmp.in  Ins. 
0 

.004 
.008 
Oil 
.015 
.018 


Lnailiii  Lbs.     Comp.  in  Ins. 


Loa<l  in  Lbs.    Comp  in  Ins. 


7,000 

S.OOO 

9  000 

10,000 

15,000 

20,000 


,021 
.024 
.028 

.o;}i 

.049 
.065 


30,000 
35.000 
40,000 
45,000 
52,000 


.081 
.115 
.132 
.148 
B.W. 


.Ans.  B.W.-5<i83  llw./sq.  in.;   E-2,.336,000  Ibs./^q.  in. 


jj.  A  steel  rod  100  ft.  in  length  has  to  Iwar  a  weight  of  4000  lbs.  If  E- 
,T,.iHM).(HK)  lbs.,  and  if  the  safe  strain  is  .0005,  determine  the  sectional  area 
.1  111.'  rod  (1)  when  the  weight  of  the  hmI  is  neglected;  (2)  when  the  weight  o( 
till'  Hid  is  taken  into  account.     Also  in  the  former  case,  determine  the  work 

,l,.n..  in  stretchinn  the  rod  ,'o  in..  ,»*  in  ,  A  in A  in.,  succe.-wivciy. 

Ans.\»M  »q.  in.;    iVA  sq.  in.;   ;«i,  133*.  m),  .  .  .  I2(X)  in.-lbs. 

33.  A  line  of  raiU  is  10  miles  in  length  when  the  temperature  is  at  32»  F. 


2S0 


THEORY  OF  STRUCTURES. 


up'r;iu';!^l '?^"'  '''""  ^^'V'TP''"*"''^  '■'  ""'  '""'^  ^-  «"^  the  work  stored 
.  ■■      ho  ra.K  ,,..r  square  nwh  of  sortion,  E  being  20,000,000  lbs.  «=.C016 

'        '   "     ^  •      ,     .        ,  l"^-   10.00(w3  miles;   338405.76  ft.-Ibs 

34.  A  wro„,m.,-,nm  bar  LV,  ft.  i„  Ien„h  an.l  1  s,.  i„.  in  secn,.:ul  area  stretches 
•0001  <4,  it.  f..,.  ,a,.h  nuTease  of  ^  F.  in  the  tomrn-rature.  If  £=29  000  0(S 
lbs.,  (leternnnf  the  w„rk  d..ne  bv  an  increase  of  '"o"  F 

w.Ih'?L'r/'  ""Vn"'""-'  "f .''^'""^i""  '"Hl^T  hoat  b.-  titilized  in  straightening 
wails  that  have  fallen  ont  of  [.luinb:'  j,,,,    -  qjj^  f.    ,.      ^ 

35.  Three  bars,  eaeh  10  ft.  long,  one  of  brass,  one  of  eopp^^'r,  and  one'of 
nui.l  steel,  have  seefonal  area.,  of  J,  J.  and  1}  sq.  ins.  res,..ethvl  -.     Conlre 

e  forees  requ.red  to  streteh  the  bars  by  the  same  amount,  ...nd  H^o  eon 
t^u.  amounts  of  work  done.  E  in  tons  jn-r  square  ineh  being  4,S(.  ,  .  bj^ 
S(MK)  lor  coiiiH'r,  and  r.>,()00  for  steel.  j„      ■   .,  ,. 

of  lelth  )'""lf  H  T''?"^'  ';''•'"'■'■  ''"'"'•'"  ""  '"♦*^"^'»>'  "f  «t.i^^/  in"a  bar 

all  .1     height  /'■  '1      r''"L-'  r  ""'  ^■^••"^'''^^^'  '^h'"^  '^^'  the  same  load 

falling  .1  height  /(  before  e.\tending  the  bar  induces  the  stress 


E  being  Young's  modulus. 


'(-^ 


,     2Eh\ 


37.  A  5-in.  e..p,K'r  bar  and  a  J-in.  mild-steel  bar  of  the  same  lenirth  Are  to 
be  .stretched  the  same  amount.     Compare  the  forces  to  do  t^    amf also  the 
amounts  of  work  ex,H.Mdc.l  in  each  ...se,  E  .x-r  square  inch  .x,in«  S.^    ton 
t">-  tile  .•<,|.|.,.r  and  1,,,000  tons  for  the  stwi.  |„,    34   13-, 

38.  .A  steel  bar  of  3  sq.  ins.  sectional  area  stores  up  .'".O  ft.'-lbs.  of  enerev 

Fz:Xmu: "  ''■"" ""'" "  '^  ^""^/^"-  "'•  '^""' "-  '-'^<''  "fir's 

!■•  oemg  i.>.(NNI  tons/S(|.  m.  , 

3Q.  Two  verti.al  bars,  the  one  of  bn.ss.  the  other  of  st^-ei;  "m.  fLd  at 
iH'.r  upper  e.Mls  and  carry  a,   their  lower  ends  a  rigid  cross-l^.r      h    h    u  , 

p.  -.s  a  weigh,  „  Its  middle  ,H.int.     W  hen  t ros.s-bar  is  horizontal  the  stt^'i 

lii.liK'cd  iM  c.ich  of  the  vert Cid  birs  is  I  ".(Ml  II..      -.    •         i ••     1     ,  . 

40.  i-,nd  ii.e  mv,.,.cs,  i,.„.,i,  „f  ,„  i,,„„  ,„,„,„.i,.„  ,..„,  ,vi,ich^riin./;.!.v'i,, 

-vn  ucu,|,    ,„e  stress  U.m,  limited  .0  4  tons  jk.  square  inch.      What' will 

.;..    \M  in,n  bar  L-Oft.  l„„g  ..,n.l  :>  ,„s.  ,„  diameter  is  s„,,chc,l  „'„  V.f ^an  inch 

''^    ■  '"■"'  "^  '  ";"■;  "■  "'' '•"'^'  ""•  --•     l'i"«l  the  i.tensitv  of  .t  esl    1   a 

'•n.-^-^     (ion,  and  the  coe(|,cie„t  of  c|,,s,ici,y  of  the  mafiial.     " 

,,    Tl      I        I      f  .'"'•  •^'♦'■'•■^^     -■-■•'•"i^:   A-     I0.70()lons/.s,,.  in. 

4..    .!,..  length  of  a  .....s,.„ ,ll..,r  is  ,limi,us|,cd  f,,„„  ,.„  f,    ,„  mIot  ft 

i.H.  .  ,.    ,.,vcM  lo..,d.     iin.l  the  strain  and  the  .ompiessiv..  unit  stress    EU-m^ 
,,     \,     ,  ,      .      ,  l"^-  .<MII.j:    -'.J...0()  Hh.  ,„.r  square  inch. 

is  stb,e:.::r;;:";'';::;;:::::i:;.''Vn^iK)  i/''  'T '  "^"" " '  ^'- -•" 

.!..■  lengih,  //being  i;:i,H.,,KKMi;s        '         ""     ''""'"""'  the  d,m„,uti.n,  of 

Ank.  .003  ft. 


|:il 


EXAMFLES. 


281 


44.  Find  the  height  from  which  a  weight  of  200  lbs.  may  be  dropped  so 
h  A  tlie  maximum  admissible  stress  produced  in  a  bar  of  1  sq.  in.  section  and 
,■)  It.  long  may  not  exceed  20,000  lbs.  jut  scjuare  inch.,  the  coefficient  of  elas- 
tj.  ity  Ix'ing  27,000,000  lbs.  Ans.  .^  ft.,  or,  more  accurately,  jVo  ". 

45.  A  bar  of  steel  4"Xl"  is  rigidly  attached  at  each  end  to  a  bar  of  brass 
■\"  <l":  the  combined  bar  is  then  subjected  to  a  load  of  20  tons.  Find 
the  Iliad  taken  by  each  bar.  E  for  steel  =  13,000  tons  jwr  square  inch ;  for 
lir:i-s^^4tK)0  tons  per  s(|uarc  inch. 

.l;/.s'.  1 7.03  tons  on  steel ;  2.07  tons  on  brass. 

46.  .\n  hydraulic  hoist-rod  .")0  ft.  in  length  and  1  in.  in  diameter  is  attached 
f  -.1  plunger  1  in.  'ii  diiiiictcr,  upon  which  the  pressure  is  SOO  lbs.  per  square 
inch.     Deterniiiu'  the  altered  length  of  the  rod,  E  being  30,000,000  lbs. 

Ails.  .0213ft. 

47.  Work  ecpiivalent  to  .">()  ft.-lbs.  is  done  upon  a  bar  of  constant  sectional 
area,  and  produces  in  it  a  uniform  tensile  stress  of  10,000  lbs.  [ler  square  inch. 
Find  the  cubic  content  of  the  bar,  E  being  30,000.000  Jbs./sq.  in. 

.Ins.  3G0  cu.  in. 

48.  A  bar  2  scp  ins.  in  area  and  11  ft.  long  is  suspended  vertically  and 
has  a  collar  .it  the  bottom  end.  It  is  provided  with  a  sliding  weight  of  0.5 
ten.  Find  the  height  from  which  the  weight  must  Ix-  dropped  in  order  to 
ju.'^t  give  a  jK-rmanent  set  to  the  bar,  the  elastic  limit  of  the  material  being 
11  tons  i>er  square  inch.     Young's  modulus  =  12,000  tons  jwr  square  inch. 

.l((.s.  .34(15  ft. 

49.  What  would  be  the  resilience  of  a  steel  tie-bar  1  in.  in  diameter  and 
4^  ins.  in  length  if  it  became  jwrmanently  atretched  under  a  load  exceeding 
10  tniis,  E  Ix'ing  1(5,000  tons  per  sf|uan»  inch?  Anx.  191  iti.-ton. 

50.  The  dead  load  uikhi  a  short  hollow  cast-iron  pillar  with  a  sectional 
are.i  of  20  s(i.  ins.  is  iV)  tons  (of  2000  lbs.).  If  the  strain  in  the  metal  is  not  to 
e\  eed  .0015,  fiiid  the  greatest  hve  load  to  which  the  pillar  might  l)e  subjected, 
L  Iwing  17,(K)0,0(K)  lbs.  Ann.  205,000  lbs. 

51.  .\  steel  susiH>nsion  rod  30  ft.  in  length  and  J  sq.  in.  in  sectional  area 
carries  3.500  lbs.  of  the  roadway  and  30(KJ  lbs.  of  the  live  load.  Determine 
the  i/roKn  load  and  also  the  e';teii8i<»n  of  the  rod,  E  l)«ing  35,0(K).t>00  lbs. 

-!'>■.  ,11,  ft. 

?;2.  .\  steel  rod  10  ft.  in  length  and  i  sq.  in.  ii'  s»'ctiona!  area  is  strained 
V>  liie  proof  by  a  tension  of  25,0(XJ  lbs.  Find  the  resilience  of  the  rod,  E  Ix-ing 
;r.,iHK).(HM)  lbs.  Aiix.  178J  ft.-lbs. 

51.  The  resilience  of  an  iron  bar  1  sq.  in.  m  section  and  20  ft.  long  ia  300 
('  -I'n^.  What  wotild  1h"  the  resilience  if  for  l»  ft.  of  its  leiisfh  it  w;is  comnrsed 
i!  mil  2  s(|.  in.  in  section,  the  remaining  foot  being  tlie  same  siye  ns  t)efore? 

Ans.  157.r>  ft.-llw. 

5(.  l)riertnine  the  .shortest  length  of  a  metal  bar  n  t^q.  in.  in  sectional  area 
■■'  I  will  safely  resist  the  shock  of  a  weight  of  IF  lbs.  falling  a  <listan<'e  uf  li  ft. 
'.  I  1\  llie  result  to  the  ca.se  nf  a  steel  bar  I  s<i.  in.  in  sectional  area, the  weight 
!>  V  :  r\)  lbs.,  the  distance  10  ft.,  the  proof-j«trai»  -L,  and  K -35,(«M),(K»()  Iha, 

Ann.      "',,...,/ Ix'ing  the  safe  unit  stress;    ';(lil"ft. 

5S.  A  pitch-pine  pile  14  in.  squan-  is  20  ft.  nlKive  ground,  a.xl  is  lieing 


2S2 


THEORY  OF  STRUCTURES. 


driven  by  a  falling  weight  of  112  lbs.    If  £  =  1,500,000  lbs.,  find  the  fall  so 
that  the  inch-stress  at  the  head  of  the  pile  may  be  less  than  800  lbs. 

Supposing  that  the  pile  sinks  2  in.  into  the  ground,  by  how  much  would 
It  be  safe  to  increase  the  fall?  Ans.  7.456  ft.;  116§  ft. 

56.  An  8"xS"  pile  IS  ft.  above  ground  is  driven  by  a  falling  weight  of 
100  lbs.  Determine  (a)  the  strain;  (h)  the  amount  of  the  compression;  (r) 
the  distance  through  which  the  weight  falls  so  that  the  stress  per  square 
inch  in  the  pile  may  not  exceed  800  lbs.  per  square  inch.  Take  £-1,600,000 
*bs-  Am.  (a)  .0005,  (6)  .009  ft.  (c)  2.295  ft. 

If  the  pile  sinks  4  in.  into  the  ground,  by  how  much  would  it  be  safe  to 
increase  the  fall?  Am.  85  ft. 

57.  A  rod  1  sq.  in.  in  section  and  5  ft.  long  carries  a  weight  of  200  lbs. 
which  drops  through  2  in.  before  commencing  to  stretch  the  rod.  Assuming 
all  the  energy  of  the  blow  to  be  stored  up  in  the  rod  and  that  the  limits  of 
elasticity  are  not  exceeded,  estimate  the  intensity  of  stress  induced  in  the 
rod.    E  =30X10'  lbs.  per  square  inch.  Am.  6345 lbs. 

58.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  has  one  end  securely 

anchored,  and  suddenly  checks  a  weight  of  W  lbs.  attached  to  the  outer  end, 

and  moving  with  a  velocity  of  1'  ft.  per  second  away  from  the  anchorage! 

Find  the  greatest  pull  upon  the  chain.  I~PH- 

Ans.  Pu\l^V\'- . 

\    Ig 

59.  Apply  this  result  to  the  case  of  a  wagon  weighing  4  tons  and  worked 
from  a  stationary  engine  by  a  rojie  .i  sq.  in.  in  sectional  area.  The  wagon 
is  running  down  an  incline  at  the  rate  of  4  miles  an  hour  and,  after  600  ft. 
of  rofie  have  been  paid  out,  is  suddenly  checked  by  the  stoppage  or  reversal 
of  the  engine.     £  =  15,000,000  lbs.  Am.  26,884  lbs. 

60.  A  chain  /  ft.  in  length  and  a  sq.  in.  in  sectional  area  has  one  end  fit- 
tachcd  to  a  weight  of  W  lbs.  at  rest,  and  at  the  other  end  is  a  weight  of  nW 
lbs.  moving  with  a  velocity  of  V  ft.  per  second  and  away  from  the  first.  Find 
the  greatest  pull  on  the  chain.  , — Z'TT' 

^n«.  PuU  =  V\l-2MiL. 
\lg{n  +  l) 

61.  A  dead  weight  of  10  tons  is  to  act  as  a  drag  upon  a  ship  to  which  it 
is  attached  by  a  wire  roi)e  l.")0  ft.  in  length  and  having  an  effective  sectional 
area  ()f  S  sq.  in.  If  the  velocity  of  the  floating  ship  is  20  ft.  per  second,  and 
if  its  inertia  is  equivalent  to  a  mass  of  390  tons,  find  the  greatest  pull  on  the 
chain.     £  =  1.VX)0,0(K)  lbs.  ilns.  208i  tons. 

62.  A  coal-wagon  weighing  .32a)  lbs.,  when  running  down  an  incline  at 
a  rate  of  6  miles  an  hour,  is  suddenly  ihecked  by  the  stopjjage  of  the  engine 
after  .")(M)  ft.  of  cable  have  l)een  paid  out.  Find  the  maximum  pull  on  the 
cable,  its  sectional  area  Ixjing  3  sq.  in.  and  its  £==15,000,000  lbs.  i>er  square 
'■"'*'•  Ans.  26,400  lbs. 

63.  A  square  steel  bar  10  ft.  long  has  one  end  fixed;  a  swMni  pull  of  40,000 
!hs,  is  cxcrf.-d  r!t  the  oth.-r  end.  Find  the  scttior.al  area  of  the  bar  consist^-nt 
with  the  condition  that  the  strain  is  not  to  exceed  .^5.     £-30.000,000  lbs. 


Find  the  resilience  of  the  bar. 


Ans.  2  sq.  in.;  533 J  ft.-lbs. 


EXAMPLES. 


283 


64.  How  much  work  is  done  in  subjecting  a  cube  of  125  cu.  in.  of  iron  to 
a  tensile  stress  of  8000  lbs.  per  square  inch?    £=30,000,000  lbs. 

Ans.  n\  ft.-lbs. 

65.  A  signal-wire  2000  ft.  in  length  and  i  in.  in  diameter  is  subjected  to 
a  steady  stress  of  300  lbs.  The  lever  is  suddenly  pulled  back,  and  the  cor- 
rpsponding  end  of  the  wire  moves  through  a  distance  of  4  in.  Determine 
the  instantaneous  increase  of  stress.     £=25,000,000  lbs.       i4ns.  Slfjf  lbs. 

66.  If  the  total  back-weight  is  ^50  lbs.,  what  is  the  range  of  the  signal  end 
(if  the  wire?  ^ns.  j{^U  ft. 

67.  A  steel  rod  of  length  L  and  sectional  area  .4  has  its  upper  end  fixed 
and  hangs  vertically.  The  rod  is  tested  by  means  of  a  ring  weighing  60  ibs. 
which  slides  along  the  rod  and  is  checked  by  a  collar  screwed  to  the  lower 
end.  A  scale  is  marked  upon  the  rod  with  the  zero  at  the  fixed  end.  If  the 
strain  in  the  steel  is  not  to  exceed  jh,  what  is  the  reading  from  which  the 
weight  is  to  be  dropped?    £=35,000,000  lbs. 

Ans.  Distance  from  point  of  suspension  =(fSJ-J^/l)L. 

68.  A  bar  I  sq.  in.  sectional  area  and  32  in.  long  is  subjected  to  a  tensile 
pull  of  10  tons.  Calculate  the  work  stored  up  in  the  bar.  £  =  30,000,000  lbs. 
]>cT  sq.  inch.  Ans.  17*  ft.-lbs. 

69.  A  load  of  1000  lbs.  falls  1  in.  before  commencing  to  stretch  a  suspending 
rod  by  which  it  is  carried.  If  the  sectional  area  of  the  rod  is  2  sq.  in.,  length 
100  in.,  and  £=30,000,000  lbs.,  find  the  stress  produced. 

If  the  rod  carries  a  load  of  5000  lbs.,  and  an  additional  load  of  2000  lbs 
is  suddenly  applied,  what  is  the  stress  produced? 

Ans.  17,827i  lbs.;   4500  Ibs./sq.  in. 

70.  Steam  at  a  pressure  of  50  lbs.  per  square  inch  is  suddenly  admitted 
U[)<)n  a  piston  32  in.  in  diameter.  The  steel  piston-rod  is  48  in.  m  length  and 
•J  in.  in  diameter,  £  being  35,000,000  lbs.     Find  the  work  done  upon  the  rod. 

What  should  be  the  pressure  of  admission  to  strain  the  rod  to  a  proof  of 
.(K)!?  .  Anc.  117.69  ft.-lbs.;    68|^    lbs.   per  square  inch. 

71.  A  Imulder-grappler  is  raised  and  lowered  by  a  wire  rope  1  in.  in  diameter, 
hanging  in  double  sheaves.  On  one  occasion  a  length  of  150  ft.  of  rope  was 
in  oinration,  the  distance  from  the  winch  to  the.upper  block  being  30  ft.  The 
I'rapplcr  laid  hold  of  a  boulder  weighing  20,000  lbs.  What  was  the  extension 
(if  tlu'  roiK-,  £  being  15,000,000  lbs.? 

The  boulder  suddenly  slipped  and  fell  a  distance  of  6  in.  before  it  was 
apain  held.     Find  the  maximum  stress  u]xm  the  rojie. 

What  weight  of  boulder  may  he  lifted  if  'he  proof-stress  in  the  rojie  is  not 
to  exceed  25,000  lbs.  per  square  inch  of  gross  sectional  area? 

Ans.  ^h  ft-;  50,452i  Ibs./sq.  in.;   78,.57U  lbs. 

72.  \  steel  Iwr  stretches  lU  of  its  original  length  under  a  stress  of  20,000 
llw.  jHT  s(iuare  inch.  Find  the  change  of  volume  and  the  work  done  \)er  cubic 
iii,h.  Ans.  ,T^;  I  ft.-lb.  per  cubic  inch. 

73.  The  steady  thrust  or  pull  upon  a  prismatic  bar  is  suddenly  reversed. 
.-'now  that  its  effect  is  trebled. 


74.  A  shock  of  N  ft.-lbs.  is  safely  borne  by  a  bar  2  ft.  in  length  and  a  sq.  in. 


284 


THEORY  OF  STKUCTUhhS. 


in  sectional  area.     Dotern.ino  the  in.roasi-.I  slu.ck  which  the  bar  will  bear 
when  the  sectional  area  of  the  last  wth  of  its  length  is  increased  to  ra. 

75.  The  r.ull  on  one  of  the  tension-bars  of  a  lattice  girder  flnctuates  from 
IS  to,.s  to  4  tons.  If  24  tons  is  the  statical  breaking  strength  of  the  metal 
lo  tons  the  pninit.ve  strength,  determine  the  se.-tional  area  of  the  bar  3  beine 
a  faHor  of  safety.         .l„.v.  L'.l.-,  sq.  in.  (Launhanlt ) ;    l.STsq.  m.  .rnuin) 

76.  I  he  stress  m  a  .liago.ml  of  a  steel  bowstring  girder  fluctuates  from  a 
tens,,,,,  of  1.>.1.>  tons  to  a  .■o„,pr,.ssio„  of  7AV.  tons.  If  the  primitive  strength 
of  h..  n,etal  ,s  1.'4  tons  and  the  vibrat,on  strength  12  tons,  find  the  proper 
sect,onal  area  of  the  diagonal,  3  being  a  factor  of  safety. 

1...  2..W  sq.  in.  (WejTauch);  1.7  s,,.  in.  (Unwin);  40  tons  per  square  inch 
be,i,g  statical  strength. 

77.  A  member  of  a  truss  is  subjected  to  tensile  stresses  varying  from  a 
max,mum  of  1.50,000  lbs.  to  a  minimum  of  .50,000  lbs.  Find  th/ proper  sec- 
tional area  of  the  member,  1st,  if  of  steel,  2d,  if  of  wrought  iron,  thrfactor 
of  safety  lK.,ng  3.  ^.,„,  -  ,,,,  ,^    ;„  .   j^.g  ,q.  ;„. 

of  T  >!;  Jw,'u  '^""!'  •''  *""'  "  '"^J^^'-'^J  to  stresses  varying  fron,  a  maximum 
of  1.0.000  lbs.  in  tens,on  to  a  minimum  of  80,000  lbs.  in  compression.  Find 
the  pro,K-r  .se.^,onal  area  of  the  diagonal,  1st,  if  of  steel,  2(1,  if  of  wrought 
iron,  o  i>e,.,g  the  factor  of  safety.  Ana.  11  sq.  in.;  18.11  sq   in 

70.  A  steel  diagonal  is  subjected  to  stress's  which  vary  from  a  maximum 
comi.ression  of  10,000  lbs.  to  a  maximum  tension  of  10,000  lbs  Find  its 
ject,o„al  area,  taking  3  as  the  factor  of  .safety  and  u=2s-60,(K)0  Ib.s. 

80.  A  WTought-iron  screw-shaft  is  driven  by  a  pair  of  cranks'set  aT  right 
ui.frlcs.  .Neglecting  the  obli,,uity  of  the  .■onnecting-rods,  an.l  a.ssuming  that 
th..  imll  on  the  cra„k-p,„  is  con.itant,  compare  the  coefficients  of  strength  (a' 
and  /)  to  l.e  used  ,n  calculating  the  diameter  of  the  shaft.  How  is  the  result 
attected  by  the  stopping  of  the  engine?  Ans.  a'  =  .904<-  a'  =  ••/ 

8i.  .Sho.v  that  the  change  of  a  unit  of  volume  of  a  solid  body" under  a  longi- 
tudinal st,ess  is,(|  -^),  whichbecon,es^if.  .4,asin  metals,  and  m/  when 
0=2,  as  in  iii(|i,i-i,il)l)er. 

82.  A  bar  of  Icn-th  /,  an.l  with  a  section  gradually  diminishing  from  .1, 
at  one  eml  to  .1,  at  tlie  other  is  subjected  to  a  tension  /'  in  the  direction  of  it.s 
axis^  .show  ij.at  il„.  work  of  deformation  is  J'L~2EVAJ„  R  l,eing  the 
eoefhcieiit  of  eJMst icily.  * 

83.  During  (he  plastic  .leformation  of  a  i,rismatic  bar,  show  that  the  change 

in  s,.ct.ona    ana  ,s  pr. rtional  to  the  deformation  calculated  on  the  altered 

length  ol  the  l)ar. 

84.  The  initial  stress  in  a  bar  fixed  at  the  ends  is  rmk/cm.\     If  its  tcm- 
,T>erature    ,.   lowered    .-,0=    C,    ^how    that    the    additional    energy  stored   up 

m  the    bar  ,s  l\  k.m. /em.',  taking  £=2xlOVcm.,  and  a  =  1  ■-:  ,si).(M)0  per 


EXA\fPLES. 


285 


8>;.    V  juTamid  upiphina;  12'»  lbs.  ikt  cu.  ft.  has  a  height  of  «iO  ft.  and  rests 
uiKiii  a  square  base  of  2.">  sq.  feet.     Find  the  amount  and  work  of  rompression. 


/•;  Iwing  the  coefficient  of  elasticity, 


.        408,7:10  ,,   , 
Ans.        '  —  ft  .-tons. 
h 


86.  In  an  indicator  the  area  of  a  piston  is  i  sq.  in.,  the  inertia  of  the  mov- 
iiiS  ));irts  attached  to  the  piston  is  equivalent  to  a  weifrht  of  .3.3  lbs.  moving 
will  the  s;imo  velocity,  and  the  jiencil  moves  4  times  as  fast  as  the  piston,  1 
iiiih  oil  the  diasnim  corresponding  to  a  pressure  of  80  lbs.  on  the  jjiston.  Find 
th>'  time  of  a  complete  oscillation.  Am.  .01453  sec. 

87.  A  right  cone  of  weight  IF  and  height  h  rests  upon  its  base  of  radius  r. 
1 111(1  the  amount  and  work  of  the  compression. 

^„,,.  Comp.=^^,;work=-^^^,. 

88.  An  elastic  trapezoidal  lamina  A  BCD,  of  natural  length  I  and  thick- 
ness unity,  has  its  upper  edge  AB  (2a)  fixed  and  hangs  vertically.  If  a  weight 
11  is  susi)ended  from  the  lower  edge  CD  (2b),  show  that,  neglecting  the  weight 

of  the  lamina,  the  consequent  elongation  =^^ -3^ log. ^.    If  an  additional 

wcijiht  is  placed  upon  W  and  then  suddenly  removed,  show  that  the  oscillation 
pot  up  is  isochronous  and  that  the  time  of  a  complete  oscillation 


■■-\ 


Wnogej; 
2gE(a-b) 


♦ 


Examine  the  case  when  a-=6. 


Wl 


Arts.  Ext.  •=]!,  •^;  time  of  oscillation- 


'^ 


W± 

VaEg 


89.  If  the  specific  weight  of  the  lamina  in  the  preceding  question  is  w, 
iiiid  how  much  it  will  stretch  under  its  own  weight,  and  also  the  work  of 
fxtciision.     Determine  the  result  when  a  =6. 

1      u*'/'   .       h     u-l'a+h    wl' 


Ana. 


b_    ivP _ 

2E{a-by  ^^a'^iEa-b'  2E' 


Work  = 


u-H' 


a' 


-b* 


-6'(a»-6»)+6Mog,^)  ; 


w'aZ' 
3£  ■ 


4£(a-6)'l      4        "  '        '  '  "'b  J 

90.  \  hollow  tower,  of  height  h,  is  in  the  form  of  a  solid  of  revolution  about 
.1  VI  it  Kill  axis.  The  hollow  portion  is  a  right  cylinder  of  radius  A*,  and  the 
1,1  !i  1^  of  tiie  base  of  tiu-  tower  is  a.  If  the  six<cific  weight  of  the  tower  is  w, 
I,ihI  the  curve  of  the  generating  line  so  that  the  stress  at  every  point  of  the 
1    .   .  may  be  /.     Also  liiid  the  loud  on  the  top  of  the  tower. 


.\i,s.  i'-]{'=(a^~Ii')c  I  ;ff/(ft' -/?'),  6  being  the  radiu.s  of  the  top  of 
til  ■  lower. 

01.  An  elastic  lamina  in  the  form  of  an  isosceles  triangle  ABC  has  its  base 
.1/)'  -^211)  fixed  and  hangs  vertically.  If  its  weight  is  U',  find  its  elongation. 
I  !-..  . ocitici.'iit  of  elasticity -i',  thickness  of  lamina-unt^y,  and  Z/=the  dis- 
taiui;  of  C  from  AB.  .        WL 


Ana. 


AaE" 


286 


THEORY  OF  STRUCTURES. 


92.  A  weight  of  1  ton  depresses  its  supports  .01  ft.  Neglecting  the  weight 
of  the  supports,  find  the  time  of  a  complete  oscillation.         Ans.  .1111  second. 

93-  A  weight  of  10  lbs.  at  the  end  of  a  spiral  spring  stretches  it  3  ins.  The 
spring  is  then  stretched  an  additional  3  in.  and  suddenly  released.  Find 
the  time  of  a  c()m{)lete  oscillation.  Ans.  .555  second. 

94.  The  weight  of  the  piston  of  an  indicator  and  its  attached  parts  is 
equivalent  to  1  lb.  on  the  piston.  The  pencil  moves  six  times  as  fast  as  the 
piston.  The  area  of  the  piston  is  .5  sq.  in.  Find  the  period  of  oscillation  wi^h 
a  100-lb.  spring.  ^ns.  .00586  second. ' 

95.  A  body  symmetrical  with  respect  to  a  vertical  plane  is  slightly  dis- 
placed vertically  from  its  position  of  stable  equilibrium.  Find  the  period  of 
an  oscillation. 


Ans.  2n 


\Ag' 


V  being  the  volume  of  the  water  displaced  and  A  the  sec- 
tional area  of  the  body  at  the  water-line. 

96.  If  the  body  in  the  preceding  example  receives  a  slight  angular  dis- 
placement, what  will  be  the  period  of  an  oscillation? 

Ans.  2>r^jy— ,  /  being  the  moment  of  inertia,  W  the  weight,  and  z  the 
distance  between  the  metacentre  and  the  C.  of  G. 

97.  A  uniform  circular  plate  weighing  4  lbs.  and  1  ft.  in  diameter  is  hung 
in  a  horizontal  plane  by  three  parallel  cords  from  the  ceiling,  and  when  set 
into  small  torsional  oscillation  is  found  to  have  a  period  of  3  seconds  A 
body  weighing  6  lbs.  is  laid  on  it,  and  the  period  is  then  found  to  be  5  seconds. 
Find  the  moment  of  inertia  of  the  body.  Ans.  ^  =  .385  ft 

98.  A  revolving  weight  of  w  lbs.  is  carried  on  elastic  bearings  which  yield 
at  the  rate  of  1  ft.  to  6  lbs.  Find  the  extent  of  the  forced  vibration  when  the 
nuinb,-r  of  revolutions  is  n  per  second;  find  also  the  value  of  n  so  that  free 
vibration  might  be  set  up. 


An...  ^:^(2.n).; 


therefore 


period    (neglecting    wt.    of    bearing) -2s  J- -i,  and 

99.  A  U  tube  a  ft.  in  length  contains  water.    If  the  water  is  slighUy  dis- 
placed from  its  mean  position  of  equilibrium,  show  i,hen  the  periodic  time  is 

^-Va,  if  3  =31.' ft. 


100.  A  weight  of  5  lbs.  is  supported  by  a  spring.  The  stiffness  of  the 
spring  IS  such  that  putting  on  or  taking  off  a  weight  of  1  lb.  produces  a 
downward  or  upward  motion  of  0.04  ft.  What  is  the  time  of  a  complete 
osoilliation,  neglecting  the  mass  of  the  spnns? 


lOI. 


A  weight  IF  is  suspended  by  a  spring,  which  it  stretches.    The  weight 


EXAMPLES. 


287 


is  further  depressed  1  ft.,  when  it  is  suddenly  released  and  allowed  to  oscillate. 
Find  its  velocity  at  a  distance  x  from  the  position  of  equilibrium^ 


Ana. 


-X') 


g_ 
W 


102.  If  a  spring  deflects  .001  ft.  under  a  load  of  1  lb.,  what  will  be  the  period 
of  oscillation  of  a  weight  of  14  lbs.  upon  the  spring?  Am.  .1315  sec. 

103.  A  mass  of  fifty  pounds  is  attached  to  the  free  end  of  a  spring, 
maile  of  wire  /j  of  an  inch  diameter  and  having  ten  free  coils.  The  mean 
diameter  of  the  helix  is  3  in.  Find  the  period  of  vibration  if  the  spring  is 
extended  axially  and  is  then  let  go. 

104.  A  vertical  elastic  rod  of  natural  length  L  and  of  which  the  maaa  may 
I*  neglected  is  fixed  at  its  upper  and  and  carries  a  weight  W\  at  the  lower 
end.  A  weight  Wt  falls  from  a  height  h  upon  Wu  Find  the  velocity  and 
extension  of  the  rod  at  any  time  t. 

9 


Ans,  V*—: 


a--»-T")-(l)" 


X  being  measured  from  mean  position  of  (PTi  +  Wi). 

105.  Determine  the  functions  F  and  /  in  Art.  4  when  Pi  is  zero,  and  also 
when  the  rod  is  perfectly  free;  i.e.,  when  P„-0  and  Pi-0. 

106.  The  centre  of  mass  of  a  body  (supposed  to  be  small)  weighing  25  lbs. 
is  20  in.  from  the  fulcrum  of  a  massless  horizontal  lever.  It  is  suJ)ported,  also, 
by  a  vertical  massless  spring  actin  the  lever  at  a  distance  of  5  in.  from 
the  fulcrum.  The  stiffness  of  the  s,  ,•  is  such  that  a  pull  of  1  !b.  elongates 
it  0.001  ft.    \\'hat  is  the  natural  timi   .f  vibration  of  the  system? 

If  the  point  of  support  of  the  spring  gets  a  vertical  simple  harmonic  motion 
iif  amplitude  0.1  ft.  with  a  frequency  of  three  complete  oscillations  per  second, 
what  is  the  nature  of  the  motion  of  the  weight,  supposing  that  the  natural 
vibrations  are  stilled? 

107.  A  metal  rod  J  sq.  in.  in  area  and  5  ft.  long  hangs  vertically  with  its 
upiHT  end  fixed  and  carries  a  weight  of  18  lbs.  at  the  lower  end.  On  striking 
tin;  rod  it  emitted  a  musical  note  of  2G4  vibrations  per  second  (middle  C  of 
l.iano-forte).  Find  the  coefficient  of  elasticity,  the  weight  of  the  rod  being 
neglected.  Ans.  30,979,160  lbs. 

108.  What  should  be  the  thickness  of  the  plates  of  a  cylindrical  boiler 
0  ft.  in  diameter  and  worked  to  a  pressure  of  50  lbs.  per  square  inch,  in  order 
that  the  working  tensile  stress  may  not  exceed  1.67  tons  per  square  inch  of 
frriiss  section?  Ans.  .42  in. 

109.  A  cylindrical  boiler  with  hemispherical  ends  is  4  ft.  in  diameter  and 
'.'.'  ft.  in  length.  Determine  the  thickness  of  the  plates  for  a  steam-pressure 
of  4  atmospheres,  the  working  strength  of  the  material  being  4000  lbs.  ./so.  in. 

Ans.  .176  in. 
no.  Find  the  thickness  of  plates  required  for  a  boiler  which  is  to  work 


2S8 


TUEOIiY  OF  STRICTURES. 


at  a  prpsstir.-  ..f  KiO  ll)s./sq.  in.,  tho  diameter  <.f  the  shell  being  S  ft.,  the 
efficiency  of  the  riveted  joint  Si)*;;,  an.l  the  working  stress  in  the  metal 
o  tons  s(i.  in.  I         -„  ■       ' 

'.     -  ,      ,    .,  •'"■'*•  -^t  "I- 

111.  A  /-ft.  holler  of  ,v,-ii;.  jilates  wiis  hurst  at  a  longitudinal  double- 
riveted  joint  hy  a  pressure  of  UiO  ll)s.  jht  square  inch.  Find  the  eoeffieient 
ot   ultimate^  strength.  ,,i„,„,    ogjcQ  ib.,. 

112.  A  ..()-i!i.  <>ylin.lrical  boiler  of  ^,-in.  plates  is  made  of  wrougRf  iron 
whose  s;ife  coefficient  of  strength  is  KXX)  lbs.  |H-r  square  inch.  Find  the  work- 
ing pressure.  Jf  /^  -  ;}0,(KJ(),0(K)  lbs.,  also  find  the  thicknes.s  and  defleetion 
of  the  end   pi  „es.  ,i„,.  50  ,,„./,^_   ;„ 

113.  Deterniine  the  thiekne.ss  of  the  plates  used  in  the  eonstruetion  of  a 
lH)il.r  10  It.  in  diameter  working  under  a  pressure  of  IL'O  lbs.  per  square  inch 

.faking  the  efficiency  of  the  joints  to  be  70  per  cent,  and  the  safe  stress  at  10  (KK) 
lbs.  ,x>r  squar..  in.h.  .j„,    j  ^3   j,; 

114.  1-iiid  the  thickness  of    plates  required  for  a  boiler-shell  to  work  at 
a  pressure  of  L'(H)  lbs.  ])er  .square  inch,  the  diameter  of  the  shell  lx>iiig  .j  ft 
f>  in.:  the  allowable  stress  on  the  i)late  may  be  taken  as  12,000  lbs.  per  square 
inch.  ,  ... 

115.  Ihe  plates  of  a  cylindrical  boiler  .T  ft.  in  diameter  are  J  in.  thick 
Find  to  what  j.ressure  the  boiler  may  be  worked  so  that  the  tensile  stress  in 
the  j.lates  may  not  exceed  li  tons  i)er  square  inch  of  grcss  section. 

.,,,.,.     ,  ,     ,  Ati>i.  ")0  Ibs./sq.  in. 

116.  (  ylindneal  boiler  of  mild  steel  'i  ft.  fl  in.  in  diameter  at  elastic  limit- 
pressure  .SOO  jbs.  [x-r  square  in.h.  What  is  the  thickness?  Joints  supposed' 
to  be  of  CO  ,H.r  cent  of  strength  of  unhurt  plate.  Replace  this  boiler  with 
tubes  ..  m.  diameter  of  same  length.  How  manv  tubes  are  needed  to  make 
u].  the  .same  volume?  What  will  Ik;  their  thickness  (no  seams)?  Their 
"■'^'"'"''  ^,  -I"*-  --l^'"-;  174;  .01']  in.:  19G  lbs.  perU.  length. 

117.  The  i>ressuro  in  a  boiler  of  1.2  m.  diameter  is  V  .atmospheres.  What 
must  be  the  thickness  of  th(^  metal  if  the  .safe  working  stress  is  3/t/mm.-'? 

a    „.,  -I"-'*.  12.4  mm. 

118.  What  must  Ix-  the  thickness  of  the  metal  in  a  cast-iron  evlinder  of 
400  ,nm.  .iianicter,  whi.'h  has  to  lx,,r  an  internal  pressure  of  7  atmosi)heres, 
tlie  s.-.fe  working  strength  of  the  metal  Ix-ing  1,5/t/uun.?  Ans.  K2fi  mm 

119.  .\  eopix-r  pii)e  is  4  in.  diain.'ter  and  g  in.  thick.  What  is  the  working 
pressure?    Take  j -^-'m)  lbs.  jxt  square  inch.  An.'^.  'm  lbs.  'sq.  in. 

120.  Wiiat  IS  the  collapsing  f)r.>ssure  of  a  flue  10  ft.  long,  3()  in.  in  diameter 
an.  .•.,inpos...l  of  \.\n.  j.lates?  Als.,  of  a  Hue  30  ft.  long,  4S  in.  in  diameter 
and   ,>,-H,.  thi.-k?  .,„,    4,j(),^4  j^,  .   ,„  _:5,,  j^, 

121.  Determine  the  tlurkn.-ss  of  a  2-in.  locomotive  fire-tube  to  supixirt 
an  external  pressure  of  5  atmosjiheres.  A^g^  .073  in. 

122.  ,V  copper  .<;team-pipc  is  4  in.  in  diameter  and  }  in.  thick.  Find  the 
working  iire.ssure,  the  safe  e.x-ffieient  of  strength  f.)r  cipiH-r  Ix-ing  1000  lbs. 
Fx;r  s,,uare   in.h.  .4....  12.^  Ihs.   jx^r  square  inch. 

123.  iMiK  the  pro|MT  head  of  water  for  a  piiic  of  3  ft.  diameter  and  1  in. 
thickness.     Also  fiu.l  the  correspon.ling  pressure  in  pounds  per  square  inch. 


EXAMPLES. 


289 


iMirrnine  what  value  of  the  working  stress  will  cause  the  ordinary  rule  for 
ilii'i  cylinders  to  give  the  same  answer. 

Ans.  316  ft.;  136.7  lbs. /sq.  in.;  2469  lbs. /sq.  in. 

124.  .\  10-in.  cast-iron  water-pipe  is  subjected  to  a  pressure  of  250  lbs. 
)n  r  square  inch.  Find  its  thickness,  the  coefficient  of  working  strength  being 
.'(RHt  Uw.   ]ycT  square   inch.  Ans.  i  in. 

125.  .\  cast-iron  main  is  36  in.  internal  diameter  and  1  in.  thick,  the  safe 
tensile  stress  Ixjing  li  tons  per  square  inch.  What  is  the  greatest  head  of 
watrr  to  whi"h  it  should  be  subjected?  Ann.  4.30  ft. 

If  the  pipes  are  of  wrought  iron,  what  should  be  the  thickness  of  the  metal, 
tile  safe  strength  per  square  inch  being  5  tons  and  the  strength  of  the  longi- 
tudinal seams  being  60  per  cent  that  of  the  solid  plate?  Ans.  J  inch. 


126.  (liven  a  wrought-iron  pipe  2  ft.  diameter,  i  in.  thick:  its  working  stress 
IS  .")  tons  to  the  square  inch,  but  strength  of  plate  is  diminished  30  per  cent 
beciiu-se  of  riveted  joint.     What  is  the  working  pressure? 

Ans.  326  Ibs./sq.  in. 

127.  A  cylindrical  boiler,  12  ft.  diameter,  is  constructed  of  f|-in.  steel 
plate.  The  test  pressure  applied  is  245  lbs.  per  square  inch.  Find  the  stress 
[irocluced  in  the  plate,  and  hence  deduce  the  stress  in  the  plate  between  the 
rivet-holos,  the  sectional  area  being  there  reduced  to  .77  of  the  solid. 

Ans.  19,465  Ibs./sq.  in.;    25,280  Ibs./sq.  in. 

128.  Sixty  tubes  of  wrought  iron,  4  in.  inside  diameter,  10  ft.  long,  }  in. 
tliiik.  l-lnd  volume,  weight,  internal  area  in  square  inches,  and  working 
pressure,  if  working  stress  =10,000  lbs.  per  square  inch.  Neglect  the  ends. 
Hnu  iii.iny  tubes  22  in.  diameter,  10  ft.  long,  will  have  the  same  volume? 
.Villi  liiid  the  thickness  suitable  for  the  working  pressure.  Find  also  the  area 
and  wight. 

Alls.  ')2.m  cu.  ft,;  6679  lbs.  weight;  12.54  sq.  in.;  1250  Ibs./sq.  in.;  2; 
1. ;{:.■)  in.;  380  sq.  in.;  6734  lbs. 

129.  X  vertical  cast-iron  pipe  is  4  in.  in  internal  diameter.  The  pressure 
at  a  eertitin  place  is  50  lbs.  per  square  inch.  At  this  place,  and  at  a  place 
:W)  feet  lower  in  level,  find  the  proper  thickness  of  the  metal  if  the  working 
stress  is  taken  as  3000  lbs.  per  square  inch.  Ans.  .033  in.;  .12  in. 

1 30.  .\  steel  spherical  shell  36  in.  in  diameter  and  f  in.  thick  is  subjected 
to  an  internal  fluid  pressure  of  300  lbs.  per  square  inch.     Find  its  coefficient 

strength.  Ans.  7200  lbs. 

131.  -V  thin,  hollow,  spherical  elastic  envelope,  whose  internal  radius  ia 
\  as  subjected  to  a  fluid  pressure  which  caused  it  to  expand  gradually  until 
radius  became  R\.    Determine  the  work  done. 

Ans.  !f(/?,«-7?')Xunit  pr. 

132.  A  prismatic  bar  of  volume  V  changes  in  length  from  L  to  L  ±x  under 
jliiiil  prcusure -p.     Find  the  eorresoonding  work.     Ans.  ipF  log.  (L±x). 


ot 

A', 

its 


the 


CHAPTER  V. 


STRESSES,  STRAINS,  EARTHWORK,  AND  RETAINING-WALLS. 

I.  Internal  Stresses. — The  application  of  external  forces  to  a 
material  body  will  strain  or  deform  it,  and  the  particles  of  the  body- 
will  \)e  in  a  state  of  mutual  stress. 

In  the  following  calculations  it  is  assumed 

(a)  That  the  stresses  under  consideration  are  parallel  to  one 
and  the  same  plane,  viz.,  the  plane  of  the  paper: 

(h)  That  the  stresses  normal  to  this  plane  are  constant  in  direc- 
tion and  magnitude: 

(f)  That  the  thickness  of  the  plane  is  unity. 

Def.  The  angle  between  the  direction  of  a  given  stress  and  the 
normal  to  the  plane  on  which  it  acts  is  called  the  obliquity  of  the 
stress. 

a.  Compound  Strain.— («)  First  consider  an  indefinitely  small 
rectangular  element  OACB  (Fig.  321)  of  a  strained  body,  kept  in 

ecjuilibrium    by  stresses    acting    as  in 
thv'  figure. 

/>  is  the  intensity  of  stress  on  the 
faces  on,  AC,  and  a  its  obliquity. 
q   is  the  intensity  of  stress  on  the 
faces  0.1,   [i(\  and  ^?  its  obli(]uity. 
OJiit  cos  n.  the  total  normal  stress 
on    OH,    is    l)alanced    by    .ICpcosa, 
the  total  normal  stress  on  AC. 

OH/}  sin  rt.  tlie  total  shear  on  Ofi,  is  equal  in  magnitude  but 
op|M)sife  ill  direction  to  .ir'/isiiia.  the  total  .shear  on  AC. 
These   two  forces,   therefore,   form   a   couple  of  moment 


0 


\     K     \ 
p     p       p 

Fio.  .321. 


\ 


OBp  sin  aOA. 


290 


COMPOUND  STRAIN. 


291 


Similarly,  the  total  normal  stresses  on  the  faces  OA,  BC  balance 
and  the  total  shears  form  a  couple  of  moment  OA-q  sin  fi  OB. 

In  order  that  equilibrium  may  be  maintained  the  two  couples 
must  balance. 

Tlierefore  OBp  sin  aOA  =  OAq  sin  ,?  OB, 

or  p  sin  a  =  q  sin  (3  =  t,  suppose. 

Hence  at  any  point  of  a  strained  body  the  intensities  of  the  shears 
0)1  any  two  planes  at  right  angles  to  each  other 
are  iqual. 

Ill)  Second,  let    it  be   required  to  find  the 

resultant  stress  upon  any  plane  MN  at  a  point 

(/  in  a  strained  body.    Consider  the  equilibrium     9 
of  an  indefinitely  small  <non^Zar  element  0^45, 
liourulod  by  a  plane  AB  parallel  to  MN  and 
tun  planes  OA,  OB,  at   right  angles  to  each 
other  (Fig.  322). 

Let  /)  he  the  stress  on  OB,  a  its  obliquity. 

Let  q  be  the  stress  on  OA,  ^  its  obliquity. 

Let  t  be  the  shearing  stress  on  each  of  the  planes  OA,  OB.     Then 

t  =  p  sin  o=g  sin  j3. 

Pn,  the  normal  component  of  p,  =  p  cos  o. 
Qn,    "       "  "        "    ?,=gcos^. 

Prndtice  0.4  and  take  0C  =  j.„0«  +  f0.4  =the  total  force  on  OB 
in  the  diiection  of  OA. 

I 'n. (luce  OB,  and  take  OD  =  qnOA+tOB -the  total  force  on  OA 
in  the  ilireetion  of  OB. 

('<ini|(lete  the  rectangle  CD. 

"/."  represents  in  direction  and  magnitude  the  resultant  of  the 
tA"  Mirees  (H\  01),  and  must,  therefore,  lie  equal  in  magnitude  and 
oi'(""ii"  in  direction  to  the  total  force  on  AB. 

Let  /v  1k'  the  stress  on  AB.    Then 


(M  ny  ^oj'P'^oc^-^  oiy = {p„0B + «7^ )» + {q^oA  4  toBy, 


292 


THEORY  OF  STRUCTURES. 


and  p,2  =  (p„  sin  r  +  <  cos  r)^  +  {q„  cos  r  +  t  sin  yy 

=  Pn2  "inz  ;-  +  q„2  cos2  r+tsin2KPn+qn)+t». 

In  the  limit  MX  and  AB  coincide,  and  p^  is  the  resultant  stress 
upon  the  plane  MX  at  O  and  acts  in  the  direction  OE. 

U  t  =  0,  then  a  and  ^  are  each=0,  and  there  is  no  shearing  stress 
upon  0.1  or  upon  OB  at  0.    The  stress  upon  each  of  these  planes 
at  0  is  wholly  normal,  and  such  stresses  arc  called  principal  strefises 
while  the  two  planes  upon  which  they  act  are  called  planes  of  prin- 
cipal KtrCSti. 

If  OA  and  OB  are  j)lanes  of  principal  stress  and  if  pi  and  p2 
are  the  principal  stresses  along  0^1  and  OB  respectively,  then  the 
resultant  stress  />,  on  any  plane  MX  is  given  by 

p,2  =  p,2sin2;-+p22cos2r. 

3.  Constant   Components  of  p,.— Equal  Princi^  .1  Stresses.— Let 

OA,  OB  be  the  princijml  axes  at  any  point  0  in  a  strained  muas.    It 


./;..- 7    ^■ 


-^x^h^ 


Fia.  324. 

is  required  to  determine  the  resultant  stress  on  any  plane  A/O.V  when 
the  two  principal  str<'ss<'s  are  like  and  equal. 

Let  cacli  principMl  stress  =  />. 

Let  tiie  aiijrje  .UO.l  -y- 

Let  the  |)oint  ff  lie  within  an  indefinitely  small  rectangular  element 
liaviiig  its  sides  pisraliei  to  O.l  and  (>B. 

I''irst.  ('(insider  the  etTect  of  the  stress  p  acting  in  a  direction 
palailtl  lo  OA,  Fig.  '62^.  (Urn  tiirvaa  may  be  due  either  to  a  pull 
or  a  push.) 


CONSTANT  COMPONENTS  OF  Pr- 


293 


I>et  q  be  the  corresponding  parallel  resultant  stress  developed  in 
tho  plane  MON.    Then 


an  1  therefore 


qMN=pCC'=pBB', 
q=p  am  f. 


^,,  8in2/' 

The  shear  stress  along  OM=q  cos  r  =  p  — 2~"- 

The  normal  stress  upon  OM = g  sin  r  =  P  sin*  y 

Second.  Let  the  stress  p  act  in  a  direction  parallel  to  OB,  Fig.  324, 
and  let  (/  be  the  corresponding  parallel  resultant  stress  developed  in 
MS.    Then 


Therefore 


^MN-pNN'. 

^"pCOSf. 


sin  2r 
The  shear  stress  along  0N=^  sm  r=P — 2~' 

The  normal  stress  upon  ON =ct  cos  ^--pcos*  y. 

Third.    Let  the  two  like  principal  stresses  now  act  simultaneously. 


Tlh'  resultant  fffect  is  tho  algebraic  sum  of  the  separate  effects,  and 
till  II  fore  tho  resultant  stress  at  O  on  tho  i)lane  MOX  is  equivalent  to 

in  I  two  Hhoar  stn'twoM  ourh  otjual  to       -     ,  but  acting  in  upposUe 

tliroi'tioiis  so  that  they  neutraliae  each  other,  and 


294 


THEORY  OF  STUUCTURES. 


(b)  two  normal  strossos,  p  s'm-  y  and  p  cos^  y,  acting  in  the  same 
dinction  so  that  they  are  together  ecjual  to  a  uornial  stress 

p  sin^  r  +  P  t'os^  r  =  P- 

Hence,  if  the  two  prmnpal  .ttresaefi  at  a  point  in  a  strained  mass 
arc  like  and  equal,  the  resultant  stress  on  any  plane  throuf/h  the  jmni 
is  also  a  like  and  equal  stress  and  is  normal  to  the  plane,  Fi;/.  326. 

Again,  let  the  two  i)rincipal  stresses  be  equal  but  unlike. 

Suppose,  forexanii)]e,  that  the  stress  parallel  toO.l  is  due  to  a  push, 
so  tl.  it  it  acts  in  the  direction  shown  by  Fig.  327.     The  resultant 

7)  Sin  ^y 

effect  on  t'le  plane  MOX  is  now  a  shear  stress     \~    acting  in  the 


c  . 


ki/ 


11 


X 


A' 

Fio.  327. 


Fio.  328. 


same  direction  as  the  shear  stress  due  to  the  stn'.'w  p  in  Fig.  324,  and 
a  normal  stress  psin^;-  acting  in  a  direction  oppimte  to  that  of  the 
corresponding  stre.s,s  in  Fig.  .{23. 

Hence,  if  the  two  unlike  but  eciual  principal  stresses  act  simultane- 
ously, they  are  eiiuivalent  to  (Fig.  32S) 

I  \  u     ^     1  .         /'  ^i"  -X    P  sin  2r 

(")  a   resultant   .shear  stress   L~-J-+'  ■-psin2r  in  the 

plane  U.V.  and 

(/•)  a  resultant  normal  .stress  p  vns^y-p  sin^  ^  =  p  cos  2^.  Tliis 
normal  stress  will  Ik>  a  |)ush  or  a  pull  according  as  r>  or  <4.5°. 

Take  OK,  I'ig.  32<>,  in  the  plane  .V.V  =  ;>  sin  '2r  and  OL  normal  to 
the  plane  =  p  cos  2;-. 


UNEQUAL  PRINCIPAL  STRESSES. 


295 


Complete  the  rectangle  OKHL.    Then  OH  is  the  resultant  stress 
oil  the  plane  MX  at  0,  and  its  value  is  given  by 


OH  =  VOL^ + OK^  =  Vp2  cos2  2r  +  f  sin^  2r = p. 


Also,  if  the  angle  LOH=^, 

„    OK    p  sin  2r    .      o 
'^    OL     pcos'Zy 

and  therefore     /3  =  2r    of     i8o°  +  2r. 

//pfwe,  i/  </i€  /wo  principal  stresses  at 
a  iiniiit  in  a  strained  mo-ss  are  equal  but 
iDilikc,    the    resultant   stress   upon   any 

pliinv  through  the  point  is  an  equal  stress  acting  in  a  direction  inclined 
at  (in  angle  2r  to  the  normal. 

4.  Unequal  Principal  Stresses.— Let  the  principal  stresses  at  any 
lioiiit  be  pi  and  p^.    Tlu'n 


Fig.  329. 


pi=^^^-^''=m+n,  suppose, 


and 


P1+P2       Pl-P3_ 

P2=— ^5 :i     -ffi-n. 


t  \VI ) 
-I  IT 

niav 

(ii: 

■ir  ! 

CmII 

-In 


"ho  principal  stresses  may  therefore  be  considered  as  composed  of 
like  and  equal  stresses  w(  =  -  y- V  and  of  two  unlike  and  equal 

,^^,^  fJ^PlZlA      xhe  effects  of  these  like  and  unlike  stresses 

l)c  olitiiiiied  sppanitoly  and  their  results  sui)rri)osed. 
I  lie  result  ant  of  the  two  like  stresses  is  a  stress  m  along  the  normal 
to  the  plane  .U.V,  Figs.  325  and  326,  and  the  resultant  of  the 

inlike  stresses  is  a  stress  n alongOF making  an  angle  2r  (I'ig.  330) 
si)°-i-2^  (Fig.  IVM)  with  th'"  nnrniul.     Take  OE  =  m  and  OF-  »i. 

plcte  the  parallelogram  OK(lF.    The  diagonal  OG  is  the  resultant 
on  MS  at  0  in  direction  and  magnitude. 


2^  THEORY  OF  STRUCTURES. 

Its  value  by  Fig.  330  is  given  by 
A  6'6'2  =  m'  +  n2  -  2mn  cos  (180°  -  2/-) 


-^Jv 


/ 


F 
N 

Fig.  330. 


-     2        — o —  ^°^  2r 

=  Pl2  00.^2  ^  +  p^2sin2^ 

Its  value  by  Fig.  331  i.s  given  by 
OG2  =  m2  +  n2-2wnco8  2r 


-       2  ^^ ^^^  2;- 

=  /)l2.sin2;-  +  P22pog2  ^^ 

as  already  obtained  in  Art.  2. 

Thas,  in  the  triangle  OEG,  the 
angle  OEG  is  either  2;-  or  180° -2r. 

Lot  6'£'  be  produced  to  meet 
0.1  and  OB  in  A'  and  L.  Then, 
evidently, 

EK  =  EO^EL^m, 
and  ff  is  the  niiddlo  point  of  KL. 

Therefore,  by  Fig.  330, 


Fig.  331. 


GK^GE  +  EK=  -\-n+m''pi, 
GL^-GE  +  EL^'-n  +  m^Pi, 


and,  by  Fig.  331, 


~GK^  -GE  +  EK^  -n  +  m^P2, 


an<J  GL  =  +  GE  hEK=+n  +  m  =  />, . 

Lt>t  t  1)0  the  arigio  I)ot\vo<<n  6'£  and  the  plane  MN.    Then 

ni?  P^~P2 

UE  cos  f  =  -^^  COS  c  IS  tho  shear  stress  in  MX  at  0. 


PL.ISES  OF  PRINCIPAL  STRESS. 


297 


This  is  greatest  when  cos  £  =  1,  i.e.,  when  £=0  or  GE  \s  parallel 
to  MX.  In  this  case  OE  is  at  right  angles  to  EG,  and  therefore 
2^  =  90°  or  ;-=45°.  Hence  the  shear  strpss  in  MN  at  0  is  a  maxi- 
mum and  equal  to  ^'"^^  when  MN  is  inclined  to  the  axis  of  prin- 

(•i])al  stress  at  an  angle  of  45°. 

5.  Planes  of  Principal  Stress.— Ellipse  of  Stress.— Let  MON  be 
a  plane  of  principal  stress,  i.e.,  a  plane  in  which  there  is  no 
shearing  stress  at  0.  The  resultant  OE,  Fig.  332,  must  be 
normal  to  MN  and  AB,  and  therefore 

^-„    OC     pnOB  +  lOA    p„t&nr  +  t 
tan^=  cot  CC/£/=pjTj=       -         --- 


troni  which 


2t 


OD     q„OA+tOB    g„  +  Uan  r' 
9  tan  j- 


qn-Pn     1    tan^r 


=  tan  a;-. 


Two  values  of  r.  viz.,  r  and  r+90°.  satisfy 
(■(juation   so   that   at  every   pomt  0  there  are  two        p,Q  332. 
planes,  at  right  angles  to  each  other,  upon  which  the 
stress  is  wholly  normal,  i.e.,  there  are  two  principal  planes. 
Again,  take  the  prmcipal  planes  as  the  planes  of  reference. 

Let  p\,  P2  be  the  principal  stresses  along  OA, 
OB,  Fig.  333,  respectively.  Consider  as  before 
the  equilibrium  of  a  triangular  element  OAB,  AB 
being  parallel  to  a  plane  MON  inclined  at  an 
angle  ;-  to  OA  at  0. 

Take  OC  =  f\OB,  OD  =  p/)A,  and  complete 
the  rectangle  CD.  Then  OE  =  pAB  is  the  re- 
sultant of  OC  and  OD. 

Take  OR  =  p„  and  let  x,  y  he  the  coordinates, 
Fia.  333.  with  respect  to  0,  of  the  point  R.    Then 


OC        pxOB 
x-p,co8  v'-p,  o^  =  /'r^;^  =  PiS«n  p 


.     ,        OD         p/)A 
md  y-pr8mv^  =  p,^,='P,  ^^-j^-pzcos-. 


298 


THEORY  OF  STRUCTURES. 


Therefore  —^  +  — 2  ==  ^'"^  T  "^  cos^  7-  =  1 ,    and    tan  r  tan  ^ 


Pi' 


Hence  the  locus  of  R  is  an  ellipse,  called  the  ellipse  of  stress, 
with  its  axes  (2;)i  and  2p2)  lying  in  the  j)lanes  of  principal  stress. 
The  stress  upon  any  plane  MOX  at  0  inclined  at  r  to  the  major 
axis  is  the  semi-diameter  of  the  ellipse  drawn  in  a  direction  making 
an  angle  y''  with  the  major  axis,  (/-  being  given  by  the  last  equa- 
tion. 

If,  further,  the  two  principal  stres.ses  are  of  equal  magnitude 
but  of  opposite  sign,  i.e.,  if  the  one  is  a  push  and  the  other  a  pull, 

the  normal    comiionent    -^--    Ijecomes  nil,  so  that  there  is  no 

normal  stress,  while  the  maxinmm  shear   stress    on    each   of  two 
planes  making  an  angle   of  45°  with  the  axes  of   principal   stress 


IS 


P1-P2 


-Pl  =  -P2- 

This  is  defined  to  be  a  state  of  simple  shear.  It  has  also 
alrea.ly  been  proved,  Art.  2,  that  if  there  is  a  s>^-ar  stress  on  any 
given  plane  at  a  point  in  a  strained  solid,  there  must  be  an  equal 
.«liear  stress  on  a.  plane  through  the  point  at  right  angles  to  the 
first. 

In  simple  shear  an  elementary  cube  becomes  distorted  (Fig.  334) 
without  change  of  volume.  One  diagonal  is  lengthened  and  the 
other  shortened  by  the  same  amount,  while  its  angles  arc  changed 
by  a  .small  quantity  <l>  to  rK)°  +  .^  and  90°-<f>.  This  small  angle 
<^  is  taken  as  a  measure  of  the  .sh  ar  stra:  ^  and  is  called  the  angle 
of  shear.  Hence,  by  Hooke's  law,  which  applies  to  shear  as  to  other 
stresses. 

q,  the  intensity  of  shear  a  </) 

or   q  =  G</>, 


/ 

/ 
/ 

< 

t 
1 
1 
1 

*. 

■ 

where  G   is   a   coefficient    called   the    modulus   of 
Fici.  ;w4.         ri(ii(Hhj.      Tlie  \-alue  of  G  is  generally  determuied 
by  torsion  ex]K'riments  and  is  usually  found  to  be 
about    fE,  E  lx>ing  Young's  modulus. 
Again, 

the  work  stored  up  by  the  distortion  =  J9!.  =  .|^. 

2        20- 


J:,)   :' 


■■'«t,M' 


COS  JUG  ATE  STRESSES. 


299 


Fio.  335. 


6.  Conjugate  Stresses.— The  obliquity  of  a  stress  is  the  angle 
botwocu  the  direction  of  the  stress  and  the  normal  to  the  plane 
iiI>on  which  it  acts.  This  angle  will  be  designated  in  the  following 
calculations  by  the  symbol  d. 

At  any  point  in  a  strained  mass  two  stresses  are  said  to  be  con- 
jiifidte  when  each  stress  acts  upon  a  plane  parallel  to  the  direction  of 
tlu'  other. 

In  Fig.  335  the  stress  acting  upon  the  plane 
M()\  at  0  is  parallel  to  the  plane  WON',  and 
its  (jl)li(iuity  is  the  angle  M'OE',  OE'  being  the 
normal  to  MON.  If  the  stress  at  0  upon  the 
piano  WON'  is  parallel  to  MON,  its  obliquity, 
viz.,  the  angle  MOB  between  OM  and  the 
luiniuil  OE  to  the  plane  WON',  is  evidently 
0(iu;il  to  M'OE'.  Hence  the  conjugate  stresses 
at  any  jwint  in  a  strained  mass  hare  equal 
ohhq'iities. 

L(>t  OA,  OB  be  the  planes  of  principal  stress  at  any  point  0  of  a 
straiiunl  mass,  and  let  MON  be  any  other  plane  inclined  at  an  angle 
r  to  OA,  Fig.  336. 

Upon  the  normal  to  MN  take  OE=m  =  ^^  J^^,  and  with  E  as 

centre  and  a  raditis=n=   '  ^     ,  describe  a  circle  intersecting  OE  at 

P  and  Q.      Draw  any  line  OSR  inter- 
secting the  circle  at  S  and  R.    Join 
.  ES  and  ER.    Then  OR  and  OS  are 

J>^  each  the  third  side  of  a  triangle  of 
Vl^  which  the  other  two  sides  are  the  con- 
stant components  m  ami  n,  while  the 
angle  ROE  is  common  to  the  two  tri- 
angles. Thus  OR  and  OS  are  conju- 
gate stresses,  and  the  obliquity  0  of 
each  is  the  angle  ROE. 

Draw  the  tangent  OT  and  join  ET. 
The  angle  EOT  is  the  maximum  value 
wliich  the  obliquity  0  can  have,  and  corresponds  to  the  case  in 
wliicli  the  two  conjugate  stresses  are  equal. 


Fio.  330. 


3^  THEORY  OF  STRUCTURES. 

Let  ^  be  the  maximiuii  value  of  0,  i.e.,  the  angle  TOE.     Then 

,     ET     n     Pi— 02 
OE    m     P1+P2 

Let  r=OR,  the  greater  of  the  two  conjugate  stresses. 
Lets  =  05,     "lesser     "    "     "  '<  « 

Draw  EV  at  right  angles  to  and  bisecting  SR. 
Then 

r  +  8=Ofl+OS  =  20F=2mcos<?  =  (p,+p2)cos<?, 

and  ra  =  0i?05=0r2=0£:2-iS;r2=m2-n2=p,p2. 

Again,  EV  =m  sin  d,  and  therefore 

VR  =  VS=V~ER2-EV2  =  Vn''~mHm^0 

/n2      TT"         .— 

=■=  "»\^  - sm2  5  =  TrtVsina  ^ -sin2  5 

=mV'cos2  6'-cos2^ 
Hence  r'=OV  +  VR  =  m(cos  0 + Vcos^^^-cos^  ^) 

and       s=Or-F5=w(cos0-V'cos2(?-cos2(^). 
Therefore         J    cos  >9+n/cos2  g-^cos^  ^ 

S       cos  tf  -  v/cOB2  <?  -  C0S2  ^' 

r  being  >  s. 

n  *i-        •  cos  d> 

1  uttmg  sin  «  =  ^^^,  the  last  equation  may  be  written 

r     I  +  cos  a  a 

8       I-COStt""*^**      2" 

or,^^  ^"  fjl  P,7"«>^  ^"•''•'•^''•'  «'  «  M«<  in  a  strained  mass  are  tensions  of 
.on  Ihs   «,,v/  171  lb.,  per  x,,uare  i,u-h.     Find  the  re,uUant  stress  and  its  obliquity 

Zess  '"""'  '""''""'^  *"  ^^°  '"  ''^  P''""'  "^  ^"^^  principal 

and  OEG^rA". 


!    1 


EXAMPLES. 


301 


Therefore 
(resultant  stress)' -?n'+n'-2mn  cos  (180° -2r) 

=213  +42' -2.213.42 -cos  54°, 
and  resultant  stress  =  191.35  lbs.  sq.  inch. 

sin  0  fl  42  sin  0 


K 


Again, 
Therefore 


sin  -sr     191.35     191  35    sin  54°* 
sin  tf  - .  1775    and    0  - 10'  14'. 


p,-v.i 


Fio.  337. 


t^s 


Ex.  2.  The  principal  Mrrssis  at  a  point  in  a  strained  nolid  are  a  tension 
uj  ;}(X)  Ibs./sq.  in.  and  a  ronipnssion  of  200  Ibs./Hq.  in.  Find  (a)  the  resultant 
stress  on  a  plane  inclined  at  30°  to  the  axis  of  greatest  principal  stress  and  its 


aP,=soo 


FiQ.  338. 


Fio.  339. 


iihliquity;  (b)  the  plane  upon  which  there  is  only  a  shearing  stress  and  the  magni' 
tilde  of  the  stress. 

300-200 
(a)  m 50, 


300  +  200     „.„ 
„=— ^— =2oO. 

Then  by  Fig.  338     (X?'=m»+n'-2mn  cos  (180°-60°), 
■=2500+62,500  +  12,500 
=77,r)00 
and  OG  =278.4  lbs.  sq.  inch. 


Also, 


nin(60°-0)       .W      1        ■    tino      ,  a     „„=  Rno 
'  =  —  =  —  =8in60  -cot  o  — cosoO  , 
2.-)0     5 


sinO 


eotfl-.8083    and     0=51°  3'. 


i.h')  I.rt  the  plane  make  an  angle  r  with  the  axis  of  greatest  principal  stress. 
The  resultant  stress  OG  Ues  wholly  in  this  plane. 


'^"-  THEORY  OF  STRUCTURES. 

Therefore  by  Fig.  339      OG'  =n»- w'  =60,000, 


and 


Again, 


0G=  245  lbs. 


C09  2r=--=.2    and    2^  =  78°  28', 


or 


r=39°14' 


Ex  3.  At  a  point  within  a  strained  solid  there  are  two  like  conjugate  streaseit 
of  70  lbs.  and  10  Ih.'^.,  the  common  obliquity  being  cos-'  .8.  Find  the  principal 
stresses  and  the  angle  bitween  the  two  planes  on  which  the  conjugate  stresses  act. 

70  +  10  =  r+s  =  (p,+p,)i. 
Therefore      p,+pi  =  \00. 
Also,  pip2=rs=700. 

Hence    pi-p,  =85  very  nearly, 
so  that  p,=92i    and    pj  =  7f 


Again, 


and 


U(j=  — ~-  =.50  =  m, 


G«=^-y£'=30N^2=n. 


Fig.  340. 


Therefore 

70'=Off'=50'  +  (30V2)'-2-50-30\  J.cos  (180°-2r), 
«•  cos  2r  =.1414    and     r=40°56'. 

Also,  10'=O5'=.W  +  (3OV2)'-2.50.30v  2-cos  (180°-2rO, 

**'  cos  (180° -2r')  =.9998    and     r'=85°55' 

)-'-r=44°o9'. 


Hence 


7.  Shaftmg.-I.Pt  a  shaft  })e  .suhjoctpd  to  the  action  of  a  couple  of 
nioiiK.nt  .1/,  ,„  a  phun-  „t  ru,lu  <n,,lcs  to  the  direction  of  the  shaft 
and  also  to  a  load  which  de\olops  stresses  in  the  direction  of  its 
length. 

A  particle  at  any  cros.s-seclion  of  the  shaft  is  acted  upon  by 
•     (a)  a  stress  />  normal  to  the  .section, 


;!i  ;i 


SHAFTIXa. 


303 


(6)  two  equal  shear  stresses  q  in  the  plane  of  the  section  and  at 
rijrlit  angles  to  each  other.  hi -'' 

Combining  one  of  these  shear  stresses 
with  /),  and  taking  OF  =  p  and  OH  =  q,     q 
t! ,.-  rcsultaut  stress  is  OG  {  =  \']i^  +  q^), 
the  diagonal  of  the  rectangle  OFGH. 

Take    OE  =  m-=^^^^   and   join    EG. 

Then 


EG^n 


P1-P2 


Fic.  341. 


The  remaining  shear  stress  is  normal  to  the  plane  OFG.  Rotate 
tlic  jilane  containing  this  stress  and  p  until  it  is  coincident  with  the 
]il:iiic  OFG,  and  take  OG'=q.  Then  OE  and  EG'  are  the  constant 
('oiiii)onents  of  06",  and  GEG'  is  evidently  a  straight  line..  Th';re- 
forc  E  is  the  middle  point  of  OF.    Hence 


the  maximum  shear  stress  =   ^  ^      =EG  =  \j-^  +  5*, 


and 


P       17^ 
the  maximum  principal  stress  =  pi=OE+ EG  =  9  +  "XT  "*■  9** 

Also,  as  proved  in  Chap.  IX,  if  the  shaft  is  round  and  of  radius  r^ 

2Mt 


9=- 


w 


3  • 


Ex.  4.  Suppose  that  a  force  P  acts  along  the  shaft,  its  line  of  action  coinciding 

iiith  the  axis  of  the  shaft. 

P 


Then 


Therefore 


iiiid 


I"x.  r».  Suppo.^  that  p  is  caused  by  a  bending  action  due  10  a  force  acting 
:it  ri<:ht  angles  to  the  shaft,  as,  e.g.,  when  a  horizontal  shaft  carries  a  heavy 
I'lilli'v  b»'tw(>en  its  bearings. 


304  THEORY  OF  STRUCTURES 

Let  -1/6  be  the  bending  moment  at  any  point  due  to  the  force.    Then 

A  Mb 


Therefore 


P'-^ 


and 


2         -r' 

p>  =  ^  (Mb + \  mT'Tm?)  • 


By  this  and  the  preredinp  example,  if  the  vahies  of  the  working  principal 
and  maximum  shear  stresses  are  given,  the  corresponding  values  of  r  can  Iw 
at  once  calculated,  the  greater  value  being  adopted  for  the  shaft  in  ques- 
tion. 

Ex.  G.  A  strd  Klinft,  in  vhich  the  working  stress  is  not  to  exceed  11^00  lbs. /so. 
in.,  is  40  ft.  between  hearing.'^  and  carries  a  30-ui.  pulley  weighing  200  lbs.  The 
effective  tangential  force  on  the  pulley  is  600  Ibs./sq.  in.  Find  the  diameter  of 
the  shaft. 

•  11200  =^  =^  \  Mb  +  -v'Mb'+Mt'  [  , 

(000+200)40 


.1/6  = 


X 12  in.-lbs.  -96000  in.-lbfc, 


Hence 
and 


Ml  =()00  >;  !.■)  =<t()00  in.-lbs. 
H  -  r_.     , ( < t( MHK)  +  \  96000' +yOOO»> 
r  =2.22  in.,  or  dia.  -4.44  in. 


/ 


dr^"^ 


"^ 


Ex.  7.  Perhaps  the  most  inifwirtant  example  of  the  application  of  the 
prinriph'8  just  enunciated  is  the  ca.se  of  a  shaft 
acted  iiiHin  by  .-i  cmnk  (Fig.  .'^42). 

\  fcnc  /'  applied  t  i  the  centre  C  of  the  crank- 
I  pin    is    resisted    by  an    equal    and    op|K)site    force 

I  at    the  Iwaring  li,   (. inning  a  couple  of  moment 

'  FVli^M. 

This  couple  may  l)e  n-solved  into  n  bending 
cimplidf  nionicnf  Mb  - /'/Irt- /'•  W  cos^-.l/coscJ, 
and  a  twisting  couple  (,f  moment  Mt-PAU- 
I'UC  sin  (J-.U  sin  -J;  J  being  the  angle  ABC. 


lA 


Fio    342. 
Therefore 
and  ihc  max.  shear 


Di --.;.!/  cos  ^  f.l/l 


4M 


X 


cot' 


2M 


(•24) 
(25) 


RASKIXES  LARTinVORK  THEORY. 


305 


8.  Rankine's  Earthwork  Theory.— A  mass  of  earth  gives  way 
bv  tlu'  sliding  or  slipping  of  the  particles  over  each  other,  and  its 
t;|;iliility  (U'lJends  partly  uj^n  friction  and  partly  upon  the  cohesion 
of  tlu-  i)articles.  This  cohesive  power  may  be  considerable  in  cer- 
t;iiii  soils,  such  as  clay,  and  esixicially  when  they  are  moist,  but  it 
i<  cMiitually  destroyed  by  the  action  of  the  air  and  by  changes  of 
t(  iii|)('rature,  so  that  the  stability  of  the  mass  must  be  considered 
;i-  ilip'-nding  upon  the  jridinnal  resistance  only.  WTien  a  mass 
(1  I. MIX'  soil  is  piled  upon  the  ground  it  will  be  in  equihbrium  so 
h'liif  as  the  surface  slojx's  at  an  angle  which  does  exceed  a  certain 
iii.^ic,  called  the  amilc  of  repose  and  usually  designated  by  the  sym- 
liiil  (/).  Tables  have  l«'en  prepared  giving  the  value  of  <^  for  differ- 
ent soils,  and  for  other  soils  exjjeriencc  may  enable  the  engineer  to 
iiili'ipolato  suitable  values  for  the  angle  of  re[)ose. 

In  Fig.  343  two  particles  are  pressed  together  by  a  normal  stress  p. 


P 


Fio.  343. 


Fio.  344. 


1 1  1/  is  the  shear  stress  which  just  makes  the  one  |)article  slide  over 

(lie  (iiIkt,  then  --==the  coefhcient  of  friction  =  «. 

111  Fig.  344  two  niius-ses  of  earth,  contiiied  as  shown,  but  with 
tiiiiliiiii  (if  movement  along  the  plane  MX,  are  pressed  together  by 
li'iiiKil  forces  /'.  If  the  two  equal  forces  Q,  acting  in  op|^)osite 
iliiiiiiotis  and  at  right  angles  to  /*,  are  just  sufKcieiit  to  make  the 
"1)1'  mass  slide  over  the  other  along  M\,  then  again 

Q 


■T^-cfX'fHcient  of  friction -/i. 


Ill  Fig.  34.')  a  stress  p,  having  an  obli(iuity  0,  is  developed  on 
till'  laces  Mi  and  CD  by  the  pn'ssurea  /'.  l'>|uilibriuni  requires 
ii>>  'hNt'Iopmciii  t)f  a  siifss  </  on  tJie  faces  AC  an<i  /?/),  having  the 


306 


THEORY  OF  STRUCTURES. 


same    tangential    component    as   p,    i.e.,  p  sin  d.     But    the   normal 
components  of  q  on  the  two  faces  neutralize  each  other,  and  the 


//A 


C       7  D 

p 
Fig.  345. 


Fio.  346. 


tangential  components  are  at  right  angles  to  MX.  Thas  the  stability 
with  res[X'ct  to  the  plane  .U.V  is  unaffected  by  q,  and  if  slipping 
is  just  about  to  take  place,  so  that  0  =  (f),  then 

P  sin  <i 

u  =  7j -,  =  tan  d>. 

'^     P  cos  0  ^ 

If  Fig.  345  i.s  now  turned  round  until  P  is  vertical,  and  if  <?  =  ^ 
then  MX,  Fig.  346,  is  inclined  to  the  horizontal  at  the  angle  of  repose, 
and  may  be  called  the  plane  of  sliding. 

If  0<  fj),  and  if  the  soil  is  dry  and  non-cohesive,  sliding  will  not 
take  j)lace  however  great  p  may  Im*;  but  if  tf>^,  sliding  will  occur 
although  p  may  Ix*  ever  so  small. 

Ilei.oe  at  any  jioint  in  a  mass  of  earth  there  is  a  tendency  to 
slide  along  .'very  plane  through  the  jwrnt,  excepting  along  the  plants 
of  principal  stress.  This  tendency  inareasen  with  the  obliquity  of 
the  H'sultant  stre.ss  but  is  unaffected  by  its  magnitude.  It  has 
been  shown  that  if  pi  and  p-j  are  the  pruicipal  stresses  at  the  point, 


sin  tj> 


Pi+Pi 


aiK 


1  therefore     ^  J±^^. 
pi     1  -  sui  0 


Hence  at  etery    jmint  the  rnnditioti   of  rqmUhrvm   requires  thai 
the  i-atio  of  the  ijrmter  to  the  It'sscr  principal  stress  shall  not  exceed 


I  +  sin  0 
I  -  sin  0' 


PRESSURE  O.V  A   VERTICAL  PLANE. 


307 


9.  Pressure  on  a  Vertical  Plane. — Suppose  that  the  earth  beliind 

a  Willi  is  spn'ucl  out  in  layers  having  the  same 
uiiilorni  slope  0.  This  angle  must  be  less  than 
the  angle  of  repose  <f),  or  the  earth  will  run  over 
the  wall.  The  stress  on  a  vertical  plane  CD, 
V\fi.  .347,  at  any  point  D  is  parallel  to  the  ground 
.-Jiirtaoe  and  is  also  conjugate  to  the  vertical 
stress  at  D  upon  the  plane  DE  which  is  par- 
allfl  to  the  ground  surface.  This  follows  be- 
cause the  vertical  stress  is  balanced  inde|5end- 
('iitl\-  of  the  shear  stresses  on  the  sides,  and 
therefore  for  any  layer  the  stresses  parallel  to 
till'  ground  surface  must  also  balance  each 
utlicr  independently  ar  I  must  be  of  equal  mean  intensity  through- 
out the  layer  in  question. 

Take  the  thickness  of  the  wall  and  of  the  earth,  perpendicular 
t(i  tlic  plane  of  the  pai)er,  to  be  unity.  Let  w  be  the  specific  weight 
of  the  earth,  and  take  CI)  =  x.  Then  if  r  is  the  vertical  stress  at  D, 
and  if  s  is  the  stress  conjugate  to   .    I  <r, 

rZ)D'  =  weight  uiwii  element  DD'  '-wxDD", 

DD" 


Fio.  347. 


or 


r''WXjjjy-=v    cos  (9, 


and  also 


Therefore 


cos  g-V'cOs'g-CUS'  (ft     s       _s 

cosO  +  V  cos^^-co**  tf> "  r  ~  wx  cos  fl' 


s  -  wx  cos  d 


COS  0  —  v^cos^  0  -  cos'  <f> 

cos  0  +  N^COS^  U  -  C08^  <!>' 

and  the  total  pressure  on  CD 


f  ,      «'J^        .cos  (?  — v'cos'fl-cos' (A     _ 


'^  cos  0  +  Vcoe*  0  -  COS*  <f> 


The  line  of  action  of  the  resultant  pressure  is  necessarily  parallel 
to  the  ground  surface  and  intersects  CD  in  the  point  T,  where 


CT'iCD~lx. 


308 


THEORY  OF    STRUCTURES. 


If  the  ground  surface  is  horizontal,  i.e.,   if  fl=0, 

1  —sin  (f) 


1  +  sin  ^ 

1         ...  w^^  1  -  sin  <i 

and         total  pressure  = —. — r. 

2    i+srn  <^ 

If  the  substance  retained  by  the  wall  has  no 
angle  of  repose,  i.e.,  if  it  is  a  fluid,  0=0.  Then 
r  =  wi  =  ,s 


I"i<i    348. 


and 


wx 
total  pressure  =  — . 


Thc.^^e  results  may  also  be  obtained  in  the  following  manner: 
Take  I)E  so  that  DE.x:  -.s-.r  (Fig.  347).     The  pressure  at  D  upon 
CD  is  then  xvDE,  and  if  from  any  other  point  K  in  CD,  KL  is  drawn 
parallel  to  DE,  and  therefore  to  the  ground  surface,  wKL  is  the 
pressure  at  A'  upon  CD. 

Hence  the  total  pre.ssure  upon  CD 

=  the  weight  of  the  triangular  j^rism  CDE 


-^IX:  DE  cos  0  = 


trx" 


cos  0  —  =  etc. 


The  resultant  of  all  the  jwssures  upon  CD  parallel  to  ED  evidently 
passes  through  the  C.  of  G.  of  the  triangular  prism  and  mast  therefore 
ii\tersept  CD  at  a  point  two  thirds  of  the  total  tlepth  CD  below  C. 
10.  Foundations  in  Earth.  -Cask  I.— Let  the  weight  of  the  s\i\)er- 
structurc  be  uniformly  distributed 
over  the  ba.sc,  and  let  /)o  be  the 
intensity  of  tlic  |)ressure  produced 
l>y  it  (Fi;:.  :i\\)}.  When  the  super- 
structure lias  just  .stopiM'd  subsiding 
and  til."  earth  on  each  .side  is  on  the  ^"""^'^^ 
point  (if  heaving  up,  the  ratio  of  the 
vertical  stress  /)(>  to  the  horixontai 
stnss  /);,  at   points    along  the  l)ot-  Vm  ;u» 

tom  of  the  su|)erstructure  must  have  its  greatest  value. 


Pa    1 


■^in  (fi 


-sin  <l>' 


il. 


FOUNDATIONS  IN  EARTH.  309 

At  the  same  level  but  at  points  in  the  natural  soil,  clear  of  the 
siiiMTstructure,  the  ratio  of  the  horizontal  stress  ph  to  the  vertical 
stress  p,  has  its  greatest  possible  value.    Therefore 


llonce 


Pe="l-sin  <p' 

Po^n+sin<f>\' 
p,^  \l-sin<i/ 


p,     \l-sin<^-^ 

If  X  is  the  depth  of  the  foundation,  and  w  the  weight  of  a  cubic 
foot  of  the  earth, 

p,  =  irx, 

p.</l+sin<^\2 
an.l  therefore  wx=\l^sin<l>)  ' 

\jcth+zhe  the  height  of  the  superstructure,  and  let  a  cubic  foot 

of  it  weigh  w'.    Then 

p„  =  v/{x+h). 

Hence  a  minimum  value  of  x  is  given  by 
w^  +  x)     /I  +  sin  A'     1_ 

w'hk* 

an(  1  therefore  x  =  ^_^^2' 

Case  II.— Let  the  superstructure  produce  on  the  base  a  uniformly 
\  aryiiiR  pressure  of  maximum  intensity  pi  and  minimum  intensity  p* 
l\v  Case  I, 


«'X~  M-sin  <i>/ 


In   the   natural   ground  the   minimum  horizontal   intensity   of 

I>rcssiirp  is 

1  -sin  <i> 

When  the  foundation  trench  is  excavated,  this  pressure  tends 
tn  rai«o  the  botton»  and  push  in  the  sides.  The  weight  of  the  siiper- 
sirneture  should  therefore  be  at  least  equal  to  the  weight  of  the 


310 


THEORY  OF  STRUCTURES. 


material  excavateil  in  order  to  develope  a  horizontal  pressure  of  an 
intensity  equal  to  p*.    Therefore 

p^^l-sin<^ 
P2""l+sin  ^* 


Hence 


P2>j 


wx 


and 


pi^/l+sin(^\» 
Pi"  M-sin  d)' 


P 

P2 


Ex.  7.  EaHh  weighing  120  Ibs./cu.  It.  and  having  an  angle  of  repose  (4>) 

of  30°  is  laid  in  horizontal  layers  behind  the  vertical  face  of  a  reiaining-wall  12 

Jt.  high.     The  ground  surface  is  level  with  the  top  of  the  wall.     Find  the  total 

pressure  on  the  vertical  face  and  the  overturning  moment  with  respect  to  any  point 

tn  the  base  of  the  wall. 

,  L    ,  ,  ~>    ,  „1  -  sin  30° 

The  horizontal  pressure  at  the  bottom  of  the  face  =  1 20  X 1 2  j  ^  ^^^^  g^o 

=  4Snihs. 
-i(480)  -240  lbs. 
-12X240 -2880  lbs. 

-2880  X  ^  - 1 1 ,520  ft.-lbo. 


"    averaRo  pressure  on  the  fac« 
"    total  pressure 


i( 


overturning  moment 


i 

Fia.  3S0. 


Ex.  8.  In  the  preceding  example  find  the  total  pressure  on  the  face  when  M* 
f/round  surface  .s/o/w.v  from  the  top  of  the  wall  at  an  angle  of  25°. 
The  total  pres-sure  purallel  to  the  ground  surface 


120     .„,  ..„ros25^-V(.(>8'25"-ro8-'30' 

.-;t-  x12'Xcos2.)°- 


t  fw  2'"t-  N  i-os'  25°  -eis'  30» 


. 60  X 144  X  .9003'"vv*!  -  •*»  0  •*»• 
1.1  H 


LINE  OF  RUPTURE. 


311 


pivertical  intensity  of  pressure  at  base  of  wall 
=  >0  X 125  -  2500  Ibs./sq.  ft. 
„;  the  horizontal  intensity  of  the  pressure 
H   the  base  of  the  wall,  is  given  by 
.2^^1_fsin30°^3^    or    pfc-833i  Ibs./aq.  ft- 
,n       1 -sin  30        '  , 

„,   the  vertical  intensity  at   the  same  depth 
•„.   the    natural    ground    clear    cI    the  waU,    is 

8'-"^^'  2._l±?ilL30:.3. 

p,    1 -sin  30 

But  pr-lOOx, 

X  being  the  depth  of  the  foundation. 


Therefore 


llctice 


pfc-3p,-300x; 


„   tin.  of  R»ptar..-Another  expression  tor  the  pressure  OB 
■'M'l  ^Hr'S' rrtTAB  (F-..  S.,,  we.  suddenly  removed 

»"»--r"°r;i\trrvt::v  «ou.d .. 

along  the  plane  CB. 

The  stability  of  ^BC  is  maintained  by  the 
reacttn  P  on  \b,  the  weight  1^  of  ABC,  and 
the  frictional  resistance  along  BC. 

Let  the  direction  of  P  make  an  angle  /?  ^ith 
the  horizon, 
r  .u.  u^.  Ut  the  angle  CBA  =  t. 

Ut  R  be  the  mutual  pressure  on  the  plane  BC. 
liesolving  along  and  perpendicular  to  BL , 


FiQ.  363. 


and 
Therefore 


-P  cos  (90°-i-.3) +  1^"  cos  i-R  tan  <^ 

P8in(90''-i-,9)  +  U'8ini  =  R. 
-P8in(i?+i)  +  W'co8i-tan.^|Pco9(^+i)  +  TV8int| 


li'. 


312 


THEORY'  OF  STRUCTURES. 


and       P=W 


But 


cos  r  — sin  i  tan  <f> 


=  TF-. 


cos  (i  +  (j)) 


sin  (/J  + 1 )  +  cos  (^  + 1)  tan  (^        sin  (/? + i  +  <p) 


W  =  u 


BABC 


sui  I  =  - 


«'j2  cos  ^  sin  1  _ 
2~cos  (0+i)' 


therefore 


j('j2  cos  fl  sin  i     (  os  (i  4  </>) 


2    cos(«/  +  i)  sin  (^  +  1  +  ^)' 


The  only  variable  upon  which  P  depends  is  the  angle  i. 

Differentiating  the  right-hantl  side  of  the  last  equation  with 
respect  to  i  and  putting  the  result  equal  to  zero,  a  value  of  i  is 
found  in  terms  of  ,9,  0,  and  ^  which  will  make  P  a  maximum. 

The  line  inclined  at  this  angle  to  the  vertical  U  called  the  line 
of  rupture. 

If  the  ground-surface  is  horizontal,  0  =  0. 

If  the  face  retaining  the  earth  is  vertical,  and  if  it  is  also  assumed 
that  the  friction  between  the  face  and  the  earthwork  is  r  -]/,  p  is  horizontal 
and/3=0.    Hence 

P  =  ~9-tan  X  cot  (i  +  0). 
This  is  a  maximum  when  2t=90°-^,  and  then 

wx^       I    n    4>\        /.,o    4>\     »j^/^~*^"2\     m21-sin<^. 

P  =  -^tan(45°-f)cot(4o°  +  |   =-^ ^    —iTinri' 

\  1  +  tan  77  / 


the  same  result  as  that  obtained  by  Ilankine's  theory. 

The  following  is  an  easy  geometrical  proof  of  the  result  2i  =  90° — <f>: 

Q N  <^)»    an.v    l'»P    KL    (Fig.    354)  de- 

^       Scribe  a  semicircle. 

Draw  KM  inclined  at  the  angle 
0  to  KL,  and  KN  inclined  at  the 
angle  i  to  KM. 

Join  \L,  cutting  KM  in  T. 
Fio.  364.  Let  0  be  the  middle  point  of  the 

arc  KM. 

Join  OL,  cutting  KM  in  Y. 


RETAINING  WALLS. 


3ia 


Draw  iVF  parallel  to  KM.    '^^j.^^^^^ 
tan  i  cot  (i  +  '^)  =  ^^  JVL     NL    VL ' 


VY 


The  ratio  ^  is  evidently  a  maximum  when  .V  coincides  with  0, 

angle  OLK. 

Hence  if  OKM=i,  OLK  must  also=t. 
ButOA'L+OLK=9O°  =  i  +  <^+i=2i  +  0- 

Therefore  *-*^      2' 

„   Retaining  walls. -Consider  the  equihbnum  of  the  section 
{BMN  of  a  wall  of  unit  thickness,  acted  «  a 

;nK)n  at  D  by  the  force  P,  vaich  tenc^  to 
overturn  the  portion  mider  consideration^ 
This  tendency  is  resisted  by  the  weight  W 
upon  MN  and  by  the  stresses  developed 
•ilone  the  division  surface  MN. 

Let  the  resultant  of  these  stresess  in- 
t.Tsect  the  surface  MX  in  the  point  F. 
Ihi.  ix)int  is  called  the  centre  of  resistance 
,or  centre  0}  pressure)  and  its  distance  FO 

V*™  XTiL  mily  and  mu»t  be  dctomined  by  expenence. 

Til  nrdinarv  practice  q  vanes  from  i  to  «•  ,      ,     ^r>      . 

"t   he  Une  of  action  of  W^  intei^ct  MAT  in  C,  and  take  CO^rL 
The  clfficient  r  is  less  than  unity  and  depends  upon  the  form  of 

'^^rranf  r^'the  inclinations  of  MN  and  P,  respectively,  to 

^^^'^;t^the  perpendicular  from  ^  "^^  the  f  ction  of  P. 
Three  conditions  of  equUibrium  must  be  fulfilled 
FhTconditiou.    r/te  Hne  0/  r.-.:.^«n..,  ^.e    the  locus  of  F,  must 
intersect  every  bed  joint  well  within  the  outer  edges  M. 


314 


THEORY  OF  STRUCTURES. 


The  position  of  P  is  determined  by  the  condition  that  the  moment 
of  W  with  respect  to  F 

>  the  moment  of  P  with  respect  to  F, 


or 


W(qtTrt)  coso>Pp, 


the  upper  or  lower  signs  being  taken  according  as  C  is  on  the  left 
or  right  of  0. 


and 


WiqtTrt)  is  called  the  moment  of  stability 
Pp  is  called  the  overturning  vmment. 


If  x',  y'  are  the  horizontal  and  vertical  coordinates  of  D  with 
resjK'ct  to  F, 

p  =  y'  cos/?-x'  sin  /?. 

The  line  of  resistance  of  a  dam,  Fig.  356,  may  be  easily  deter- 
mined graphically  in  the  following  manner: 

Let  ABFG  be  the  cross-section,  the  thickness  perpendicular  to 
the  plane  of  the  paper  being  unity.  Divide  the  figure  into  con- 
venient portions  ABCL,  etc.,  and  find  the  centres  of  gravity  Gi, 
(?2,  <''3,  ('4  of  the.<?e  portions.  Calculate  the  -^ight.s  Wu  W2,  W3,  W4 
of  the  several  portions  and  lay  off  al,  12,  23,  34,  Fig.  357,  on  a 
vertical  line  to  represent  these  weights.  Take  any  pole  0  and 
draw  the  polygon  bcdf,  Fig.  358.  Then  Ri,  the  resultant  of  Wi,  W2, 
passes  through  7;  R2,  the  resultant  of  Wi,  W2,  W3,  through  h;  and 
i?3,  the  total  resultant,  through  k.  The  intersections  of  TFi,  ffj,  R2,  R3 
with  the  planes  LC,  RD,  HE,  and  GF  determine  the  line  of  resist- 
ance V'Q'R'S'T'  when  there  is  no  water  pressing  on  the  dam. 

Let  the  water  now  rise  to  ttie  top  of  the  dam.  The  water-pres- 
sures on  the  several  portions  are  represented  in  magnitude  by  the 
areas  AML,  M'LKN,  etc.,  and  in  direction  by  the  lines  Pi,  P2, 
P:i,  Pi  drawn  normally  to  the  faces  AL,  LK,  etc.,  and  through  the 
centres  of  gravity  of  the  above  figures.  Lay  off  al,  Im,  mn,  np  parallel 
to  Pi,  P2,  P3,  P*,  respectively,  and  proportional  to  the  areas  ALM, 
^PLKX,  etc.  Draw  the  polygon  qr)(t.  Fig.  359.  Then  /?/  dra\vn 
through  u  parallel  to  MQ  gives  the  line  of  action  of  the  resultant 
of  Pi  and  P^-  Similarly  R2  and  R3  give  the  lines  of  action  of 
the  resultants  of  Pi,  P2,  P3,  and  of  Pi,  P2,  P3,  Pt,  resix-ctively. 


RETAINING  WALLS. 


315 


N,.v  n  is  the  resultant  of  P.  and  W,.    Throughthe  -ter^^^jj 
/'"and  \\\  draw  a  line  parallel  to  II  cutting  LC  m  Q.    Sumlarly 


i        »    «   -JS  8 


through  the  intersection  of  R>  and  /?>'  draw  a  l;""^  P^f.^J  ^^  "f 
o2g  KD  in  K.    Then  Q  and  «  are  the  points  at  which  the  result- 


310 


THEORY  OF  STRUCTURES 


IIP 


ants  of  the  forces  acting  above  their  respective  planes  cut  those 
planes.  The  points  *S  and  T  are  obtained  in  the  same  manner  and 
QRST  is  the  line  of  resistance  reciuired. 

Second  condition.  The  orKjle  between  the  direction!^  of  the  resultant 
prensure  on  MX  and  a  normal  to  MX  mud  be  /ess  than  the  angle  of 
friction,  Fig.  355. 

Let  <f)  be  the  angle  of  friction  and  R  the  mutual  normal  pressure. 


Then        i2  tan  ^  >  P  cos  a  +/?-  ir  sin  a 

>|/'cos/?-(Psin/9  +  lF)  tanajcosa, 


and 


Therefore 


or 


R=P  sin  a  +(3  +  W  cos  a 

=  (P  cos  /3  tan  a  +P  sin  ji  +  W)  cos  a. 


P  cos  fi  -  (P  sin  ;9  4-  W)  tan  a 
tan  9>  p  -g-3  tan  a-fP  sin  ^  +  W' 


tan  ^  + tan  «         P  cos /? 


1  -  tan  (^  tan  a  '  P  sin  ^  +  \\' 


Hence 

If  MX  is  horizontal, 


P  cos  /? 


a=0, 


and  then 


P  cos  /? 

**"'^>p^hr^TTr 


If  the  ground  surface  is  also  horizontal, 
and  then 


P 
tan  <^  >7ip. 


Four  fifths  of  ^  is  usually  taken  as  the  limiting  value  of  the 
angle  of  friction. 

Third  condition.  The  maximum  intensity  of  pressure  on  the  sur- 
face MX  must  not  exceed  the  safe  icorking  strength  of  the  material. 


RETAISING  WALLS 


317 


The  load  upon  any  surface  (or  bed)  MN  cannot  be  said  to  be 
ovorunifornJ-.  distrib'uted,  but  this  third  condition  .s  pract.caUy 
0,  ured  by  the  assumption  that  the  intensity  of  the  pressure nor- 
::,  ;  the  surface  diminishes  uniformly  from  the  outer  or  most 
compressed  edge  inward^ 

lx>t  R  be  the  total  pressure  normal  to  MN. 

lot  /  be  the  maximmn  normal  intensity  of  pressure  at  M. 

Draw  MG  at  right  angles  to  MN,  Fig.  360.  and  take  A/G-/. 

There  are  three  cases  to  be  considered. 

Case  a— Let  the  normal  intensity 
vary  from  MG=f  at  M  to  nU  at  A'. 

Join  GN. 

The  intensity  of  tho  pressure  at  any 
point  in  MA'  is  the  intercept  between 
MS  and  GN  of  a  line  drawn  through  p,o  ggo. 

tlu>  point  parallel  to  MG. 

Hence  R  =  area  of  triangular  prism  MGN==^. 
Also,  the  resultant  of  all  the  stres^s  parallel  to  MG  Pa^«  through 
tlu   C.  of  G.  of  MGN  and  intersects  MN  m  the  pomt  F,  where 

'  t    t    t 


Therefore 


^^=^^  =  2  "3"  6 


and 


•i^e- 


Thus  F  is  one  of  the  limiting  points  of  the  middle  tkird  of  the 

'"'cvlET-I^t  the  normal  intensity  vary  froni  MG=^/  at  M  to 
,nl  -It  a  Doint  A'  Iving  inside  the  edge  A.    Join  Gh. 

T  le  nS  intensity  at  any  point  m  MK  is  the  intercep  between 
MK  and  GK  of  the  line  .lrav«i  through  the  pomt  parallel  to  MG 
•'  m    resultant  of   all   the   stresses  on  MK  P--  th-gh    h 
(■    of  G.  of  the  triangular  prism  GMK  and  inter^ct^  MA    in  the 
point  F',  where  MF'-^MK,  MK  Ix^ing^now  the  offootive  w,,  th. 

In  this  case  OF' =qt>OF>-^. 


and  therefore 


I 


31S 


Also, 


THEORY  O't    STRUCTURES. 
/?  =  aro;i  of  triangular  prism  GMK 

==\fMK  =  l  f3MF'  =|/(0M  -OF') 


=|/(|-.). 


Hence 


I      2  R 


Over  the  distance  A'.V  the  stress  is  either  nil  or  negative,  i.e.,  a 

tension. 

C.\sEc.— Let  the  normal  intensity  vary  from  MG=f  at  M  to 
a  minimum  S'S^f^  at  X.     Join  GS. 

The  normal  intensity  at  any  point  of  MN  is  now  the  inter- 
cept between  MX  and  GS  of  a  line  dravvTi  through  the  point  parallel 
to  MG. 

The  resultant  of  all  the  stresses  on  MX  passes  through  the  C.  of  Q: 
of  the  trapezoidal  prism  GMXS  and  intersects  MN  in  a  point  F" 
between  F  and  O,  so  that  in  this  case 

t 


qt'OF"<OF<^, 


and  therejore 


q< 


6* 


Draw  ST  parallel  to  MX.    Then 

area  GMXS XOF"  =  n-ctangle  TMXSxO 


or 


and 


But 


+  triangier7rs(^-0, 
f+fo.,    f-hfl 


/+/"- 


2    6' 

2/  . 


"        2     '     {)q  +  l 


RETAINING  WALLS. 


319 


lie  nee 


l=Kl'')- 


Ex  10.  A  masonry  wall,  Fig.  361,  of  rectangular 
..action,  X  ft.  high,  4  ft.  mde.  and  weighing  125 
;;,.  /cu  ft.  is  buiU  upon  a  horizontal  bed  and  retains 
a;,Ur  on  one  side,  level  mth  the  top  of  the  wall.  Find 
s  Ui)  when  7=i;  ('')  if  the  safe  working  strength  of 
,/„  ,nalerial  is  10,000  Ibs./sq.  ft.  How  much  (c)  of 
the  mill  can  he  removed  in  case  (a)  without  changing 
the  moment  of  stability  f 


kCD   B 


M  F 


moment  of  staouuy  r  ^ 

The  horizontal  water-pressure  on  wall>62i|  lbs.,  acting  3  ft.  above  the 


1h'(1. 


125 


Therefore  the  oveHuming  moment '■62\  ■ -^  ■  ^  "  12  ^  ft.-lbe. 
The  weight  of  the  wall -4x.  125 -500x  lbs.,  and  the  distance  r.    its  line 
„f  a.ti..n  from  the  centre  of  resistance  F  -  iq. 

Therefore  the  moment  of  stabUUy-SOQxXiq-iOOOqx   ft.-lbs.    Hence  for 

wiuilibrium 

i^x'<2000gx, 

ori'<1929. 

(a)  If  g-»,  x'<64,  and  therefore  x<8  ft.,  bo  that  the  height  of  the 

wall  must  not  exce-  J  8  ft. 

({.)  Again,  if  /  -10,000  lbs., 

I  /ft      \     l./J0000^_  \     L/'§?_iV 
/irs<  assume  9 -j7(^^-l^-g\^500x  '     6Vx       /* 


Then 


..<m(5..)<3.(^.) 


fi;r,?'nvj«^°'!„i'',>..  .na  .wo«  .h.  -«n.p.io»  ^  <»- 

liic  value  of  q  is  incorrect. 

.■WOT      _\_ 
■3  10000X4^  "2 


1      2  R     1      2 
Second.  Assume  9"2  "3  /("'i     3 


120" 


Then 


x«-t-|x<96. 
Therefore  x< 9.03  ft.. 

so  that  the  height  of  the  wall  must  not  exceed  9.03  (t. 


or 


320 


THEORY  OF  STRUCTURES. 


Also, 


903      ,,„       J  .       1 
-I^=--*247andi8>-, 


showing  that  the  value  assumed  for  7  is  justified. 

(r)  The  triangular  portion  MAIi,  Fip.  3fil,  may  be  removed  if  its  C.  of  G. 
is  vortieally  above  the  centre  of  resistance  F.  Let  the  vertical  FG  intersect 
A B  in  ('.     Also  let  MG  meet  .1 B  in  I) ;  />  is  the  middle  point  of  AB.    Then 


Therefore 


.4C=§  l/)  =  Ji.lfl. 
.lB=3.tr=3X!t=2ft., 


so  that  B  is  the  middle  point  of  IF,  and  thus  one  fourth  of  the  wall  may  be 
removed. 

Again,  let  the  resultant  of  the  horizontal  water-pressure  (P)  and  the  weight 
uiwn  the  base  make  with  the  normal  to  the  ba.se  an  angle  0  before  the  portion 
MAB  has  been  removed  and  an  angle  d'  after  its  removal. 

If  ir  is  the  weight  on  the  base  in  the  fiM  ease, 


and  in  the  second  case. 


P 

tanfl=-, 


tanfl'-; 


iW 

Therefore  tan  0:tan  9':  :4:3. 
Ex.11.  Fig.  362  represents  a  slice  of  a 
irnll  of  unit  thickne.s.'i,  retaining  earth  whir' 
has  an  angle  of  repose  <}>  ami  slo/ws  away  from 
the  top  of  the  wall  at  an  angle  0.  Consider 
the  equilibrium  of  the  whole  mass  in  front  of  a 
rertieal  plane  through  the  inner  edge  N. 

The  weight  11'  on  the  l)a.se  is  then  made 
up  of  the  weight  of  the  slii'c  of  the  wall  and 
the  "eight  of  the  triangular  |)rism,  also  of 
unit  thickness,  of  earth,  viz.,  NVD. 

I/>t   the  line  of  action  of  W  intersect  the 
base  in  G. 
Let  F  \w  the  centr«'  of  resistance  in  the  ba.se  and  let  O  be  the  middle  point 
of  M\.     Then 

the  moment  of  KtnhililifWiOFT  (HI)  cos  a, 
^W'lgT  r)  t  cos  n, 
n  iM'iiig  the  slofie  of  the  base. 

Ag.iiti,  if  X  is   the   vertical  distance  Ix-twecn  X  and  the  ground  .<<U)rface, 
i.e.,  .Vr,  then 


,                           ,.         ^,,,     tc-r'          „  cos  fl- V'cos'fl-cos'A 
the  ri'sultant  pressure,  /',  on  (A  -         cosfl :^—  ..        —, 


and  ads  parallel  to  the  ground  surface  at  a  distance  ,-  above  N. 


EXAMPLES. 


8S1 


Therefore  the  overturning  moment  is  Pp,  p  being  the  perpendicular  from 
f  upon  the  direction  of  P.    Hence 

W(q'fr)t  cos  a>.Pp- 
For  exampU,  if »  -  0,  i.e.,  if  the  ground  slopes  at  the  natural  angle  of  repose, 

P-     -cos  4>. 
If  0  -0,  i.e.,  if  the  ground  surface  is  horizontal, 

2   1  +sin  <^  3 


since 


P  is  also  horizontal.    Therefore 


W'(5Tr)tana>  — j^:-^^. 


If  9  -0  -  i',  i.e.,  if  the  substance  retained  by  the  waU  has  no  angU  of  rtpou 
the  case  becomes  one  of  fluid  pressure,  and 

lF(5Tr)<co8a> -^. 

Ex  12.  GraveUy  earth  with  an  angU  of  repose  of  WT  and  weighing  96  H«./ru. 
/,.  ulpread  out  in  horizontal  layers  and  i.  retained  fry  a  waU  ofreOmgvl^ 
icon  0  ft.  high  and  weighing  120  Ibs./cu.  ft.  The  earth  rues  to  the  top  of  the 
unll.     Find  the  xvidth  t  of  the  wall,  q  being  i. 

Moment  of  stability  -«■  12o|  -12a'  ft.-lbs. 


Overturning  moment 


Therefore 


or 


""  2    l+8in37      3 

12a'>859, 

«'>7.16 

(>  2.08  ft. 


and 

Fx  13  The  water-faee  AC  of  a  t/HiH,  Fig.  363.  has  a  hatter  of  1  in  10;  the 
.rml^l  ;S/T«!  the  top  is  6  ft. :  ..e  r^r  of  the  waU  DEF  ^^^-^^^^ 
l>t:.  having  a  batter  of  2  in  10.  and  EF.  a  batter  of  78  xn  100;  the  ma^n^ 
,rn.,hs  \2aibs.  per  cubic  M.  and  the  maximum  compressum  '"'«""'<  ^'^ 
S,-,  lb,,  per  square  inch.     Find  the  safe  heights  of  the  two  portums  AE  and  EC. 

First.  To  find  the  height  of  the  u|)pcr  portion: 

Consider  the  equilibrium  cf  the  whole  ma«.  of  masonry  and  water  in  iront 
of  a  vertical  plane  through  the  edge  S'. 


322 


lltEOUY  OF  STRUCTURES. 


Taking  lOx  ft.  as  the  height,  the  other  dimensions  of  the  section  are  aa 

shown  in  the  figure. 

2xl0x 


Wt.  of  triangle        AM'C  =  125 


■1250x»lb8. 


II     <(     << 


K         <I         << 


B.V'D-125— ?^-625x' 


BA''£  of  water        -312ix'    " 
"     "     rectangle  ylB/)r  =  125 -6  lOx-TSOOx      " 
Total  R  on  base  =2187 ix'  +7500x. 

Horizontal    pressure    on    vertical 

(lOx)' 
face   Ar'£=62i-^r--=3125x',  acting 

-A  X  ft.  above  the  base. 

Let  X  ft.  be  the  distance  Wtween 
M'  and  the  line  of  action  of  the  ver- 
tical resultant  pressure  on  M'\". 

\     2   R  *. 

Assume  that       9~2~3    If 

where 

<  -width  M'N'  of  base  -3(x  +  2)  ft. 

2  R 
Then  also  5-  -,    is   the    distance    be- 

tween  M'  and  the  centre  of  resistance  Fig.  363 

F  in  the  base. 

Taking  momenta  about  F, 

3125x'.V^-«(.-|f) 

-ffi    ^^ 
-^'-3    / 

- 1 2.50x'  •  Jx  +  7.'»00x(3  +  2x) 
+  625x'(0  +  Jx)  +3124x'(6  +  !x) 

-rr44W^'«74x'-.7500x)'. 

which  reduces  to 

193776X'  +612.')x'  -412704*  +528768, 

and  X  lies  between  28i  and  29  ft. 

Take  1-284  ft. 


.«31jHc> 


-*— «> 


.t_ 


Then 


Also, 


RETAINING  WALLS. 
R  -39142.97  lbs. ;  t  -width  of  base  -14.55  ft. ; 
Wt.  of  ma«)nry/lin.  ft.  of  thicRness -i(6  +14.55)28i  Xl25 
=  36605  lbs. 

„.L_2_- 39142.97  353^^ 

9    2     3   144X85X14.55 


323 


1  Of? 

an.l  therefore  the  assumption  q-^'^Jt  '^  i"^^'^^' 
If  ^  is  the  distance  between  M'  and  the  vertical  through  the  C.  of  G.  of 

the  wall, 

x'(36605)  =12.-50x'  +  Jx  +  7.500x(2x  +  3)  +625x'(ix  +  6). 

Substituting  in  this  equation  the  value  of  x, 

x* -7.89  ft. 

""  tS  Too,  »  .h»  height  of  th.  lo«er  portion;   the  other  dim.nio™  - 
as  shown  in  the  figure. 
The  total  R'  on  the  base 

-36605  +  3900y'  X 125  + 14.55  X  lOOy  X 125  +  500y'  X 125 

^.§|i(10j,  +  2.85)(10y+28.5). 


1     2^R^ 
9'""2~3  /«" 


Assuming,  as  before,  that 

'  '2~3 /?"' 

•iid  using  similar  symbols, 

flii  ,10m +28.5     p,/->     2  _  y     N 

Y^^*^  ■*■  "^^-^^         3""  "'^A       3  144  X86/ 

-3900y  X 125  X52!/  +  l455y  X  125(7.275  +  78j/) 

/  10  \ 

+  500v'  X 125  {7»y  + 14.55  +  -3^; 

+  i-f(10»+2.85)(1002/  +  28.5)(78y  +  l4.55  4-W^^2:85) 

2       «" 
+36605(7.89  +  787)  -  j  144^85' 


324 

which  reduces  to 


THEORY  OF  STRUCTURES. 


y«  -  .27047*  +  .18461/' -  .12261/ -  .000146  =0. 

Hence  y  lies  between  .45  and  .46  and  is  approximately  .457. 
Thus  the  height  of  the  lower  portion  is  about  45.7  ft. 


Also, 


and 


i?' =235314  lbs., 
2  /?'     23.5314 


3  //'     991440 


=  .237, 


1      2  ft' 

and    therefore  fl'=.5-.237  =.2<)7  is>i,  and   the  assumption    ^-.J  "3    jf 

is  justified.    The  width  of  MN  =78i/  + 14 .  55  +  lOy  =54.77  ft. 

13.  Practical  Rules —When  the  surface  of  the  earthwork  is 
horizontal  and  tlie  face  of  the  wall  against  which  it  abuts  ver- 
tical, the  pressure  on  the  wall  according  to  Rankine's  theory  is 

_  '^  1-sin  <l> 
■'  ^Tl+sin^' 

and  the  direction  of  P  is  horizontal. 

This  result  is  also  identical  with  that  obtained  in  Art.  11,  on 
the  assumption  of  ('(-ul()nib"s  wedge  of  niaxinmni  pressure  (Pon- 

celet's  Theory). 

Exi)orience  has  conclusively  proved  that  this  theoretical  value 
of  /'  is  very  much  greater  tlian  its  real  value,  so  that  the  thick- 
ness of  a  wall  designed  in  accordance  with  theory  will  be  in 
excess  of  what  is  rtHiuired  in  practice.  In  the  deduction  of 
the  fornmla,  indeed,  the  altogether  inmlmissible  assumption  ia 
made  that  there  is  no  friction  between  the  earthwork  and  the* face 
of  the  wall.  This  is  (•(niivaleiit  to  the  supposition  that  the  face 
is  prfectlv  smootii  and  tiiat  therefon'  the  pressure  acts  normally 
to  it.  Boussines(iue,  Levy,  ami  St.  \enant  have  demonstrated  that 
the  hypothesis  of  a  nciriiial  pressure  only  hohls  true, 

cither,  fird,   if    tlie   ground-surface    is  horizontal  and  the  wall- 
face  inclined  at  an  angle  of  45° -7,"  t<>  the  vertical, 

or.  xecnnd,  if  the  wall-face  is  vertical  and    the  ground-surface 
incUneil  at  an  angle  0  to  the  horizon. 


PRACTICAL  RULES. 


325 


\\l.en  the  surface  of  the  ground  is  horizontal  and  the  face  of 

ti„.  wall  vertical,  and  when  ^  =  45°,  the  above  formda  gives  the 

,orn.c-t  magnitude  of  P.     Its  direction,  however   is  no    ^ouzoni^, 

r.nakesTn  angle  with  the  vertical  equal  to  the  angle  of  friction 

;....n  the  earth  and  the  wall.     The  wall-face  is  gene''^y  ^.'^ffi" 

;  .,lv  rough  to  hold  fast  a  layer  of  earth,  and  in  all  probability 

Bous^inesque's  assumption  that  the  friction  between  the  wall  and 

h„    earth    is    equal    to    that    inherent    in    the   ear  h    is    a    near 

;,.o;imation  to  the  truth.  The  direction  of  P  will  thus  be  con- 
.  Irably  modified,  leading  to  a  smaller  overturning  momen  and 
',    .orresponding    diminution   in    the    necessary    thickness    of    the 

"'"in  practice  the  thrust  P  may  always  be  made  small  by  carrying 
„„  the  backing  in  well-punned  horizontal  layers.  .    ,       ,  , 

In  order  to  neutraUze  the  very  great  thrust  often  induced  by 
.H.rnate  freezing  and  tha«-ing  and  the  consequent  swelling  a  most 
;L.tive  expedient  is  to  give  a  batter  of  about  1  in  1  to  the  rear 
ll„,.  of  the  wall  extending  below  the  line   to  which  frost   pene- 

'''rho  greatest  difficulty  in  formulating  a  table  of  earth  thrusts 
ari.s  from  the  fact  that  there  is  an  infinite  variety  o^  ^^^h  work^ 
V  an  example  of  this,  Airy  states  that  he  has  found  the  cohesive 
j,ow,.r  of  clav  to  vary  from  168  to  800  pounds  ^r  square  foot, 
,h,.  crresponding  coefficients  of  friction  varying  from  1.15  to  .36 
ana  that  even  this  wide  range    is  less  than   might  be   found   m 

'""('correct  theorv  for  the  .lesign  of  retaining-walls  is  as  yet  wanting. 
Xnurding  to  Baker,  experience  has  shown  that  with  good  backing 
ana  a  „K,d  foundation  the  stability  of  a  wall  will  be  insured  by  nuikmg 
„^  thickness  one  fourth  the  height,  and  giving  it  a  front  batter  of 
1  ,„.  .  in.  per  foot,  and  that  under  no  conditions  of  ordinary  surcharge 
,„   h"-avv  backing  need  its  thickness  exceed  one  half  the  height. 
U.k..r-s  "usual  practice  in  ground  of  average  character  .s  to  make   he 
.i.ukncss  one  third  the  height'from  the  top  of  the  footings,  and  if 
.  .natcTial  is  taken  out  to  form  a  face  panel,  three  fourths  of  it  is 
int  Lack  in  the  form  of  a  pilaster. 

(icneral  Fanshawe's  rule  for  brick  walls  of  rectangular  seetion 
r.  taining  ordinary  material  is  to  make  the  thickness 


326 


THEORY  OF  STRUCTURES. 


2A^c  of  the  height  for  a  batter  of  1  in    5; 


25% 

<  ( 

"  1  in    6; 

26% 

( ( 

"  1  in    8; 

27<'c 

( ( 

"  1  in  10; 

28^^ 

( 1 

"  1  in  12; 

30^; 

( < 

"  1  in  24; 

32'^ 

( < 

for  a  vertical  wall 

.\s  a  general  rule,  a  batter  of  1  in  24  to  1  in  12  is  usually  adopted 
in  practice,  an-J  should  not  exceed  one  sixth  of  the  height  of  the  wall. 
The  stability  of  a  wall  with  a  straight  or  a  curved  batter  in  compression 
IS  always  greater  than  that  of  a  wall  with  a  vertical  face.  This  is 
»'>i|H'cially  the  case  if  the  joints  are  at  right  angles  to  the  face,  when 
the  tendency  to  slide  forward  is  largely  diminished. 

Again,  if  the  wall  turns  round  the  toe  ever  so  little,  the  wall  with 
a  vertical  face  overhangs  and  gives  an  idea  of  instability,  while  it 
also  offends  the  eye.  When  the  face  has  a  batter  a  slight  rotation  is 
not  noticed. 

By  the  use  of  counterforts  it  can  be  shown  that  there  is  a  small 
.saving  of  maj^onry  in  a  vertical  wall  of  rectangular  section.  Brunei 
rurved  the  face  of  a  retaining-wall  and  made  its  thickness  one  fifth  or 
one  sixth  of  the  height.  He  also  used  counterforts  30  ins.  wide  at 
intervals  of  10  ft. 

l'Ap<'rieiic('  indicates  that  counterforts  should  be  stepped.  The 
width  of  tlio  top  of  a  wall  should  be  as  small  as  possible  consistent 
with  .sound  construction. 

The  thickness  at  the  footing  adopted  by  \'auban  for  walls  with 
ti  front  batter  of  1  in  .">  or  1  in  0  and  plumb  at  the  rear  is  approxi- 
mately given  by  the  em|)irical  formula 

thickness -.19// +4  ft., 

//  U'ing  the  height  of  the  wall  above  the  footing.  Counterforts 
wtrc  introduced  at  intervals  of  lo  ft.  for  walls  above  Sf)  ft.  in  height 
and  at  intervals  of  12  ft.  for  walls  of  less  height.    The  counterfort 

proj«'«i,s  fiuiii  tilt'  wall  a  distance  of  T"^^  ft.  approximately,  and  the 


PRACTICAL  RULES. 


327 


1;) 


in 


approximate  width  of  the  counterfort  is  ^  -f  3  ft.,  diminishing  to 

f2ft.  . 

The  great  importance  of  the  foundation  will  be  better  appreciated 
,,v  1    'ring  in  mind  that  the  great  majority  of  failures  have  been  due 
native  foundations.     If  water  can  percolate  to  the  founda  ion 
n  ng  action  begins  and  a  consequent  settlement  takes  place, 
,        kmost  rapid  in  the  region  subjected  to  the  greatest  pressure, 
:  :'a     "r   ZoZ  to  counteract  this  tendency  to  settle  the  toe 
v       supported  by  raking  piles,  the  rake  being  given  to  dimmish 
bncUng  action  of  the  thrust  on  the  piles.    It  is  also  advisable  to 
:,  ;;::e  L  weight  as  uniformly^  possib^  T^'^lZ!  Z, 
.li.ion  which  is  not  compatible  with  large  front  batters  and  deep 
•    Vas  thev  tend  to  concentrate  weight  on  isolated  PO'nts.    In 
::o  of  dock  walls,  too,  a  large  front  batter  will  keep  a  ship  far  ^er 
.  V  f  om  the  coping  and  will  necessitate  thicker  fenders,  as  well  a. 
;  0    wkh  wider  throws.     As  an  objection  to  offsets  Bernays  urges 
U^  in  settUng,  the  backing  is  liable  to  hang  upon  then,    forming 
;  ;  Les  und  rneath.     He  therefore  favors  the  substitution  of  a 
H  .r  fo'  the  offsets     On  the  other  hand,  if  water  stands  on  both 
•r ^f  the  watthe  hydrostatic  pressui.  on  the  offsets  will  greatly 

'"'■  i::k' wS:'t'hable  to  far  greater  variations  of  thrust  tha. 
.,,..1  nuv  retaining-walls.    The  water  in  a  .lock  with  an  .mpermeable 
:  Anlv  stand  at  a  much  higher  level  than  the  water  a   the  back 
.,    U    ,vali:  and  its  pressure  may  thus  even  more  than  neu  ralize  the 
,  Lt  due'to  the  backing.    With  a  porous  bottom  ^^^^^ 
,  ,11  n.av  be  greatlv  dinnnishod  by  an  upwanl  ,.ressure  on  the  ba^. 
•1    :  ,.xn,.rience  of  dock-wall  failures  has  led  to  the  conclusion  tha 
.      ,.  ;  nonu.nt  of  stability  is  not  of  so  much  importance  as  "weigh 
;  :   lood  grip  on  the  ground."     Many  authorities,  both  pract.ca 
..:„.,  theoretical  have  urged  the  ^-at  advantages  in  economy  ^^^^^^^ 
.uvngth   atten.hng  tl.p  employment   of    counterforts.    The  use  of 
r  rtl  uid  cement,  or  cement  concrete,  will  guard  against  the  break  ng 
Jlv^f  the  counterforts  from  the  main  bo<^  of  the  wall,  as  h^o   en 
lnn,H.ned  in  the  case  of  the  ohler  walls.     But  a  umforn.  d  ^tnbution 
:fi:Z^^  well  as  of  weight  is  important,  and  it  therefore  seems 


328 


THEORY  OF  STRUCTURES. 


more  desirable  to  introduce  the  extra  weight  of  the  counterforts  into 
the  main  wall.  Besides,  the  building  of  the  .'ounterforts  entails  of 
itself  an  incretiseil  expense. 

14.  Reservoir  Walls. — Let  /  he  the  maximum  safe  pressure  per 
square  foot  of  horizontal  base,  at  inner  face  of  a  full  reservoir,  at 
outer  face  when  empty. 

Let  u"  be  the  weight  of  a  cubic  foot  of  the  ma-^onry. 
Assume  that  the  wall  is  to  be  of  uniform  drengih,  i.e.,  that  the 
section  of  the  wall  is  of  such  form  that  in  passing 
from  any  horizontal  section  to  the  consecutive  one 
below,  the  ratio  of  the  increment  of  the  weight 
to  the  increment  of  the  surface  is  constant  and 
equal  to  /. 

Let  AB,  Fig.  364,  be  the  top  of  the  wall. 
Take  any  point  0  as  origin,   and  the  vertical 
through  O  as  the  axis  of  x. 
Let  OA  =<i,  0B  =  t2,  and  let 

T  =  ti-¥h^AB. 


(/jN   G   ffi 
Fio.  364. 


For  the  profile  AP  consider  a  layer  of  thickness  dx  at  a  depth  x. 
Then 

xvydx 


or 


therefore 


dy  ■    /' 

dx^l% 
w  y 

■r  =  - log.  1/+C, 


(1) 


0  being  a  constant  of  integration. 
When  x=Q,y  =  ti; 

f 


tlicrefore 


0  =  -logefi+c, 


and  hence  x=-\oe,^, (2) 

which  is  the  equation  to  -IP  and  is  the  logarithmic  curve. 


RESERVOIR  WALLS.  329 

It  may  be  similarly  shown  that  the  equation  to  BQ  is 

-i'*s ®' 

Equations  (2)  and  (3)  may  also  be  written  in  the  forms 

y  =  tie7' (4) 

and  y^tjr' (5) 

Corresponding  points   on   the  profiles,   e.g.,  P  and  Q,  have  a 
common  subtangent  of  the  constant  value  ^  for 

JVr=PiVtaniVPT(=2/^)=^.    .    .     .    :    .     (6) 

AreaP.V0A  =  />  =  <i(^e7'-Q=i(7,-M.   .    •     (7) 
where  PAT  =  71. 

kK&QN0B  =  jJydx=^Ux2-t2), (8) 

where  QN  =  Y2. 

Therefore  the  area  QP4B={;(ri  +  72-M^)  =  ^(7"-!r),  •     (9) 

whereP(?  =  ri  +  r2  =  r'. 

Thus  the  area  of  the  portion  under  consideration  is  equal  to  the 
l^roduct  of  the  subtangent  and  tlje  difference  of  thickness  at  top 
anil  bottom. 

Jjnes  of  Resistance  with  Reservoir  Empty.— l^t  gi  be  the  point  in 
wliich  the  vertical  through  the  C.  of  G.  of  the  portion  OAPN  intersects 
P.V.    Then 

Ngi  X  area  OAPN  =  J  ydx^,    and  therefore 

^9^iY^-U)i;-lifJ\dy  =  \!;;iY^^-t^')> 
or  %i  =  "4 — • 


:330 


THEORY  OF  STRUCTURES. 


So  if  32  be  the  point  in  which  the  vertical  through  the  C.  of  G.  of 
the  portion  OBQN  intersects  QN, 


A'j/2  = 


l'2  +  <2 


Let  G  be  the  point  in  which  the  vertical  through  the  C.  of  G. 
of  the  whole  mass  ABQP  intersects  PQ.    Then 


or 


NG  X  area  ABQP  =  A'j/ 1 X  area  AOXP  -  Xg2  X  area  BONQ, 


Therefore 


NG  = 


4{Yi-ti  +  Y2-t2)~' 


The  horizontal  distance  between  G  and  a  vertical    through  the 
middle  point  of  AB 

x-^    1.,      .^     (yi-hy--(Y2-t2)^     (Y,-U)-{Y2-t2) 

=^(^-2^ii-(2)-  4{y,-ti  +  y2-t2)  "  "       4 

=  one    half    of  the   horizontal  distance  Ix^tween    the  verticals 
through  the  middle  points  of  AB  and  QP. 
The  locus  of  G  can  therefore  1k'  easily  plotted. 
Lines  oj  Resistance  xrith  Reservoir  Full. — Let  7?  be    the  centre 
of  resistance  in  PQ  (Fig.  36.")). 

Draw  the  vertical  QS.  and  consider  the 
equilibrium  of  the  mass  QSAPQ. 

Let     tr'  =  weight     of     a     cubic     foot  of 
water. 


BOA 


or 


— -=moment   of    water-pressure   against 

QS  about  R 
moment   of    weight    of   QBS    about 
/?  + moment   of  weight   of  QPAB 
about  72, 


^-^= moment  of  QBS  about  R^^^<J" -T)wGR. 


RESERVOIR  WALLH. 


331 


The  first  term  on  the  right-hand  side  of  this  equation  is  gener- 
ally very  smaU  and  may  be  disregarded,  the  error  being  on  the 

safe  sidft. 

In  such  case 


"6  /  T'-r 

Also  the  mean  intensity  of  the  vertical  pressure 
wX&reaAPQB^  L_][\ 


=  po  = 


FQ 


and  the  maximum  intensity  of  the  vertical  pressure 


=  Pi-(|_35)r     ^'    l-2q 


2R 


or 


R 
T 


?(l+6g)=/(l+69)(l-y7). 


General  Case.-Ut  the   profile  \ye  of   any  form,  and  consider 


any  portion  ABQP,  Fig.  366. 

Take  the  vertical  tiirough  Q  as 
ttio  axis  of  X,  and  the  horizontal  line 
coincident  with  top  of  wall  as  the 
axis  of  ?/. 

The  horizontal  distance  (y)  hot  ween 
I  lie  axis  of  X  and  the  vertical  through 
the  ('.  of  G.  of  the  portion  under  con- 
-idcration  is  given  by  the  equation 


yp'^'^P''^'' 


Qr:i 


-Y -^ 

Fig.  366. 


/  hoing    the   width,   dx   the   thickness, 

im,l  y  the  horizontal  distance  from  OQ  of  the  C.  of  G.  of  any  layer 

TV  at  a  depth  x  from  the  top. 


832 


THEORY  OF  S'^RLCTURES. 


When  t^<  1  jir  is  et7)pty,  the  deviation  of  the  centre  of  resist- 

ance from  Ifie  centre  of  base 

=qT=Y-y<l. 

^Vhpn  the  rqsor\oir  is  full,  let  q'T  be  the  deviation  of  the  centre 
of  resistance  from  the  centre  of  the  base,  and  disregard  the  moment 
of  the  weight  of  the  water  between  OQ  and  the  profile  BQ.    Then 

. m  ^  moment  of  water-pr.  ±  moment  of  wt.  of  OBQ 
^  weigiit  of  ABQP  ^  ^ 


Hence 


1     w'j*        _     ,, 
«'  /    tdx 


1     u/x' 


w  /    tdx 


15.  Relations   between   the   Elastic  Constants.— In  an  isotroi 
body,  i.e.,  in  a  body  whose  properties  are  the  same  in  all  directioiu* 
if  three  simple  stresses,  p,  of  the  same  magnitude  and  sign  are 
applied  in  three  different  directions,  there  is  no  distortion  of  form 
and  the  volume  only  is  changed.     If  dV  is  the  change  in  the  volume 

V,  the  strain  is  measured  by  the  ratio  y,  and  if  A' is  a  coefficient 

such  that  J.  --^.  then  A'  or  the  ratio  of  the  stress  to  the  strain  is 

called  the  hulk  modiiliix  or  the  modulim  of  cubic  comprexnhiliiy. 

The  Uuear  ditiiension.s  of  tin-  IkmIv  change  wjually     Let  a  be 
one  of  these  dimensions.     Then  Vcta^, 


% 


and 


SV    :i(i'  da       da 


V         a 
if  the  cubic  strain  is  small. 


Hence  the/.««ir..^rain---l^-i?. 

a     3  1       3  ik 


RllLATlOSS  BETWEEN  THE  ELASTIC  CONSTANTS. 


333 


Relation  between  G  and  E.—A  shear  stress  q  along  the  face  AC, 
Fig.  367,  of  the  square  ABDC  distorts  the  sqtiare 
into  the  rhombus  BA'C'D. 

Draw  CE  perpendicular  to  BC.    Then  the  strain 
ulong  the  diagonal  BC 

C'B-CB    CE'  \CC' 
"      CB      ''CB  ~2CD 

•  —\  (shearing  st  -am), 


Fio.  367. 


or 


2 


Again,  it  has  already  befn  shown  that  a  pair  of  equilibrating 
.stresses  q,  Fig.  368,  develope  an  equal  tensile  stress  q  along  one  diago- 
nal and  an  equal  compressive  stress  q  along  the  other.    Hence 
the  total  longitudinal  strain 

1,±    i.    1  1 
"E^aE     2     2  6" 

q  q  q  q  ^    (7  being  Poisson's  ratio,  Chapter  IV. 


Therefore 


2\«T+1/ 


Relation  between  G,  E,  and  A'.— Let  a  pull  of  in- 
lu).  .368.       tensity  p  act  upon  the  upper  and  lower  faces  BD 
and  FH  of  an  elementary  cube,  Fig.  369,  and  let  this  stress  be 

<li\  ided  into  three  jwrtions  each  equal  to  t>-    Without  changing  the 

cDtidition  of  stress  it  may  Ix?  assumeil  that  a  pull  and  a  push,  each 

of  intensity  r,,  act  upon  the  remaining  faces. 

o 

Combining  the  puU  ^  on  the  faces  BD  an«l  FH,  with  the  push  ^ 

on  the  faces  BE  and  CH,  the  re.sult  is  a  simple  shear  along  a 
plane  making  an  angle  of  45*»  with  the  axis  of  the  pull. 

Combining  again  a  second  pull  j-  on  the  faces  BD  and  FH  with 


334 


THEORY  OF  STRUCTURES. 


the  push  on  the  faces  AH  and  BG,  the  result  is  also  a  simple  shear 
jj  in  a  plane  at  right  angles  to  the  first  plane 


^ 


tlr;- 


and  making  an  angle  of  45°  with  the  axis  of 

the  pull.    There  is  now  left  a  pull  ^  on  each 

^£  face  of  the  cube,  producing  a  cubic  dilata- 
tion.   Then 


TTq 


x'H 


Fio.  369. 


the  lineal  strain  due  to  the  bulk  strain. 

3\k)    9  K' 
the  lineal  strain  due  to  the  first  shearing  strain 

2  \aj   6  ^' ' 

the  lineal  strain  due  to  the  second  shearing  strain 

J-(k\At 
2\G7     6(r" 

Thus  the  total  longitudinal  strain  =  k  i?  +  nr;-^7^7; 

;>  A      O  (/      O  (/ 

3\3A'^«/     E' 


and  therefore 


K- 


GE 


SUS-iE 


and    E- 


9KG 
•,iK  +  G' 


Again,  the  lateral  strain  =  -  ^  -  L  ^  -  ?  ^  J--  _  -^,|  and  therefore 
Poisson's  ratio  is  given  by 


«- 


from  which 


:i\:\K^G} 

3\2G    3A7 


6A*-|-2r;        2G 


0- 


•iK-2G     E-20' 


iK 


2(ff-H)' 


RELATIOSS  BETWEEN  THE  ELASTIC  CONSTANTS. 


335 


and  therefore 


K  = 


aE 


3(a-2)' 


The  preceding  results  may  also  be  obtained  in  the  following 
manner: 

Ix't  a  solid  body  be  strained  uniformly,  i.e.,  in  such  a  manner  that 
lines  of  particles  which  are  parallel  in  the  free  state  remain  parallel 
in  the  strained  state,  their  lengths  being  altered  in  a  given  ratio 
wiiich  is  practically  very  small.  Lines  of  particles  which  are  oblique 
to  each  other  in  the  free  state  are  generally  inclined  at  different 
angles  in  the  stnuned  state,  and  their  lengths  are  altered  in  different 
ratios. 

Ijoi  the  strain!  IK  of  the  body  convert  a  rectangular  portion 
AliCD  (Fig.  370)  into  the  rectangle  AB'C'Ly,  where  AB'  =  (1  +a)AB 
iim\  AD' =  {1+^) AD. 

Now  a  and  ^  arc  very  small,  so  that  thtir  joint  effect  may 
1)0  considered  to  be  equal  to  the  simi  of  their  separate  effects. 
Hence: 

First.  Let  a  simple  longitudinal  strain  in  a  direction  parallel  ta 
Mi  convert    the    rectangle  ABCD  into  the 

rectangle  AB'ED,  where  BB'^^aAB.  o'l 

.\  line  OF  will  move  into  the  position  af 
OF,  when  FF-aZ>F,  and 

.        ^r,    OF' -OF  0> 

the  strain  along  OF  = 


FF'  cow  9 
OF 


aPFcmd 
OF 


OF 
a  COB*  6, 


Fra.  370. 


9  being  the  angle  OFD. 

Also,  the  "digtoriion  or  deiriation  from  rectangularity  " 

-  angle  FOF'  -      -  - — rrm c  cos  0  sm  0. 


OF 


ur 


fWond.  liPt  a  simple  longitudinal  strain  in  a  direction  parallel 
to  AD  convert  the  rectangle  ABCD  into  the  rectangle  ABKty, 
v\\vtv  Diy  - ?  AD. 

The  line  OF  will  move  into  the  position  O'F",  where  00* -^^40 


!     1^^ 

--      -.V 


^^  THEORY  OF  STRUCTURES. 

and  F"F  =  Diy=pAD;  therefore 

the  strain  along  OF^  '     . 

Ut 

Draw  O'M  parallel  to  OF.    Then 

O'F"  -  OF  =  OT"  -  O'M  =  /^"Af  sin  0  =  (F"i!'  -  FiV)  sin  0 
='{Diy-(Xy)  sin  <?=/3(.4  0-vlO)  sin  0 
= ^  CD  sin  0;  therefor 

....      ,       ,,Ei    i3  01)  sin  0 

the  strain  along  OF  =       ^^^ =^  gina  ^. 

The  (fis(or/ion  =  the  angle  F"0'M 

F"M  cos  tf    ,3  0Dcosd 
"^      Q/i'       =       ^ =  /?  sin  ^  cos  fl. 

//enrc  when  the  strains  are  simultaneous,  the  line  OF  will  take 
the  position  O'F'"  between  O'F"  and  OF',  anil 

the  to<«/  .s/mtH  along  OF=-a  cas^  tf+^  sin*  tf* 
the  tota/  distortion  =  (a  -,9)  sin  tf  cos  d. 

Again,  draw  a  line  OC,  perpendicular  to  OF. 

The  angle  06'.4  ={K)°-<?,  and  hence,  from  the  above, 


and 


the  total  strain  along  0(1=  a  sin"  fl+,9  cos"  (?, 
the  corresponding  di.stortion  =-  (a  -/?)  sin  <?  cos  0. 


Denote  the  strain  along  OF  by  f,,  that  along  OG  by  cj,  and  each 
of  the  equal  distortions  by  ~.    Thr-n 

ei+ej -«+/?. 

Again,  if  OF,  0*7,  Fig.  371.  are  the  .sides  of  a  rectangle  enclone<l 
in  the  n'ctangle  .l/i(7).  the  straining  will  convert  the  rectangle  int<. 
an  obii.iiH'  figuJT  with  its  opj«>.itr  ^\k\v>  parallel.  The  lengths  of 
adjacent  sides  are  altered  by  the  amounts  f,  and  e^,  antl  the  angle 


RELATION  BETWEEN  THE  ELASTIC  CONSTANTS.  337 

0  by  <f).  The  above  results  may  also  be  considered  to  hold  true  if 
the  straining,  instead  of  being  uniform,  varies  continuously  from 
poitit  to  point. 

Consider  a  unit  cube  ABCD  subject  to  stresses  of  intensity  pi 
ai.tl    7)2    upon   the   parallel   faces  AD, 
liC  and  AB,  DC.      By  Art.  1,  Chap. 
IV, 


a 


Pl_p2 

"E     aE' 


•*Pt 


^    aE^  E' 
and  the  strain  perpendicular  to  the  face 

at     atd 

If  the  stresses  are  of  equal  intensity  but  of  opposite  kind,  i.e., 
if  the  one  is  a  tension  and  the  other  a  compression, 
/'I  =  —  Pa  •"  P)  suppose ;  then 

a=-/j=-iM+-j,  and  the  third  strain  is  nil. 

Thus  the  volume  of  the  strained  mWd 

-(1  +a)(l  -a)(l)  =  1  -0"  =  1,  approximately, 

HI  that  the  volume  is  not  sensibly  changed. 

.Mso,  if  OGHF  is  an  enclosed  square,  0  being  the  middle  po\nt 
of  AD,  e-450,  and 

/v  4-  9 

«!  -  ca  -  — :~  --  0  -  strain  along  OF  or  OG, 
aiil  the  distortion  -  change  in  angle  0 

-*-^^-2«-|'(l4). 
riiis  result  may  he  at  once  deduced  from  the  figure.    For 


tan 


F(Ki    <)l)     l-^fi     la 


ri)     1 


+[i      la  /ft>°-^\ 

— -  - ,— —  -  tan  I      .,  -  I , 

4  at      1  fo  \     2     / 


338 
or 


THEORY  OF  STRUCTURES. 


4>       i> 

1+a    ,  ,  .     4>    ■,  ,  i*' 
l  +  tan-^    1+2 


since  <f>  is  very  small.    Hence 


^ 


-a. 


As  already  shown  in  Art.  2,  shearing  cannot  take  place  along 
one  plane  only,  and  at  any  point  of  a  strained  solid  the  shears  along 
planes  at  right  angles  are  of  etjiial  intensity.  The  effect  of  such 
stresses  is  merely  to  produce  a  distortion  of  figure,  and  generally 
without  sensible  change  of  volume. 

Thus  shears  of  intensity  q  along  the  parallel  faces  of  the  unit 
square  ABCD  will  merely  distort  the  stjuare  into  a  rhombus  ABCW 
_i^__  (Fig.  372).     Denoting  the  change  of  angle  by 

^-,-C'  <f>,  and  a.<isuming  that  the  "stress  is   propor- 
tional to  the  strain," 

q-G<f>, 

where  G  is  a  coefficient  called  the  modulus  of 
riijidity.  and  depends  upon  a   change  of  form. 
It  is  stated  in  the  same  units  as  are  employed 
to  specify  the  stress. 

Consider  a  section  along  the  diagonal  BD. 

The  stre.><i<es  on  the  faces  .1/i,  .1/),  and  on  CB,  CD,  re.«iolvod 

parallel  and  |X'rjx'iidicular  to  BI>,  are  evi<lently  equivalent  to  nil 

and  a  normal  forc«'  f/N  2  n-spj-ctively.      Thus   there  is  no  sliding 

tendency  along  BI),  but   the  two  [wirtions  ABD  and  CBl)  exert 


u|)on  each  (ttlier  a  pull,  or  tension,  of  intensity 


7^ 


7^-'_. 


BI)        V  2 

■'similarly  it  may  U'  shown  that  then'  is  no  tendency  to  slide 
along  AC,  but  that  the  two  i)ortioiis  ABC  and  .l/X'  exert  u|)on 
eacli  other  a  pres.Min'  of  intetisii y  r/.  i he  straining  due  to  the  .stii-ar- 
ing  stresses  is,  therefore,  identical  with  that  produced  by  a  thrust 


GENERAL  EQUATIONS  OF  STRESS. 


339 


iiiid  tension  of  equal  intensity  upon  planes  at  45".    Hence,  as  proved 

above, 


g=G^=G^^(l+i), 


and  therefore 


G  = 


Ea     _q_ 


Now  a  rarely  exceeds  4,  and  hence  G  is  generally<  ^E. 
Again, 


K-- 


jE 


2*7  +  1 


3((7-2)     3<7-2 


G. 


Therefore 


<7  =  7 


6A'+2G 
JA-2G- 


/ 

X 

i6.  General  Equations  of  Stress. — liet  x,  y,  zhe  the  co-ordinates 
witli  r('si)oct  to  three  rectangular  axes  of  any  point  O  in  a  strained 
liody. 

Consider  the  equilibrium  of  an  ele- 
iiiciit  of  the  body  in  the  form  of  an 
indefinitely  small  parallelopiped  with 
its  edges  OAi=dx),  0B{  =rfj/).  0C{  =dz) 
|t!irallel  to  the  axes  of  x,  y,  z.  It  is 
Mssimicd  that  tiie  faces  of  the  element 
■ire  sufficiently  small  to  allow  of  the 
li'^irihution  of  stress  over  them  lx>ing 
rcgiirdeil  as  uniform.  The  resultant 
tunc  on  each  face  will  tlu'refore  be  a 
Millie  force  acting  at  its  middle  point. 
Li  t  A'l,  V'l,  Z\  be  the  comixnients  parallel  to  the  axes,  x,  y,  z  of  the 

resultant  force  \^r  unit  area  on  the  face  BC. 
' '    A\.,  Yj,  Z2  Ix"  the  corres|«)nding  conifwnents  for  the  face  AC. 
"   A3,  Y3,  Z3  l)e  the  corresjKniding  components  for  the  face  AB. 
These  com|Kinent8  are  functions  ot  x,  y,  z,  and  therefore  become 

-(.v,.f .),   -iy,.q^4  -(z,.f.) 


Fro.  373. 


340  THEORY  OF  STRUCTURES 

for  the  adjacent  face  AD; 

-(•^•^-f*).  -('-'-f^*).  -(z=-f*) 

for  the  adjacent  face  BD; 

.(.v,.^^,   -(r,.^»,   -(z,.'j^) 

for  the  adjacent  face  DC. 

Hence  the  total  stress  parallel  to  the  axis  of  x 

^Xidydz-  \Xi  +  ~dxjdydz+X^zdx-(x2  +  --^yjdzdz 

+ X^^xdy  -  (a'3  +  ^dz)dxdy 

VdF^if^-db/'^y^'- 

Similarly,  the  total  stress  parallel  to  the  axis  of  y 

'-y-di^-d^^-dir'^y'^' 

and  the  total  stress  parallel  to  the  axis  of  z 

Ki^^i^^-dir'^yd'- 

Let  p  1)0  the  density  of  the  mass  at  O,  and  let  P„  P^,  P,  be  the 
(•oiniMjnents  parallel  to  the  axes  of  j,  y,  z  of  the  external  force,  per 
unit  mass,  at  (). 

(nhdydzP,  is  the  component  parallel  to  the  axis  of  x  of  the 

external  force  on  the  element; 
fnlnii/ilzP,  is  the  comiK)iient  parallel  to  the  axis  of  y  of  the 

external  force  on  the  element ; 
fMh,l>i,ur,  is  the  com|Mrtient  parallel  to  the  axis  of  z  of  the 

external  force  on  the  element. 
The  clemciit  is  in  etjuilibrium. 


II' 


GENERAL  EQUATIONS  OF  STRESS. 


341 


Therefore 


dA'i    dA'2    dXz      D  . 


dz 


dW    dY.dYi  . 

-W^-d^^-di-^f^"' 

dZ\  ,  dZs    dZz       „ 
'di^l^^-dk^f'^'- 


(1) 


These  are  the  general  equations  of  stress. 

Again,  take  moments  about  axes  through  the  centre  of  the  ele- 
ment parallel  to  the  axes  of  co-ordinates,  and  neglect  terms  involving 
(difdydz,  dx(dy)^dz,  dxdy{dzf.     Then 

Yi^Zi,    Zi=X3,    and    X2  =  Yi.       ...    (2) 

Adopting  Lame's  notation,  i.e.,  taking 
Si,  N3,  N3  as  the  normal  intensities  of  stress  at  0  on  planes  per- 
pendicular to  the  axes  of  x,  y,  z; 
Ti  as  the  tangential  intensity  of  stress  at  0  on  a  plane 
perpendicular  to  the  axis  of  x  if  due  to  a  streas 
parallel  to  the  axis  of  y,  or  on  a  plane  perpen- 
dicular to  the  axis  of  y  if  due  to  a  stress  parallel 
to  the  axis  of  x;  and  T2,  T3  similarly,  equar 
lions  (1)  become 


dNi    dT.dTi      _ 


dz 


dTidNidTr  . 

dTa     dTi    d\i 


(3) 


dx      dy       dz     ^  '' 

Next  consider  the  equilibrium  of  a  tetrahedral  element  hav- 
ing three  of   its  faces  parallel  to  the  co-  " 
ordinate  planes.     Let  I,  m,  n  be  the  direc- 
t'utn-cosines  of  the  normal  to  the  fourth 
f;ici>. 

Also,  let  A',  1',  Z  be  the  components 
paralk'l  to  the  axes  of  x,  y,  z  of  the  in- 
tensity of  stress  ft  on  the  fourth  face. 

X  -  l\i  +mT3+nT2+ yPJidx. 

Rut  the  last   term  disappears  in  the    '  Fro.  874. 

iiiiiit  when  the  tetrahedron  is  indefinitely  small,  and  hence 


S42 


THEORY  OF  STRUCTURES. 


X=^L\i+7nT3  +  nT2; 
Y^lT3  +  ni\2  +  nTi; 
Z==lT2  +  inTi  +  nX3. 


(4) 


Those  throe  equations  define  R  in  directiop  and  magnitude  when 
the  stresses  on  the  three  rectanguhir  planes  are  known. 

Let  it  l)e  re(iuired  to  detertnine  the  planes  upon  which  the  stress 
is  wholly  normal.     Wo  have 


X  =  IR,     }'  =  w/?,    Z  =  nR. 


(5) 


Substituting  the.se  values  of  A',  1',  Z  in  eqs.  (4 J  and  eliminating 
I,  m,  n,  we  obtain 

R^  -  RH.\  1  +  -Va  +  -Vs)  +  RiXiXo  +  X2X3  +  -V3.V, )  -  Ti^  -  T^^  -  T^ 
-{Xu\,X3-XiTr'-X2Ti^-XaT3'  +  2TiT2T3)=0,.     .     .     (6) 

a  eubic  (Equation  giving  three  real  \aluos  for  R,  and  therefore  three 
sets  of  values  for  /,  ni,  and  n,  showing  that  there  are  three  planes 
at  O  on  eacli  of  which  the  intensity  of  stress  is  wholly  normal.  The.«e 
planes  are  at  right  angles  to  each  other  and  are  called  principal 
jilariefs,  the  corresponding  stresses  iK'ing  jrrincipal  stresses.  They 
are  the  principal  planes  of  the  <iuadric 


.V,  j2  +  X2if  +  X:iZ^  +  •iT.yz  +  2T2ZX  +  2T3xy  =  c. 


(7) 


For  the  equation  to  the  tangent  plane  at  the  extremity  of  a  radius  r 
whose  ilirection-cosines  are  /,  m,  n  is 

Xrx  +  Yry+Zrz  =  c, (8) 

and  the  equation  of  the  j)arallel  diametral  plane  is 

Xx  +  Yij  +  Zz  =  0 (9) 

The  direction-cosines  of  the  jxTix-ndicular  to  this  plane  are 


i". 


}'  Z 


so  tliat  the  resultant  stress  R  must  act  in  the  direction  of  this  per- 
pendicula  . 


;!!' 


ttmL 


GESERAL  EQUATIOSS  OF  STRESS. 


343 


Hence  the  intensities  of  stress  on  the  planes  perpendicular  to 
(lie  axes  of  the  quadric  (7)  are  wholly  nomiai. 

Refer  the  quadric  to  its  principal  planes  as  planes  of  reference. 
All  the  T's  vanish  and  its  equation  becomes 


Also,  the  general  equations  (3)  Ix^come 


(10) 


dx 

dX2 
dy 

dXj 
dz 


=pPr; 


=pPy 
=pP.. 


(11) 


Again, 


Consider  A',  Y,  Z  as  the  co-ordinates  of  the  extremity  of  the 
Ptiaifrlit  line  representing  R  in  direction  and  magnitude.  Equation 
(IL'  is  tlien  the  equation  to  an  ellipsoid  whose  semi-axes  are  JVi,  JV2, 
X.i.  As  a  plane  at  0  turns  around  0  as  a  fixed  centre,  the  extremity 
(if  a  line  representing  the  intensity  of  stress  R  on  the  plane  will  trace 
(i\it  an  ellipsoid.    This  ellipsoid  is  called  the  ellipsoid  of  stress. 

Xiite  1.  The  coefficients  iii  the  cubic  equation  (6)  are  invariants. 
Tims,  A^  +.V2  +  -V3  is  constant,  or  the  sum  of  three  normal  intensities 
(if  stress  on  three  planes  placet!  at  right  angles  at  any  point  of  a 
St  ruined  l)ody  is  the  .same  for  all  positions  of  the  three  planes. 

Xotc  2.  The  perpendicular  p  from  0  on  the  tangent  plane, 
fi [nation  (8), 

.     .     .    (13) 


■'wr^'' 


and  therefore    R= — . 


:it 


Xotc  3.  Let  the  stress  be  the  same  for  all  positions  of  the  plane 
(K  Then  Ni^N^^X^,  and  the  ellipsoid  (12)  becomes  a  sphere. 
'lir  stress  is  therefore^  cverywlicre  normal,  and  the  body  must  be  a 
jMiiiMi  iluid.  ConvciM-ly,  if  ihe  stress  i.s  everywhere  normal,  the 
Idily  must  be  a  perfect  fluid,  the  ellipsoid  becomes  a  sphere,  and 
tliiTefore.Vi-A'a-iVa. 


344 


THEORY  OF  STRUCTURES. 


17.  Relation  between  Stress  and  Strain. — In  Art.  15  it  was  shown 
that  when  the  size  and  figure  of  a  body  are  altered  in  two  dimensions, 
there  is  an  ellipse  of  strain  analogous  to  the  ellipse  of  stress.  If  the 
alteration  takes  place  in  three  dimensions,  it  may  be  similarly  shown 
that  every  state  of  strain  may  be  represented  by  an  ellipsoid  of  strain 
analogous  to  the  ellipsoid  of  stress.  The  axes  of  the  ellipsoid  are 
the  principal  axes  of  strain,  and  every  strain  may  be  resolved  into 
three  simple  strains  parallel  to  these  axes. 

It  is  assumed  that  the  strains  remain  very  small,  that  the  stresses 
,z  developed  are  proportional  to  the  corre- 

sponding strains,   and   that  their  effecta 
may  be  superposed. 

Consitler  an  element  of  the  unstrained 
body  in  the  form  of  a  rectangular  par- 
allelopiped  having  its  edges  PQ{=h), 
PRi=k),PS{  =  l)  parallel  to  the  axes  of 
co-ordinates. 

When  the  body  is  strained,  the  element 
becomes  distorted,  the  new  edges  being 
FQ',  P'R',  P'S'. 

Let  x,y,zhe  the  co-ordinates  of  P. 

Let  x+u,  y+v,  z  +  w  be  the  co-ordinates  of  P*. 

By  Taylor's  Theorem  the  co-ordinates  with  respect  to  P*  of 


375. 


(?'are^l+5^j,    A^,     A^; 


Therefore 


«'-4:.  *('*i). 

dw 
''dy' 

«'-'s.  4^  '(■-'^)- 

i»0.  =  »(.43; 

''«'=K'4„)^ 

ps-.^..^).] 

(14) 


i'  1," 


ISOTROPIC  BODIES. 

FQ'-PQ      dv     „ 
Hence  strain  parallel  to  axis  ol  x= — pg — °°dx°°     '' 


<(    ((    (< 


y- 


FR'-PR    dv 


<(      <(     <<       2  = 


FS'-PS    dw 


^■5^=^'"' 


PS 


=5F=^'- 


345 


(15> 


Again,  cos  Q'FR' 


(      du\du    (      dv\rfv    dt£^ 
V  -^TxlTy^'V^dyldx^dydx 


In  the  limit  this  reduces  to 


Similarly, 


^.^„.    du    dv 
cosQ'FR=^+^. 

„.„„,    du     dw 
cosQ'FS'=5^+5^; 

„.„,„,    dw    dv 
cos  ft'W =5^+3^. 


(16) 


Volume  of  unstrained  element =fcA;i; 


Volume  of  distorted  element    ^hkl (l +^)  (l  +^  (l  +^) 


multiplied  by  the  cosines  of  small 
angles 


Therefore 


Difference  of  volume 


du    dv    dw 


Vol.  of  unstrained  element     dx    dy    dz 


(17) 


=  B(e,+  ey+c,) 

=  the  volume  or  cubic  strain. 

18.  Isotropic    Bodiei,  i.e.,  bodies  possessing  the  same  elastic 
properties  in  all  directions. 


346 


THEORY  OF  STRUCTURES. 


r 


A  normal  stress  of  intensity  .Vi  parallel  to  the  axis  of  x  producer 
a  simple  longitudinal  strain  '  ',  and  two  simple  lateral  strain. 

each  =  — ^g,  parallel  to  the  axes  of  y  and  z,  E  being  the  ordinary 

modulus  of  ehisticity  and  n  l^)is.son's  ratio  (Art.  1,  Chap.  IV). 

\ormal  stre-sses  .V2,  A'3  parallel  to  the  axo.^  of  y  and  z  may  be 
similarly  treated. 

U't  the  three  normal  stresses  act  simultaneously  and  superpose 
the  results.     Then 

total  strain  parallel  to  axis  of  x  ^ '-— -  '  ^    *  ' 


it        i<         <  <     i  i 


\'i     -V.-t^Vi     dr 
y~  E~      oE      ^\hj 


=  Eey; 


ti     (< 


<•     .<      ..    <<    ,     -^  «     -Vi+A^    fiw 


E 


oE 


dz 


(18) 


The  form  in  which  thest>  e(|uations  are  given  is  due  to  Grashof. 
Solving  for  Aj,  A'^,  A'.i, 


'  '"(«T+T}T..  -2)*^'"^(<»+l)(ff-2)^'''"*'*''^' 
which  may  U'  written  in  the  form 


A'i=",4AV,  +  iiA>^-t-e,), 
S\  =  AEey¥me,^e,), 
A'l  -  A  El',  +  kE(e, + f  „), 

»(<i-\)E 


(19) 


(20) 


wtiere 


(ff+l)(«-2)' 
.  <iE 

('»+I)(r»-2)' 

and  k  is  the  coefficient  of  dilalntum. 


ISOTROPIC  BODIES. 


347 


Apain,  the  straining  changes  the  angle  RPS  by  an  amount 
""'  I  'IL  producing  two  tangential  stresses,  each  equal  to  ^ '  T7  +  dJ  j ' 
|,:irallcl  to  the  axes  of  y  and  z.    Therefore 


Similarly, 


(21) 


\dy     dx I 

(1  is  called  the  coefficient  of  rigidity  or  trami^erse  elasticity.    It 
i<  designated  n  in  Thomson  and  Tait's  notation,  and  ft  in  Lamp's 

iidliition. 

liHntion  betuven  A,  i,  and  G— Equations  (20)  and  (21)    prr- 
w  rvc  the  same  forms  whatever      ctangular  axes  may  lie  chosen. 

K.'rp  the  axis  of  z  fixed  and  turn  the  axes  of  x  and  y  through 

ail  angle  a. 

bt  A'l'  be  the  normal  stress  parallel  to  the  new  axis  of  x.    Then 

Si'  =  A',  cos2  „  +  ,\  J  i,in2  a  +2T3  sin  a  cos  a.  .     .     (22) 
bt  y,  y'  and  u',  r'  l)e  the  new  co-ordinates  and  displacements. 


(23) 


Iho  values  of  A','  given  by  etjs.  (22)  and  (23)  must  be  identical. 


Now 


lltllCC 


i-ar' oo«tt-y' sina,  y-x' sin  a +]/ cos  a,  1 
m'-u  cos  a  +  r  sin  a,     r'  =  -u  sin  o  +  t'  cos  a-  j 


(24) 


du'    du 


dr 


-p-.  =^  -7-;  cos  o  +  -J-,  sm  tt 
dx'    dx'  dx' 


du  dr  /du     dr  \ 

dx  dy  ^y     dx  f 


du 


~ct)(?  a  +  V"  sill'  «  '■  7?  HUi  a  co«  o , 
dx  dy  O 


348  THEORY  OF  STRUCTURES. 

and  by  eq.  (23), 

Ai  ^{A-A)  [;i^f:os'a+^sin'a  +  -^  sin  a  cos  aj  +iff.     (26) 
Also,  by  eqs.  (20)  and  (22), 

.>i  '{A-A)\^c(m'a+^sma+  _^_^smaco8aj+i0.     (26) 
Eqs.  (25)  and  (26)  must  be  identical. 
m.r,.f„re  ,,-.-l^=  _;Sj^.,.„ ^^^ 

Adding  together  oqi^.  (20), 

Ai  +  A'2  ^  .V,  =  (.4  +  2X)E(e,  +  Cy  +e,) 

It  may  be  easily  showii  tlmt  the  normal  stn-sses  can  each  he 
si'parated  into  a  fluid  pn-snure  />  and  a  distorting  .stress. 
Hence,  putting 

the  tubic  elantiritv- 1 ? .  _«'  /ofi\ 

•     dutiv    dw    3(<T-2)    '^^  •     •    ^*' 

dx^  (ly    lit 

10.  AppHc«tloni.-l.  rr." /urn.— (hie  rud  of  ^i  rvhndrical  Iwr  of 
isotropic  material  ,«  fixed  arxl  the  bar  is  stretehed  in  the  direction 
of  Its  h.ngih.  The  axis  of  the  bur  is  the  only  lin,.  mt  move<i  laterally 
by  contraction. 

Take  this  line  as  the  ax  s  of  t. 

The  .lisplttcements  «,  ,■,  w  of  any  ,K,int  r,  y,  z  may  be  e.presse«J 
in  the  form 

wax,     v~:iy,     u liz (2«) 

By  eqs.  (20)  an<l  (2?)), 

A'i-«i4-2,?i f3oj 

A',- -/M +;(,?  + a)-, V, (31) 


APPLICATIONS. 


349 


By  eqs.  (21)  and  (29),  all  the  tangential  stresses  vanish. 

Hence,  since  iVi,  iVj,  N3  are  constant,  and  since  the  equations 
of  internal  equilibrium  contain  only  differential  coefficients  of  the 
strosses,  the  hypothesis,  eq.  (29),  satisfies  these  equations. 

First.  Let  iVa-O-iVa;  i.e.,  let  no  external  force  act  upon  the 
curved  surface.     Then 


or 


a     A+^ 


(32) 


Tlm.<»,  the  coefficient  of  contraction  is  less  than  the  coefficient  of 
«'X|mnsion. 

Again,  by  eqs.  (30)  and  (32). 


a  a  <J 


(33) 


:Wnnd.    If  the  bar,  instead  of  bping  free  to  move  laterally,  has 
Its  surface  actcti  upon  by  a  uniform  pressure  F,  then 


A'  -A'  -p. 


By  eqs.  (31)  and  (32), 


I         jAP-kNx 


(35) 


For  example,  let  P  l)e  suthcient  to  prevent  lateral  contraction. 
Tlicii  ,V-0  and,  by  e<jH.  (31)  and  CIS), 


a 


.4 -A', 


AP 


-(-»-l)P. 


_'  Torsion.— {(i)  I/'t  a  rirruiar  cylinder  (hollow  or  solid)  of 
i.iiirth  /  uiulergn  torHKUi  around  it^  axis  (the  axi-s  of  x),  an<l  let  t 
I .  thf  angle  through  which  one  end  iw  twistcnl  relatively  to  tiie other. 
\  I  mint  in  a  trauMverw  wn-tion  distant  x  from  the  latter  will  b« 

!      ttti  through  an  angle  /-.-. 


350 


THEORY  OF  STHVCTUUES. 


The  displacements  m,  r,  w  of  the  point  x,  y,  z  in  this  section  may 
Im'  oxjiressed  in  the  form 

M=0,     r=-2Xy,     w=->ryx-. 


By  eqs.  (20)  and  (21), 


and 


A',=0  =  .V2  =  .V3, 


axis 


r,=.o.   r2=+(;//-J,   7^3= -g«|. 

The  algebraic  sum  of  thf  moments  of  T^,  T3  with  respect  to  the 


r  being  the  distance  of  the  jH)int  (x,  y,  z)  from  the  axis. 

Hence  the  moment  .!/=/>  (Chap.  IX).  of  the  couple  producing 
torsion 


-0^~^fr-dS^G-^J=.r,OJ, 


rIS  f)eing  an  element  of  the  area  at  (x,  //.  z),J  the  polar  moment 
of  inertia,  and  0  tix'  t(.rsi(.n  jmt  unit  of  length  of  the  cylinder,  or  the 
rale  of  hn'st. 

The  torsiontd  rigidity  of  a  solid  cylinder 

R  b<'ing  the  ra<llus  of  the  cylinder. 

('>!  'lursinrj  of  a  l»ar  nf  rllijitic  section. 

The  displacements  u,  r,  w  may  now  Ik'  exprcsseil  in  the  form 


Tlieref<  n . 


n  =  F{ !/,:),     r  -  - Oxz,     w  =  Oxy. 

(hi  »/(•     dir 

<lx  lUr  dz' 


APPLICATIONS  351 

r.=0,     T,^^''(i^Oy),    T,=Gi^^-ez).         .     (36) 

Hence,  by  the  general  eqs.  (3), 

dhi    <Pu    ^  ,.,- 

rf^+J^-0 ^'" 

Also,  the  surface  stresses  are  zero,  and  therefore 

Hence,  by  eqs.  (35), 

'^dz-'^dy='Oizdz  +  ydy) (39) 

aij        dz  " 

This  pqujvtion  must  hold  true  at  the  surface. 
I^t  the  equation  to  the  elliptic  section  be 

ifl     z^ 

Sr+^-> («» 

dz         r-i/  .,,, 

T'-™  d-y'-^ '    ''    '     ^^^^ 

.m\.  by  eq.  (38), 

f2l/^  +  622^|--%2(52-02) (42) 

'i    di/z  satisfies  this  last  Pijuation  and  also  etj.  (30),  if 

''—\rS ^^^> 

Anain.tho  algel)raic  sum  of  the  moments  To,  T3  with  resjiect 

to  the  axis  of  x 

/ilu         \        ,  idu     ,.  \ 

^a\idA-o)if-{d-0)^\ 


352  THEORY  OF  STRICTURES. 

The  total  moment  (A/)  of  the  couple  producing  torsion 

and  the  torsional  rigidity 

=  y=^'b.  +  ,. (45) 

ic)  Torsion  of  a  bar  of  rcctuiigular  section. 
As  in  case  (h),  u  must  satisfy  the  o(|uation 

d„2  +  ^-^ (46) 

Also,  the  e(iuations  of  condition  corresfxinding  to  eq.  (38)  are 

^-fl«  =  0    when    y^±b, (47) 

and 

dz^  J"^    when    ?=  ±r; (48) 

26  and  2c  (h<c)  l)eiiig  the  sides  of  the  rectarjglc.     The  total  moment 
of  torsion,  \\i., /(Ttu  -TiZ)>iS,  is  then  found  to  Im- 

'i     r-'>r_^  (2n  +  l)»       J  ■     ■     ■ 

If  6»f,  i.e.,  if  tlic  s«'«-tion  is  a  s.|uiin>,  cj.  (48)  IwH-omea 

Sf-MMWUCn (50) 

/i=f6'(  iK'inif  the  |N)l»r  moment  of  inertia  with  re8i>ect   to  the 
:(\fs.     I  Chap.  IX.) 

,,6    . 

If  -  IS  ver>-  small,  et].  (48)  iHH'ornes 


APPLICATIONS. 


353 


The  torsional  rigidity  of  a  rectangular  section  is  sometimes  expressed 
l)V  the  formula 


M     5    bM 


0      1862+c^ 


G. 


(52) 


For  tho  further  treatment  of  this  subject  the  student  is  referred 
to  St.  Vonant's  edition  of  Clebsch,  ami  to  Thomson  and  Tait's  Natural 
I'liilosophy. 

;]  Work  done  in  the  s  mall  xtrn  in  of  a  body  (Clapeyron  's  Theorem) .  — 
Multiply  eqs.  (3)  by  udxdydz,  vdjcdydz,  wdxdydz,  and  find  the 
tri|)l('  integral  of  their  sum  throughout  the  whole  of  the  solid. 

riic  terms  involving  the  components  P,,  Py,  P,  may  be  dis- 
regarded, as  the  deformations  due  to  their  action  are  generally 
iiiMppreciable. 

Also, 


J  J  f'-d^''  '^^  ^y  ^' 


du 


''//(N/u/-Nr'W)dyd2-///N,^dx  dy  dz; 

A',',  .V/'  being  the  values  of  .Vi  at  the  two  points  in  which  the 
liiii'  [larallel  to  the  axis  of  x  cuts  the  surface  of  the  bo<ly,  and  «,' 
Us"  the  corresponding  values  of  u. 

I-et  dS,  dS'  lie  the  elementary  areas  of  the  surface  at  these  points 
an.l  /',  /"  the  cosines  of  the  angles  between  the  normals  to*  these 
tltiiicnta  and  the  axis  of  x. 

Tiie  double  integral  on  the  right-hand  side  of  the  last  '-quation 
tliiMi  lM>come8 

/JXN;i'u/dS  -  N /'ruJ'ilS')  -  KN^ludS). 

Tnatiiig  tiie  other  terms  similarly, 

0    :i  \( S'll  +  T3m  +  T2n)u  +  (TJ  +  X..m  +  Tin)r 

+  (TJ  +  Tim  +  N»n)w\dS 
r  -/>  I    J    J    \  XT  du     .,  dr     .,  du> 


/dr    du 


/dir    du\ 


/(iv    (lu'\  /ftir    tlu\      _  /du     dr\ 

'■^^id-z^d^)^HTx'-^)^^<dy^di) 


364  THEORY  OF  STRUCTURES. 

Hence  the  work  dofw  =  JJ(A'u  +  Fv+Zii.)dS 

-i///dx  dy  dz  j (7^;|^^+^vr/-V.  +  A'» +iV3)» 


'y/fdxdydz 


(A'i+A\.  +  A'3)2 
A' 

G 


}' 


E  being  the  ordinary  modulus  of  elasticity. 

20.  Transmission    of    Energy. — C'oJhsider   a  small  parallelepiped 
of  the  nuiU'riul,  moving  in  any  direction  OR  with  a  velocity  v,  the 
coiniK)nents  of  r  parallel  to  the  axes  OX,  OY, 
OZ  Ix'ing  Vj,  Vy,  anil  r„  respectively. 

I-et  /j.j.,  fyy,  },„  .  .  .be  the  mean  intensities 
of  stre.>ss  over  t!ie  faces  wliich  are  at  right 
angh-s  to  the  axis  defined  by  the  first  suffix, 
and  in  tlie  direction  of  the  axis  defined  by 
the  secojid  suffix. 
Fio.  376.  '-^'t  '^^'  Tuy  Tf  ^^  the  energies  transmitted 

across  tiie  three  faces  perpendicular  to  OX,  OY, 
and  OZ,  r(>s|)ectively.     Then 

Tx  "dy  dzifjjV^  ^hyt'y-¥j„r,)  +  energy  of  resilience  +  energy  of  laotion 

Disregarding  the  energies  of  resilience  and  motion, 
T,'-dydzi]rA-,+f^ur^-^f„v,). 

So   also  Ty^dZ  dX(fyrl\,+lyy'-y+fy,V,) 

and  T.^di  dy(f,,v,  +  f,yVy+},.r,). 


Ex.  1.').   Find  Ihr  Inist  iliamrt,  r  <■/  n  Khali  which  will  trnnnmit  1000  W. P. 
at  a  surface  vtloril'i  »/  10  /  .*.  f  /  ...     nixnniim  wnrkimj  »(rf.is  of  10,000  Ihs. 


THICK  CYLINDERS. 


355 


Let  /r,  vr  f /8.  be  the  stress  and  velocity  at  any  radius  r,  and  let  9  be  the 
iiK  liiiution  of  the  radius  to  the  axis  Oy.    Then 

/.»-0,      /.»--/r8in#,      /«-/rC08ff; 

r»  "0,     »»  —  —  Vr  sin  tf,     »i — vr  coa  fl. 
Therefore    Tz^dydx  (frVr  sin'  9  +  frVr  cos'  0)  -dy  dz  l,vr. 

If  /;  is  the  radius  of  the  shaft,  /  the  surface  stress,  and  V 
the  surface  velocity, 


Fio.  377. 


/' 


■5/ 


and     "'"o^' 


IV 
Therefore  T.-dydt'jp^r*, 

anil  the  total  work  transmitted  across  the  section 

IV  IVicR* 


Hence, 


10000X10 


11 R'- 550X1000, 


7 

or      ft' -3.5 

and    ft -1.87  ins. 

21.  Thick  Cylindeii.— If  the  thickness  of  the  cylinder-shell  is 
hirRo  as  compared  with  the  radius,  it  cannot  be  assumed  that  the 
Muss  is  uniformly  distributed  over  this  thickness.  As  in  the  case  of 
thin  cylinders,  however,  it  will  be  assumed— 

(ii)  that  the  metal  is  homogeneous  and  free  from  initial  strain; 

ih)  tliat  the  |)re8sures  are  uniformly  distributed  over  the  internal 
.III' I  ••xtcrnal  surfaces. 

It  will  iilso  1)0  iw-sumed — 
r)  that  the  cylinder-ends  are  free; 

di  that  the  annulus  forming  the  section  of  the  cylinder  Is  com- 
!""W'd  (»f  an  infinite  numlK'r  of  concentric  rings. 

riulcr  these  conditions  the  straining  of  the  cylinder  cannot  affect 
it-  ivliiidrical  form.  Hence  right  sections  of  the  cylinder  in  the 
M!i-traiiM'il  state  remain  planes  after  the  straining,  so  that  the 
l^I,^tit^lt^maI  strain,  i.e.,  the  strain  in  the  direction  of  the  cylinder's 
liiijlth,  is  the  same  at  every  point. 


356 


THEORY  OF  STRUCTURES. 


Consider  an  elementary  ring  bounded  by  the  radii  r  and  r-^-dr. 
At  any  point  of  this  ring, 
Let  q  \w  the  normal  (i.e.,  radial)  stress; 
''  /  Ik>  the  hoop  .str('s.s,  \x\,  the  stress  tangential  to  the  ring; 
"   s  be  the  stress  at  tight  angles  to  the  plane  of  the  ring; 
a,  ,?,  ;-  ho  the  corrcsiKMidirig  .strains; 
E,  aE  1h'  cfX'fHcieiits  of  (iji-cel  and  lateral  elasticity; 
Oo,  /o  and  9,.  /,  be  the  values  of  7  and  /  at  the  outer  and  inner 
surfaces  of  the  cylinder  resjx'ctively. 

I'hen,  since  q,  f,  and  s  are  principal  stresses. 


<  I 


E     aE'    '^-E'aE' 


aE 


1  «     f+Q 


But  r  is  constant.  Also,  since  the  ends  are  free,  the  total  longitu- 
dinal pressure  on  a  transverse  section  is  nil,  and  hence  it  may  be 
inferred  that  s  is  zero  at  every  point.    Adopting  this  value  of  a, 

f+q  =  &  constant  =  c . 

The  radial  stress  q  diminishes  from  the  inner  to  the  outer  surface 
and  may  be  represented  by  the  ordinate's  of  some  curve  7077,,  in  which 
Tf/lo,  rq,  and  ri7i  define  the  radial  stres.ses  at  the  outer  surface  of 
radius  ro,  at  the  radius  r,  and  at  the  inner  .surface  of  radius  ri.  Then 
''yan«a  ro7o7iri=the  total  rc-siscance  10  sepa- 
ration between  the  two 
halves  of  the  cylinder 

=  2(7,r, -7oro), 
area  raqoqin  =7,ri  -qoTo. 

So,        area  r77iri  =  7,ri  - tt  =»  - /Vdr, 


or 


Fi<i.  37s 


the  sign  iM'ing  negative  as  /  and  7  are  opposite  kinds  of   stresbes,  i.e., 
one  is  a  tension  and  the  other  a  i)ressure. 


Therefore 
Take 


i(r)-f. 


B+C 


(2) 
(3) 


and 


THICK  CYLINDERS. 


357 

(4) 


The  two  principal  stresses  /  and  q  may  now  be  regarded  as  con- 
sisting of  a  pair  of  equal  stresses  of  the  same  kind  and  of  intensity 

lii      .^-^j ,  and  a  pair  of  equal  stresses  of  opposite  kind  and  of  intensity 

'■(-'?)■ 

First.  Assume  that  only  the  stresses  ±C  are  acting  and  that 

therefore 

B=0='-^,    or    f=.-q. 


Then 


or 


and  therefore 
Thus 


da 
gr*  =  a  constant  =  A . 


f+q 


=  -g=-3. 


(6) 


ore 


Second.  Assume  that  only  the  stresses  B  are  acting  and  that  there- 
C-oJ~^,    or    f-q. 


Then 


.  '^ ,    ^  dq 

f'q-^{qr)^q+rj^, 


or 


dq 
dr 


=0. 


:iii<l 


Therefore  g-a  constant  -/-B (6) 

Third.  Assuming  now  that  the  two  pairs  of  stresses  act  together 
■^ujM'rposing  the  results  just  obtained, 


,-B--^ 


(7) 


MICROCOPY    RESOLUTION    TEST   CHART 

iANSI  end  ISO  TEST  CHART  No    2l 


1.0 


I.I 


If  I-  1^ 

■^  !■■     112  2 

;-  J^   I™ 

140     12.0 


1.25  III  1.4 


1.8 


1.6 


^ 


^     APPLIED  IM/1GE 


■''■'^     ti]5»     Mil-r-       M-l.^' 
-forr^^t,,       >^p^     y„,^  '  *feiiy  USA 

i  ''61    *8<'   -  0  50u  -  P»>o"e 


358 
and 


THEORY  OF  STRUCTURES. 


D      A 

9-B+-.. 


But  q=qo  when  r=ro,  and  q  =  qi  vvhf>n  r=ri. 

Therefore  qo=B  +  ~    and    9i=fl  +  4, 

Ironi  which  it  follows  that 


Hence 


n2-rt.r.2 


^-     ro^-r.2 


and 


qor(?-q\ri^    qi-qa    rihg 


^       ro^-n^ 


2r.2 


r2      ro2-r,2- 


When 


r  =  ri,    /=/i;    and  when    r  =  ro,    /=/o. 


Therefore         /.  =  ^Jorl^qj(rJ  +  r^ 


and 


;  __<lo(ro'+ri')-2qiri^ 


Eti8.  (11)  are  employed  in  the  design  of  gun-barrels. 
Again,  take  ?o=0.    Then 


and 


Therefore 


,  2r,2 

'0  -ri' 

/i     ro2.fr,2 
/o°^    2r,2     >!• 


(8) 


(9) 
(10) 


(11) 


(12) 


K^.  (12)  are  used  in  the  design  of  the  cylinders  for  hydraulic 
l..e.s.<.   an-un.ulators,  etc.     Al.s<,.  .ince  7,  =  -/.  '^'^^''^^^^    ^,    „„„t 
Jilway.s  1„.  i,..s  than  /,  whatever  the  thickness  of  the^cvlinder  mav  \^ 
and  If  i\  18  the  sale  working  stress  of  the  nmterial,  9,  is  the  maximum 


'ii'i 


THICK  CYLINDERS. 


359 


intensity  of  pressure  to  which  the  cylinder  can  be  subjected.    This 
result  is,  cf  course,  based  on  the  assumption  that  qo=0. 
By  eq.  (10),  since  /=/i  when  r  =  ri,  ther 


9oV-9in='     (<?i  -qoW    2qor(?-qi(ri'  +ro') 


ro2-ri2 


.2_r.2 


ro^-n 


Therefore 

ri2    fi+q-2qo    fi+q 


r6' 


Hence 


Therefore 


^-1  = 

To 


,  - ,       ,  approximately,  if  go  is  small. 

h-q      fi-q 

^r:)'-(-'7)'('-'7)"' 

°v'^2j~s  fvV'^2  f^sfy 

=  1  +  J  +2  ^,  approximately. 


thickness  of  cylinder    9  /,  ^]_q 
To 


(13) 


The  result  given  byecj.  (13)  may  bo  used  as  a  second  approxima- 
(imi  for  the  thickness,  a  first  approximation  being  that  given  by  eq.(l) 
in  Chap.  IV,  Art.  8. 

Kx.  10.  The  barrel  of  n  gun  in  nimle  vp  of  three  rings  A,  B,  C,  the  bore  o/ 
ih'  iimi  nnd  the  ihmiirt.rs  nj  the  rings  hcing  in  the  ratio  o]  2:4:5:6.  The  ring 
II  />  shrunk  upon  A,  and  the  ringC  upon  H,  the  rrsulting  preHsuren  at  the  surfneen 
'/  niiiliirl  lieing  4  tons/mi.  in.  Ij  the  firing  of  the  gun  produces  an  internal  pra- 
^•iri  nj  M  tonn/sq.  in.,  find  the  xtresKe.'*  intluerd  in  the  gun. 

Let  r,,  r,,   rt,   ri  be  the  radii  of  the  four  surfaces; 

71,   (/.,   q^,  (]t   "     "    radial  strosNps  at        " 

/i,   /..,   /j,   A   "    "   hoop  Mtn'sses  "        "        before  firing; 

]\,  I'\  I\  F, "        "       after      " 

Ik  fore  firing.    For  ring^,^, -0,7i-4  t./sq.  in.    Therefore 


8r-'        32  .  .     ,       i  r,'  +  r,"     20  , 


^^°  THEORY  OF  STRUCTURES. 

For  ring  B,  5.  -4  t/sq.  in.  =53.    Therefore 

For  ring  C,q,^4  t./sq.  io.,  ^,  =0.     Therefore 


'— *s:-^:=-T?-"^-  -^  /.-,^--^.,.,. 


in. 


A  anf/"'-     "^^  "^  ^  ''^^  •^'•'^"'•^  P^*-  ^1"'^'^  -^••^  «**  the  surface  between 

Let  Q  be  the  pressure  per  square  inch  at  the  surface  between  B  and  C 
FoTimgA,q,=20,q,^P.    Therefore  "ei ween  «  and  C. 

p     2Pr,'-20(r.'  +  r,')     8  „     100        ^  5        40 


Fio.  379.— Initial  Stresses  in  Metal  before  Firing. 
For  ring  B,  9,  -P,  j,  -Q.    Therefore 

r.>_r.i  ~  nV — :r 


ri'  -  n 


9"     9 


■ad 


p^_<?(r,'  +  r.»)  ~2Pr,'_41        32 


Aasuming  that  the  F.  for  ring  .4  is  equal  to  the  F.  for  ring  B,  then 


339''     9    ' 

^  28P-2oQ-G0. 

For  ring  C.  g,-0  and  7,-0.    Therefore 


r,» 


50 


F.--2V   ;'    .--■:?'y. 
f.'-r.'       ir" 


W,  I. 


I 


THICK  CYLINDERS.  361 

.\ssuining  that  the  Fa  for  ring  B  is  equal  to  the  Ft  for  ring  C,  then 

or  1000Q-352P, 

or  Q=.352P. 


fflTfe 


»— SS2^^^2t 


Fio.  380. — StresBcs  In(lu:«d  by  Firing.  Fio.  381. — Combined  Streoaes. 

But28P-25Q-60. 

Therefore  P-3it/8q.  in.    and    Q-1.1  t/sq.  in. 

Hence  the  stresses  after  firing  are 

F,- -25  t/sq.  in.,     Ft- -8J  t/sq.  in., 
Fi  -  - 6.1  t/sq.  in.,    F»  -  - 5  t/sq.  in., 
and  the  total  resultant  stresses  are 

/i+F,-V-25--14i  t/sq.  in., 

/.  +  F.- V  -8*  -  -il  t/sq.  in., 

/.  +  F,--V,*-0.1--28,>,Vt/8q.  in., 

/,  +  F,  -  -  »,7  -  5  -  -  V,*  t  /sq.  in. 

7i(!8.   379,  380,   and    381     represent    respectively  the    stresses    in    the 
ni  iKriiii  nf  the    gun  dcvcioiicd   before  firing,  by  firing,   and    these    streaaes 

(I'Mliiiifd. 


362 


THEORY  OF  STRUCTURES. 


exa:\iples. 

1.  The  principal  stresses  on  twd  planes  an'  and  hb',  at  a  point  0,  are  thrusts 
of  94  and  20  lbs.  jx-r  square  inch.  Find  the  kind,  intensity,  and  obliquity 
of  the  stresses  on  a  third  jilane  inclined  at  ((5°  to  an'  and  hi/. 

A)iK.  -Ifi.-J  lbs.   (thrust);    34°  W. 

2.  At  a  point  within  a  .strained  solid  tiie  stress  on  one  plane  is  a  tension 
of  120  lbs.  i)er  square  inch  with  an  obliquity  of  30°,  and  on  a  second  plane 
a  compression  of  00  lbs.  with  an  obliquity  of  00°. 

Find  (1)  the  principal  stresses;  (2)  the  jjlancs  of  principal  .stress;  (3)  the 
plane  on  which  the  stress  is  wholly  tanKential. 

Ans.  (I)  \  tension  of  127.7')  lbs.  and  a  thrast  of  47.1 1  Ibs.Asq.  in  • 
(2)  r=21°40':  (3)   r  =31°  10' and  stress  ^77.570  Ibs./sq.'in. 

3.  At  a  point  within  a  solid  there  is  on  some  one  plane  a  thrust  of  200  lbs, 
per  .square  inch  of  obliquity  l.j°,  and  on  another  a  thrast  of  80  lbs.  per 
square  inih  of  obliquity  30°.  Find  the  principal  stresses  at  that  point. 
the  position  of  the  axis  of  greatest  principal  stress  relative  to  the  first  plane! 
and  the  inclination  of  the  two  planes  to  each  other. 

Ans.  213.028  and  oS.l.-,2  Ibs./sq.  in. 

4.  At  a  point  within  a  strained  .solid  the  stresses  on  two  planes  at  right 
angles  to  each  other  are  a  thrust  of  30\  2  lbs.  and  a  tension  of  m  lbs.  jx-r  square 
inch,  the  obliquities  In-ing  4.")°  and  30°  resjiectively.  Determine  'i)  the  principal 
stresses;  (b)  the  ellipse  of  stress;  (c)  the  intensity  of  stress  upon  a  plane  inclined 
at  GO"  to  the  major  axis. 

Ans.  (a)  A  tension  of  01.70  lbs.  and  a  thrust  of  39.80  lbs.; 
(c)  A  thrust  of  .■)7.00  lbs. 

5.  At  a  point  in  a  plane  the  stress  on  one  plane  is  a  tension  of  l.'jO  lbs.  per 
square  inch  with  an  obliquity  of  1 5°,  and  on  another  plane  the  stress  is  a  thrust 
of  90  lbs.  with  an  obIi()uity  of  30°.  Find  the  principal  stresses,  the  position 
of  the  axis  <.f  greater  principal  stress  relatively  to  the  first  plane,  and  the 
inclinations  of  the  two  planes  to  each  other. 

Am.   l.">0.1  and  8.").8  Ibs./.sq.  in.;  9°  :?0';  40°  30'. 

6.  At  a  FHiint  within  a  strain<'d  solid  there  are  ;wo  conjugate  stresses  viz. 
a  tension  of  2tK)  lbs.  and  a  thrust  of  \r,0  lbs.  jx^r  .square  in.-h,  the  common 
oblKpnty  being  30°.  Find  (n)  the  principal  strcssi-s:  (/-)  the  maximum  shear 
and  the  direction  and  magnitude  of  the  corresponding  resultant  stress;  (r) 
the  resultant  stress  uik)ii  a  plane  inclined  at  ,30°  to  the  axis  of  greatest  principal 
stress. 

Arm.  («)  A  tension  of  204.0.",  lbs.  and  a  thrust  of  140.9.-,  lbs.  per  .square  inch; 
(b)  17.J.H  lbs.  iK-r  square  inch.;    177.9.-,  lbs.  in  a  direction  making  an 

angle  of  9°  20'  with  the  axis  of  greatest  principal  stress; 
(r)    103.3  lbs.  JHT  s(|uare  inch. 

7.  From  extiTnal  conditions  the  stresses  on    two  planes  at  a  point  in  a 
....  .1  f    .    .  ,.  .     .         "are  inclined  at  10° and 


solid  .ire  thrustsof  .-(land.JOlbs.  jxt  ,x(|uare  inch  and 


20°  rcsiK'ctively  to  the  normal  of  these  pi 
at   the   point,   the   positiiin   nf  the  axis  (>f 


ines.      Find  the  principal  .stre.s.s(v 


•rrcntc-it 


to  the  first  plane,  and  the  inclination  of  the  two  pi 


prinrip-il  -fr 


ril.tti 


planes  to  each  other. 
Am.  55.93  and  20.97  Ibs./sq.  in.;    !(!<>  1.-,';     4S 


EXAMPLES. 


363 


8.  At  a  point  within  a  strained  solid  tho  stress  on  one  plane  is  a  tension  of 
■)0  lbs.  per  square  inch  wth  an  obliquity  of  30°,  and  upon  a  second  plane  is 
a  compression  of  1.10  lbs.  per  square  inch  with  an  obliquity  of  4.5°.  Find 
(<;)  the  prinripal  stresses;  (h)  the  anple  Isetwoen  the  two  planes;  (c)  the  plaae 
upon  which  the  resultant  stress  is  a  shear,  and  the  amount  of  the  shear. 

Ans.  (a)  p,  =i70.9S  lbs.  (comp.),    7)2= -10.12  lbs.  (tens.); 
(6)  61°  31'; 

(c)  91.11  lbs.;    r=26°5r. 
Q.  A  shaft  is  subjected  at  the  most  dangerous  section  to  shearing  intensity 
(ii  :i(({)A-/cin.'  and  to  a  normal  stress  of  400Ar/cni.  .     Find  the  maximum  and 
iiimiinum   principal   stresses  and  the  maximum   shear.     What  will  be  the 
iii:i\iinum  direct  stress  if  Cauchy's  method  is  adopted? 

.Ins.  577.2,-277.2,  427.2fc/cm.';  646.5,  -421.5,  534*;/cm.'. 

10.  The  total  stress  at  a  point  O  upon  a  plane  AB  is  60  lbs.  j)er  oquaro 
imli,  and  its  obliquity  is  30°;  the  normal  component  upon  a  plane  CD  at  the 
point  O  is  40 lbs.  per  square  inch;  CD  is  periiendicular  to  AB.  Find  (a)  the 
total  stress  upon  CD,  and  also  its  obliquity;  (l>)  the  prin<'ipal  stresses  at  0; 
(.1  the  equal  conjugate  stresses  at  0. 

Ans.  (n)  tan-'(?);   oO  lbs.; 

{!,)  70..')7  lbs.  and  1.5.39  lbs.; 

(r)  34..32  lbs.;    obhquity  =41°  42'. 

11.  At  a  point  in  a  strained  body  the  stn^sses  are  a  tension  of  2.55  lbs.  and 
171  lbs.  per  square  inch.  Find  thr-  stress  on  a  planr-  in('lined  at  27°  to  the 
plane  of  principal  stress,  and  its  obliquit}'.  -l/.'s.  240  Ibs./sq.  in. ;  8°  8'. 

12.  The  principal  stresses  at  a  jwint  in  a  straiii(>d  mass  are  a  tension  of 
:{(«)  lbs.  and  a  thrust  of  KK)  lbs.  [)er  square  inch.  Find  (a)  the  obliquity  and 
i'lirnsity  of  stress  on  the  plane  at  .30°  to  the  plane  of  greatest  stress;  (h)  the 
iitcM>ity  of  the  tangential  stress  on  the  plane  upon  which  that  stress  is  greatest; 

■     till'  angle  to  the  plane  of  greatest  stress  of  the  ])lane  upon  which  the  stress 
i-  cn'irelv  a  shear.     .\lso  find  the  intensity  of  the  shear. 

".l«s.  („)47°7',  272lbs.;(M  2.30  lbs.;  (r)  .5.3°  .52',  219  lbs. 

13.  If  the  principles  of  the  ellipse  of  stress  are  applicable  within  a  mass 
(  f  oiirth,  and  if  at  any  iniint  of  the  mass  t!ie  stress  upon  a  plane  is  double 
ii  loiijugate  stress,  the  angle  l)etween  the  two  stresses  being  20°  28',  show 
ill  ,1   the  angle  of  ie|K)se  of  the  earth  is  27°  .58'. 


1*. 

i-   -III 
■    ih( 
1.  „.ti 

'5. 

In  llw. 


At  a  jwint  in  a  strained  solid  the  intensity  of  shear  on  two  pianos  at 
iiigtcs  is  24  lbs.;  the  obliquity  of  the  resultant    stress    on    one    ]>lane 

'S,  and  of  the  resultant  stress  on  the  other  plane  issin'-ti.  Find 
'  magnitudes  of  the  two  forces.  Alsf)  liml  (>'<)  the  stress  upon  a  plane 
iig  the  two  planes  in  question;    (r)  the  principal  stresses  at  the  p<iint. 

Alls.  (<i)  40  lbs..  .30  lbs.;  ih)  40.5  lbs. 

At  a  [loint  within  a  solid  there  is  a  pair  of  i  .)njugate  stresses  of  '.iO  and 

[H'r  s(]uare  inch.  Their  common  obli(iuity  is  10°.  Find  tlie  principal 
>  and  the  angle  which  the  normal  to  the  plane  of  conjugate  stresses 
with  the  plane  «)f  principal  stres.ses. 

Am.  43 J  and  27 J  lbs./s<i.  in.;  30°  29'. 


364 


THEORY  OF  STRUCTURES. 


*  It  u  *  P*""*  ''■'**""  ""  ^'^'■''''""'  **"'■•'  ♦^^'^  «*''*^'''«  ""  «"e  plane  is  a  tension 
of  100  lbs.  per  square  ineh  with  an  obliquity  of  30°,  and  on  a  seeond  plane 
a  compression  of  M  lbs.  with  an  obliquity  of  (50°.  Find  (a)  the  angle  between 
the  planes;  (b)  the  plane  ui)on  which  the  stress  is  whoUv  a  shear-  (c)  the 
I)lanes  of  principal  stress;  also  (,/)  find  the  conjugate  stresses  which  have  a 
common  obuquity  of  45°. 

Ans.  (a)   -,0°.y;   (I,)  64.6  lbs.;  r=31°16'; 

(c)   /;.  =100.46  (tens.),  pj=  -39.26  (compr.)- 
(rf)  92.63  lbs.  (tens.),  45.11  lbs.  (compr.). 
17.  At  a  point  wnthm  a  strained  mass  the  principal  stresses  at  a  given 
pomt  are  m  the  ratio  of  3  to  1.     Find  the  ratio  of  the  conjugate  stres^s  at 
the  same  ponit  havmg  the  common   obliquity  30°.     Also  find  the  inclination 
of  the  axis  of  greatest  principal  stress  to  the  horizontal. 

,a    .  .         ,  .  ,  ^"s-  Equal;  60°. 

IB.  A  bar  of  iron  is  at  the  same  time  under  a  direct  tensile  stress  of  5000 
lbs.  per  square  inch  and  a  shearing  stress  of  3500  lbs.  per  square  inch 
What  would  be  the  resultant  equivalent  tensile  stress  on  the  material? 

.„   T  b-        .u        ,  ..  ^"«-  6S01  lbs. 

19.  laking  the  safe  tensile  stress  of  wrought  iron  to  be  5  tons  per  square 
inch,  determine  whether  it  would  be  safe  to  subject  a  piece  of  wrought  iron 
to  a  tensile  stress  of  3.92  tons  per  square  inch  together  with  a  shear  stress 
of  3.36  tons  per  square  inch.        Ans.  Unsafe,  since  max.  stress -5.85  tons 

JO.  At  a  point  in  ^  solid  there  is  a  tensile  stress  of  6  tons/sq.  in.  and  a 
shear  stress  of  4  tons/sq.  in.    Find  the  principal  stresses. 

,.    T,.         ...  ^n«-  Sand  -2  tons/sq.  in. 

31.  The  principal  stress  at  a  point  in  a  plane  section  of  a  strained  soUd 
IS  /  tons.  If  the  shear  and  normal  (tensile  or  compressive)  stresses  at  the 
same  point  are  equal,  find  their  values.  j^^a     618/ 

«.  The  shearing  stress  at  a  point  in  a  section  of  a  strained  mass  "is"  4  tons, 
l-ind  the  normal  stress  (tensile  or  compressive)  at  the  same  point  if  the  max 
principal  stress  is  r»ot  to  exwed  0  tons.  a^s   ^  tons     * 

23.  At  a  point  in  a  plane  section  of  a  strained  mass  there  is  a  shear  stress 
q  and  a  normal  stress  p  equal  tj,  one  half  of  the  max.  principal  stress  at  the 
same  point.    Show  that  q=pV-2. 

24.  A  round  shaft  is  in  torsion,  and  the  shear  stress  pro<luced  across  the 
section  lu^r  the  .■ircumforeiu-e  is  8000  lbs.  ,K>r  squar,-  inch.     At  the  same  sec- 
tion the  shaft  IS  subjected  to  U-nding,  and  a  compressive  stres.s  of  6000  lbs 
fHir  square   inch   is   produced.     What   is  the  greatest   compressive  stress  in 
the  matenal  there?  ^««.  11,544  lbs. 

25.  A  timber  cantilever  AB,  100  cm.  long  and  20  cm.  in  diameter,  has 
one  end  .1  built  into  a  wall.  At  B  a  fixed  horizontal  arm  00  cm.  long  is  loaded 
at  the  end  with  a  weight  of  lOOOfc.     Find  the  max.  an.l  min.  tensile  and  shear 

^""TV  ,u,      ,  ^"■^-  n2.4(t.).  -30(c.)and74.2Jfc/em,. 

26.  The  crank-shaft  of  an  engine  is  5  in.  in  diameter;  the  distance  from 
the  centre  of  the  bearing  to  the  point  opposite  the  centre  of  the  crank-pin, 
AB  in  tig.  342,  is  12  ins.;    the  half-stroke  AC  in  figure  is  10  ins.,  and  the 


t : ,' 


EXAMPLES. 


365 


]irpssure  applied  to  the  crank-pin  in  the  direction  of  BC  is  5000  lbs.  Find 
tlu-  greatest  intensity  of  thrust;  tension  and  shearing  stress;  and  the  angle  0 
made  by  the  line  of  principal  stress  with  the  axis  of  the  shaft. 

Am.  6530  lbs.;  815  lbs.;  4080  lbs.;  27°. 
iT.  A  round  bar,  of  2  ins.  diameter,  is  to  be  adjusted  to  a  pull  of  10,000  lbs. 
|)er  square  inch  by  the  application  at  a  turnbuckle,  at  the  middle  of  the  bar, 
of  a  load  of  280  lbs.  with  a  leverage  of  20  ins.  Assuming  that  one- half  of 
the  turning  moment  affects  each  half  of  the  bar,  determine  the  maximum 
und  minimum  tensile  and  shear  stresses. 

Am.  10,192.3  (t.),  192.3  (c),  and  5192.3  Ibs./sq.  in. 
a8.  A  propeller-shaft  8  in.  in  diameter  is  subjected  to  a  thrust  of  100  tons 
uniformly  distributed  oyer  its  two  ends,  and  a  twisting  moment  of  30  foot- 
tons.    Find  the  greatest  intensity  of  thrust  and  shearing  stress. 

Ana.  4,71  and  3.715  tons/sq.  in.;  37"  15'. 

29.  A  model  for  illustrating  shear  consists  of  a  frame  of  four  equal  bars, 
each  16  iu.  in  length,  pivoted  at  the  ends  and  connected  at  diagonally  opposite 
corners  by  springs.  With  one  side  fixed  and  the  opposite  side  subjected  to 
stress  in  the  direction  of  its  length  the  following  observations  were  obtained : 


Extension  of  Ist  Diagonal 

Load  in  Pounds. 

Movement  of  Side. 

and  Compression  of  2d 
in  Inches. 

0 

0 

0 

2 

.44 

.30 

4 

.88 

.61 

6 

1.31 

.89 

8 

1.77 

1.19 

Deduce  the  modulus  (G)  for  this  frame,  and  show  that  the  longitudinal 
strain  is  one  half  the  shear  strain. 


30.  A  wall  12  ft.  high  and  3  ft.  thick  weighs  120  lbs.  per  cubic  foot.  It 
is  built  in  earth  weighing  100  lbs.  per  cubic  foot  and  having  an  angle  of  repose 
of  30°.  Find  the  depth  of  the  foundation  consistent  with  the  equilibrium 
of  the  earth.  An*.  1.846  ft. 

3 1 .  A  wall  is  built  up  in  layers,  the  water  face  being  plumb  and  the  rear  stepped. 
If  /  l)e  the  thickness  of  the  nth  layer  and  y  the  depth  of  water  above  its  lower 
face,  .show  that  width  of  laj'cr  X thickness  of  layer  =  v'-O '  +  6. Wi  +  mly* - 2.4 ; 
-1  l)oing  the  sectional  area  of  the  wall  above  the  layer  in  question,  t  the 
liDrizontal  distance  Iwtween  the  water  face  and  the  hne  of  action  of  the 
resultant  weight  ab<n'e  the  layer,  I  the  layer's  thickness,  and  m  the  ratio  of 
tiic  si)ecific  weights  of  the  water  and  masonry. 

32.  .\ssuming  Hankine's  theory,  find  the  pressure  on  the  vertical  face  of 
:i  rctaining-wall  30  ft.  high  which  retains  earth  sloping  up  from  the  top  at 
tlic  angle  of  repose,  viz.,  30°. 

(Weight  of  masonry  ="128  lbs.  per  cubic  foot;  weight  of  earth-120  lbs. 
IKT  cubic  foot.)  Am.  46,764  lbs. 


366 


THEORY  OF  STRUCTURES. 


COB 


33.  A  wall  8  ft.  high  supports  an  embankment  having  a  slope  of  20°,  the 

weight   of   the  material  IxMn;;  120  Ihs.  per  rubie   foot,  and  the 

"E      *     ^  angle  of  repose  2r>°.     Find  the  average  intensity  of  the  thrust 

-    on  the  wall,  the  total  thrust,  the  horizontal  component  of  thrust, 

and  the  overturning  moment. 

Am.  1<»4.S.S  lbs. ;  1,559.04  lbs. ;  140.-)  lbs.;  .'WOf).:,-)  ft.-lbs. 

34.  \  wall  retaining  wafer  on  the  vertical  face  .IB  overhangs 

\w  3S2       ^^'  '"*"  "'""""*  ^^'-     If  «''  =  i-l«,  and  if  the  deviation  of  tiio 

-      cr-ntre  of  resistance  in  liC  from  the  middle  point  0  is  <  IHC, 

show  that  the    afe    maximum  value  of  I)E  is  .1732.5.1B,  the  s[)ecific  weight 

of  the  wall  being  twice  that  of  the  water. 

35.  -V  vertical  retaining-wall  is  strengthened  by  means  of  vertical  rect- 
angular anchor-plates  having  their  upper  and  lower  edges  18  and  22  ft.  respec- 
tively below  tile  surface.  Find  the  holding  i)ower  per  foot  of  width,  the  earth 
weighing  130  lbs.  per  cubic  foot  and  having  an  angle  of  rejwse  of  30°. 

.1««.  27,733J  lbs. 

36.  .\  vertical  rectangular  anchor-plate  of  depth  h  ft.  has  its  horizontal 
centre  line  ;  ft.  b<>low  the  ground  surface,  which  is  horizontal.  Find  the 
maximum  holding  |X)wer  of  the  plate  i)er  foot  of  width. 

4w2h  sin^i 


Ans.  — 


cos'  (j) 


,  4>  being  the  angle  of  reiwse  of  the  earth  and  w  its  specific 
weight. 

37.  -A.  ditch  is  (')  ft.  deep,  is  cut  in  clay,  and  has  vertical  faces  shored  up 
with  boards.  A  horizontal  strut  is  placed  at  intervals  of  5  ft.,  2  ft.  from  the 
bottom.  Taking  sin  ^■.  =.27n,  find  the  thrust  on  the  strut  and  also  the  greatest 
thrust  which  may  bt?  put  on  the  strut  before  the  earth  without  will  heave  up. 

„    „.  A,,.^.  0120  lbs.;  19,03.ilbs. 

38.  The  weight  of  a  building  is  .5000  tons.  The  area  of  the  concrete  bot- 
tom of  the  foundations  is  2000  square  feet.  At  what  depth  ought  it  to  l)e 
below  the  level  of  the  soil,  if  the  soil  is  such  that  ,,6  =42°,  and -if  the  soil  weighs 
100  lbs./cub.  ft.?   ,  ^„«.  2  ft 

39.  A  wall  of  n-<-tangular  section,  IS  ft.  high  and  8  ft.  thick,  weighs  12.5  lbs. 
per  cubic  foot.  Find  the  maximum  intensify  of  pressure  at  the  base  when 
the  wall  retains  (a)  wafer  level  with  fiie  top,  (h)  earth  level  with  the  top; 
the  angle  of  repose  in  flic  latter  case  being  30°  and  the  earth  weighing  100  lbs. 
per  cubic  f(.of .  j^ns.  (a)  19,200  lbs. ;  (6)  5273  lbs. 

40.  A  brickwork  pier  IS  ins.  square  supports  a  load  of  4  tons;  the  result- 
ant presssure  a.ts  at  a  distance  of  4  ins.  from  the  centre  of  the  pier.  Cal- 
culate the  niaximum  and  mininiuni  stresses  in  the  brickwork. 

Ans.  4.15  tons/sq.  ft.  in  cotnp.;  ..59  toii/sq.  ft.  in  ten.sion. 

41.  The  total  vertical  [wssure  on  a  horizontal  section  of  a  wall  of  n.asonry 
is  1(K)  tons  per  foot  length  of  wall.  Tiu-  thickness  of  the  wall  is  4  ft.,  and  the 
centre  of  stress  iis  ti  ins.  from  fiie  centre  of  thi.kne.ss  of  the  wall.  Determine 
the  intensity  of  stress  at  the  opjwsite  edges  of  the  horizontal  joint. 

Aim.  43.75  and  0.25  ton/sq.  ft. 


EXAMPLES. 


367 


42.  The  slope  of  a  cutting  is  1  in  If  The  earth  weighs  120  lbs.  per  cubic 
foot  and  has  an  angle  of  repose  of  36°.  Find  the  average  intensity  of  pres- 
suii'  on  a  vertical  plane  extending  3  ft.  below  the  ground  surface.  Also  find 
tlic  total  pressure  on  the  vertical  plane  and  the  overturning  moment  at  the 
f,H,i  of  the  plane.  Ans.  93.191  lbs.;  279.o73  lbs.;  223.G33ft.-lbs. 

43.  .\  wall  20  ft.  high  and  6  ft.  thick  retains  earth  on  one  side  level  with 
till'  top,  and  on  the  other  the  earth  rises  up  the  wall  at  its  natural  slope,  viz., 
l.r,  to  the  height  of  5  ft.    Will  the  wall  stand  or  fall? 

^W eight  of  masonry  per  cubic  foot  =130  lbs.;    of  earth  =  120  lbs.) 
Find  the  locus  of  the  centres  of  pressure  of  successive  layers. 
.4  ns.  Overturning     moment  =  4128  .  ft.-lbs.;      moment   of  stability  - 
<);5()00</  +  750(V  -dq)  =36912J  ft.-lbs.  if  7  =i.    The  wall  is  stable. 

44.  A  wall  12  ft.  high,  2  ft.  wide  at  the  top  and  3  ft.  wide  at  the  bottom, 
i-j  (•(instructed  of  ma.sonry  weighing  120  lbs.  jx;r  cubic  foot.  The  overturn- 
iiif;  force  on  the  rear  face  of  the  wall,  which  is  i)lumb,  is  a  horizontal  force  P 
acting  at  4  ft.  from  the  base.  Find  P  so  that  the  deviation  of  the  centre  of 
])r(ssnro  in  the  base  may  not  exceed  J  ft.  The  centre  of  pressure  being  fixed 
at  J  ill.  from  the  middle  of  the  base,  show  that  |  of  the  section  may  be  removed 
without  altering  its  stability,  and  find  the  increase  in  the  inchnation  of  the 
nsult.int   pressure  on  the  base  to  the  vertical  consequent  on  the  removal. 

Ans.  3G0  lbs.;  tangents  of  angles  are  in  ratio  of  o  to  3. 

45.  A  reservoir  wall  is  4  ft.  wide  at  top,  has  a  front  batter  of  1  in  12, 
a  rear  batter,  of  2  in  18,  and  is  constructed  of  masonry  weighing  125  lbs.  per 
culiic  foot;  the  maximum  compression  is  not  to  exceed  10,526i''j  lbs.  per 
s(|iian'  f(jot.     Find  the  Hmiting  height  of  the  wall.  Ans.  16  ft. 

46.  A  masonry  pier  12  ft.  high  and  9'X9'  in  section  weighs  125  lbs.  per 
iiihic  foot  and  carries  a  platform  hinged  above  its  centre.  The  end  of  the 
platform  3  ft.  from  the  hinge  is  secured  to  the  pier  by  a  bolt  extending  to 
the  bottom  of  the  pier.  At  the  other  end  of  the  platform,  33 J  ft.  from  the 
liiiisic,  a  weight  11'  is  placed.  Find  W  so  that  the  deviation  of  the  centre  of 
resistance  at  the  base  of  the  pier  from  the  middle  point  of  the  base  may  not 
<.'wvd  31  ft.  Ans.  20,2.50  lbs. 

47.  A  wall  4  ft.  wide  at  the  top,  with  a  front  batter  of  1  in  8  and  a  rear 
hatter  of  1  in  12,  is  30  ft.  high.  Will  the  wall  be  stable  or  unstable  (1)  when  it 
retains  water  level  with  the  top;  (2)  when  it  retains  earth?  (Weight  of 
masonry  [Kjr  cubic  foot  =125  lbs.;  of  earth  =  112  lbs.;  angle  of  repose=30°; 
aii(i7  =  i.) 

Ans.  (1)  Moment    of  wcip,ht  =  125,850  ft.-lbs.;    overturning  moment 
=281,250  ft.-Pjs.,  and  the  wall  is  therefore  unstable. 
(2)  Moment  of  weight  =  140,719  ft.-lbs.;    overturning  moment 
=  168,000  ft.-lbs.,  and  wall  is  therefore  unstable. 

48.  A  dock  wall,  plumb  at  the  rear  and  having  a  face  with  a  batter  of  1 
ill  24,  is  20  ft.  high  and  9  ft.  wide  at  the  base.  Counterforts  are  built  at  intrt-- 
vils  of  12  ft.,  projecting  3  ft.  from  the  rear  and  6  ft.  wide.  Determine  the 
tlii(kncss  of  an  equally  strong  wall  without  coimterforts,  with  the  same  face- 
bat  ler  and  also  plumb  in  the  rear. 

!f  the  walls  are  founded  in  earth  weighing  112  lbs.  per  square  foot  and 


368 


THEORY  OF  STRUCTURES. 


having  an  angle  of  repose  of  32°,  find  the  least  depth  of  foundation  in  each 
case,  the  masonry  weighing  125  lbs.  per  cubic  foot. 

.4ns.  10.95  ft.;  2.72  ft.;  2.71ft. 

49.  Determine  the  limiting  depths  of  foundation  for  (a)  a  wall  of  rect- 
angular section  20  ft.  high;  (b)  for  a  wall  of  traixjzoidal  .section  having  plumb 
rear  and  front  faces  4  and  20  ft  high  respectively,  (.\ngle  of  repose  of  earth 
-30°;    weight  of  earth  =  U2  lbs.  jjcr  cubic  foot;    of  ma.sonry  =  140  lbs.) 

Ans.  (a)  3.22  ft.;    (h)  1.93  ft. 

50.  .\  wall  with  one  face  vertical  is  built  up  as  shown  by  the  figure.  If 
w  is  the  s|)ceific  weight  of  the  wall,  n  the  number  of  sections,  and/  the  max- 
imum stress  in  the  material,  show -that 


^e) 


w  n(2a  +  nx) 
"/       a  +  x 


Fig.  383. 


SI.  A  wall  of  rectangular  seciion  20  ft.  high  and  8  ft. 
wide  weighs  125  Ibs./cu.  ft.  and  retains  earth  level  with  the 
top  of  the  wall  and  weighing  100  Ibs./cu.  ft.  The  angle  of 
repose  of  the  earth  is  30°.  Find  the  distance  of  the  centre  of 
resistance  from  the  middle  of  the  base  and  the  maximum 
material  of  the  wall.     How  will  these  results  be  modified  if 


stress  in  the 

water  is  substituted  for  the  earth? 

Ans.  21  ft.,  7500  Ibs./sq.  ft.;  lA  ft.,  5470}?  lbs.  sq.  ft. 

53.  A  wall  has  a  front  and  a  rear  batter  of  1  in  12,  is  24  ft.  high,  and  weighs 
125  Ibs./cu.  ft.  The  water  on  one  face  rises  to  the  top  of  the  wall.  Deter- 
mine the  width  of  the  base  (o)  if  5  -i,  (6)  if  the  stress  in  the  material  is  nowhere 
to  exceed  12,000  Ibs./sq.  ft.  Ana.  14.16  ft.;  12.08  ft. 

53.  A  wall  of  height  h  has  a  horizontal  base  and  a  top  width  of  a  ft.  The 
rear  and  front  faces  have  batters  of  nb  and  b  respectively.     Show  that 

hh(3n-l)+a 
^^  hb(n+i)+a' 


ir 

r 

54.  Find  the  total  pressure  per  foot  of  breadth  against  a  retaining-wall 
12  ft.  deep,  when  loose  earth,  weighing  120  lbs.  per  .  ubic  foot,  presses  against 
it,  the  top  layer  being  horizontal  and  of  the  same  height  as  the  wall.  Also 
calculate  the  overturning  moment.  (^-30°.)     Ans.  2880  lbs.;  11,520  ft  .-lbs. 

55-  The  front  face  of  a  wall  is  plumb,  and  the  rear  face,  which  retains  water 
level  with  the  top  of  the  wall,  has  a  batter.  The  density  of  the  wall  is  twice 
that  of  tlie  water.  If  the  width  of  the  ba.se  is  N  times  the  width  of  the  top, 
find  the  deviation  of  the  centre  of  pressure  in  the  base  from  the  middle  of  the 
base,  and  if  this  deviation  is  i  of  the  thickness  r'  the  ba.se,  show  that  the 
height  ()f  the  wall  is  (iV»  + 1)'  times  the  width  of  the  top,  and  find  the  maximum 


intensity  of  j)ressure  in  the  base. 


Ans.  62Jx 


3iV  +  l 


'A''  + 1  Ibs./sq.  ft. 


56.  Earth  slopes  up  from  the  top  of  the  vertical  faces  of  a  wall  at  an  angle 


i:XAMPLES. 


369 


of  '.'0°.  The  earth  weighs  130  lbs.  per  cubic  foot,  and  has  an  an.  >.  repose 
of  .JO".  Find  the  average  intensity  of  pressure  and  the  total  prtosJ. »  on  the 
wall.     Also  fi.id  the  overturning  moment. 

.Ins.  26.<)3x  lbs.;  'M.QSx'  lbs.;  8.434x'  tt.-lbs.,  x  being  height  of  wall. 

57.  A  wall  with  a  plumb  rear  face  is  to  bo  30  ft.  high  and  4  ft.  wide  at  the 
t(>i>:   the  earth  slopes  up  from  the  inner  edge  at  the  angle  of  20°,  30°  being 
the  angle  of  repose.     Assuming  Rankine's  theory,  dctermint  the  proper  width* 
of  the  base,  the  masonry  weighing  144  lbs.  per  cubic  foot  and  the  earth  110 
ll)s.  ^718.  If  7=},  <  =  10.94  ft. 

58.  A  wall  6  ft.  wide  at  the  bottom,  plumb  at  the  rear,  and  with  a  front 
batter  of  1  in  12,  retains  water  level  with  the  top.  Find  (a)  the  limitin;  s'- 
tiiin  of  the  centre  of  pressure  at  the  base  so  that  the  stress  may  be  nowntre 
nepative. 

How  (b)  high  may  the  wall  be  built  when  subjected  to  this  condition?  (A 
ruble  foot  of  masonry  — 125  lbs.) 

Ans.  (a)  12  in.  from  middle  point  of  base;    (6)  height —S.Q  ft. 

59.  A  wall  of  rectangular  section,  weighing  125  lbs.  per  cubic  foot,  retains 
earth  on  one  side  level  with  the  top  of  the  wall.  The  earth  is  laid  in  horizontal 
layers  and  has  an  angle  of  repose  of  25°.  The  height  of  the  wall  is  6  ft.  Find 
the  average  intensity  of  pfe.ssure  on  the  wall,  the  total  pressure,  and  the  over- 
turning moment.  Also  find  the  thickness  of  the  wall  if  the  centre  of  resistance 
deviates  from  the  middle  point  of  the  ba.se  J  of  the  thickness. 

Ins.  158.29  lbs.;  949.7  lbs.;  1899.42  ft.-lbs.;  3.9  ft.;  weight  of  earth- 130 
Ibs./eu.  ft. 

60.  The  section  of  a  retaining-wall  is  a  parallelogram,  the  upper  and  lower 
faiis  being  horizontal  and  one  of  the  diagonals  being  vertical.  The  wall 
retains  water  level  with  the  top.  If  the  thickness  of  the  wall  is  6  feet,  find 
the  height  (1)  so  that  the  stress  may  nowhere  exceed  12,000  lbs.  per  square 
foot;  (2)  so  that  q  =  i.     (Weight  of  masonry  =  125  lbs.  per  cubic  foot.) 

Ans.  (1)  23.02  ft.;  (2)  20.78  ft. 

61.  A  wall  01  rectangular  section  supports  water  at  one  side  level  with 
the  top.  How  much  of  the  wall  can  be  cut  away  by  a  plane  through  the  toe 
if  I  he  distance  of  the  centre  of  pressure  s*  the  base  from  the  toe  is  ^  of  the 
width  of  the  base? 

\\  ith  the  section  of  the  wall  thus  modified,  show  that  its  height  must  be 
dmililo  the  width  of  the  base,  and  also  show  that  the  tangents  of  the  angles 
which  the  resultant  pressures  at  the  base  make  with  the  vertical  are  in  the 
raiio  of  4  to  3.  The  weight  of  the  masonry  per  cubic  foot  is  twice  that  of 
the  water.  Ans.  One  quarter  of  the  wall. 

02.  i^how  how  to  aesign  the  section  of  a  brick  wall,  25  ft.  in  height  and 
w(M^'hinf»  110  lbs.  per  cubic  foot,  which  is  to  retain  earth  weighing  100  lbs. 
per  cubic  foot  level  with  the  top  of  the  wall.      The  angle  of  repose  of  the 

arih  Iwhind  the  wall  =30°;  of  the  earth  below  the  wall  =25°.  Resistance 
'ii  brickwork  to  crushing  =40  tons/sq.  ft.  Factor  of  safety  =8.  Pressure  of 
<  -rtn  below  wall  is  not  to  exceed  IJ  tona/sq.  ft.  The  coefficient  of  friction 
for  the  brickwork  —  .74,  and  factor  of  safety  against  sliding  —  1 .2.    Front  batter 

=^^1  in  12.    The  rear  face  is  vertical. 


370 


THEORY  OF  STRUCTURES. 


63.  The  front  and  rear  faces  of  a  wall  retaining  water  level  with  the  top' 
have  a  batter  o.  1  in  12.  The  height  of  the  wall  is  12  feet.  Find  the  thick- 
ness of  the  wall  (a)  if  q  is  .25;  (6)  if  the  stress  at  the  base  is  nowhere  to 
exceed  12,000  lb:<./sq.  ft.     (Weight  of  masonry  =125  lbs.  per  cubic  foot.) 

Ans.  (a)  5.06  ft.;  (6)  3.99  ft. 

64.  A  vertical  rectangular  retaining-wall  of  height  h  has  a  row  of  rect- 
angular counterforts.  The  width  of  a  counterfort  is  equal  to  the  distance 
between  two  consecutive  counterforts.  Find  the  thickness  of  a  rectangular 
wall  giving  the  same  moment  of  stability.  If  V,  is  the  volume  of  the  counter- 
forted  wall,  Fj  the  volume  of  the  equivalent  unifor.n  wall,  and  Vj  the  differ- 
ence between  the  volumes  of  the  portions  of  the  wall  with  and  without  counter- 
forts, show  that  V,'  -  Fi'  =  Vj'. 

65.  A  BCD  is  the  section  of  a  masonry  wall  retaining  water  level  with 
the  top  and  weighing  125  lbs.  per  cubic  foot.  The  water  face  BC  is  vertical 
and  40  ft.  in  height,  the  base  CD  horizontal  and  twice  the  width  of  the 
horizontal  upper  face  AB,  and  ths  rear  of  the  wall  slopes  uniformly  from 
.4  to  D.  Find  the  width  of  the  base  so  that  the  stress  in  no  part  can  exceed 
10,000  lbs.  per  square  foot.  Ana.  20  ft. 

What  must  the  value  of  9  so  that  the  triangular  portion  of  masonry 
ABD  may  be  removed  without  altering  the  stability  of  the  wall?    Ana.  0. 

66.  The  upper  half  of  the  section  v..  a  masonry  wall  is  a  rectangle  4  ft. 
wide,  and  the  lower  half  a  rectangle  6  ft.  wide,  one  fice  being  plumb.  Fmd 
the  height  of  the  wall  so  that  the  stress  on  the  base  may  nowhere  exceed 
10,000  lbs.  per  square  foot  when  the  wall  retains  water  (a)  on  the  plumb 
face;  (6)  on  the  stepped  face.     (Masonry  weighs  125  lbs.  per  cubic  .oot.) 

Ans.  (a)  13.08  ft.;  (6)  12.2  ft. 

67.  A  masonry  dam  h  ft.  high  is  a  right-angled  triangle  ABC  in  section, 

and  retains  water  on  the  vertical  face  AB.    Show  that  the  thickness  (  of  the 

4h* 

base  BC  is  given  bv  t'=-z— ,  ,  qt  being   the  deviation  of  the  centre  of 

a(oq  +  l) 

pressure  in  the  base  from  the  middle  point. 

Also  show  that  the  thickness  will  be  given  by  <  — ^ 


4h* 


■„,.    ,  ,v  if  the  rock 
3(<>9  +  l) 

upon  which  the  wall  is  built  is  seamy,  and  if  it  is  assumed  that  the  commu- 
nicntion  between  the  water  in  the  seams  and  that  in  the  reservoir  pro<luce3 
an  upward  pressure  upon  the  base  BC,  varying  uniformly  from  that  equiva- 
lent to  the  head  at  B  to  nil  at  ('.  If  7 -J,  show  that,  in  order  that  the  wall 
nmy  slide,  the  rfM-fTieient  of  friction  must  be  less  than  63  per  cent  in  the  first 
and  HI  |>er  cent  in  the  second  ea.se.  (Weight  of  a  cubic  foot  of  masonry - 
2JX\veight  of  cubic  fiMit  of  water.) 

68.  A  wall  .10  ft.  high  is  of  triangular  section  ABC,  the  face  .iB  beitig 
plumb,  and  watei  iH'itig  retained  on  the  side  .It'  level  with  the  top  of  the 
wall;  the  iniwonry  weighs  12.")  lbs.  [x-r  cubic  fmtt.  Hiid  the  thickness  of 
the  i)a.so  liC  in)  wiieri  q  ^l;  [in  wii'Mi  .'t-evs  in  masonry  is  not  to  cxired 
10,0OU  lbs.  per  square  foot;    (r)  when  (/-§  and  the  wall  also  retains  earth 


EXAMPLES. 


371 


oil  tho  side  AB  level  with  the  top,  the  angle  of  repose  being  30°.    The  weight 
(if  tlie  earth  per  cubic  foot  — 144  lbs. 

Ans.  (a)  17.69  ft.;  (6)  19.72  ft.;  (c)  8.52  ft. 

69.  The  faces  of  a  reservoir  wall  4  ft.  wide  at  top  and  40  ft.  high  kive 
the  same  batter,  and  water  rises  on  one  side  to  within  6  ft.  of  the  top.  Find 
the  width  of  base,  assuming  (a)  that  the  pressure  on  the  horizontal  base  i» 
to  t)o  nowhere  negative;  (6)  that  the  pressure  varies  uniformly  and  at  no 
fHiiiit  exceeds  5068  lbs.  per  square  foot.  (Weight  of  masonry -125  lbs.  per 
cubic  foot.)  Ans.  (a)  24.32  ft.;  (6)  30  ft. 

70.  A  wall  of  an  isosceles  triangular  section  with  a  base  36  ft.  wide  has 
to  retain  water  level  with  its  top.  How  high  may  such  a  wall  be  built  con- 
sistent with  the  condition  that  the  stress  in  the  masonry  is  nowhere  to  exceed 
10,">4f)J  lbs.  per  square  foot?    (Weight  of  masonry  per  cubic  foot  =  125  lbs.) 

Ans.  54  ft. 

71.  The  faces  AB,  AC  of  a  wall  are  parabolas  of  equal  parameters  having 
their  vertices  at  B  and  C;  water  rises  on  one  side  to  the  top  of  the  wall. 
Determine  the  thickness  of  the  horizontal  base  BC,  (a)  for  a  wall  50  ft.  high; 
(/))  for  a  wall  100  ft.  high,  so  that  the  pressure  on  the  base  may  at  no  point 
exceed  10,000  lbs.  per  square  foot.  Also  (c)  compare  the  volume  of  such 
wall  with  the  volume  of  an  equally  strong  wall  of  the  same  height,  but  with 
a  siction  in  the  form  of  an  isosceles  triangle  with  its  vertex  at  ^1.  (Weight 
of  masonry  — 125  lbs.  per  cubic  foot.) 

Ans.  (o)  32.44  ft.;  (6)  90.35  ft.;  (c\  In  case  (a)  ratio-7:\/ll4; 

"  (6)  "  -\/136:21. 
^2.  The  section  of  a  wall  48f  ft.  high  and  weighing  125  Ibs./cu.  ft.  is  a 
paniliiila  with  the  vertex  at  the  highest  point.  The  wall  retains  water  on 
one  side  level  with  the  top,  and  the  stress  is  nowhere  to  exceed  10,000  Ibs./sq. 
ft.  Find  the  width  of  the  base,  which  is  horizontal.  Find  the  width  of  a 
n  ill  of  rectangular  section  and  of  the  same  height  which  might  be  substituted 
fur  the  above,  and  compare  the  volumes  of  the  two  walls. 

Ans.  3U  ft.;    39.04  ft.;  .53. 

73.  How  will  the  results  in  the  preceding  example  be  modified  if  the  walls 
:irc  r..,  h  (it  ft.  in  height?  Ans.  54.58  ft.;  92.1  ft.;   .4. 

74.  rhc  section  A  BCD  of  a  retaining-wall  for  a  reservoir  has  a  vertical 
fail  //('  and  a  parabolic  water-face  AD,  with  the  vertex  at  D.  The  width  of 
till  liise  />C'-4xwidth  of  the  top  AB.  If  AB-6  ft.,  find  the  height  of  ths 
wall,  .ind  trace  the  curves  of  resistance  (o)  when  the  reservoir  Ar 
i^  lull ;  (/()  when  empty.  (Cubic  foot  of  masonry  -2  X cubic 
foot  iif  water.) 

.1'  .  ;{!'  ft.  if  f/-J,  and  then  max.  comp.  -8000  llw.  per 


01 
ih 

of   7.1 

ft.  f.. 


11. 

7S.   riie  figure  represents  the  section  of  the  upjier  jjortion 

.1  inasimry  dam  which   has   to    retain  water  level  with 

'|||'  <if  tlie  dam.     The  face  AC  is  plumb  for  a  depth 

7!  ft.     The  width  of  the  section  is  constant  and  -22 J 


B 


X 


I  depth  AB-40  ft. 


Fio.  384. 


I  "111   tho  maximum   stress  in    the   masonry  at  the  horizontal  bed  BF. 


372 


THEORY  OF  STRUCTURES. 


With  the  same  maximum  stress  show  that  the  width  of  the  horizontal  bed 
CO,  FG  being  straight,  is  7.').3.Sfl  ft.     (Masonry  weighs  130  lbs.  [wr  cubic  foot.) 

Ai's.   14,0491,, 'is  lbs.  pcrs(i   ft. 

76.  When  a  cyliiidrical  bar  is  twisted  show  that  it  is  sul)jecled  to  shears 
along  transveisi^  and  radial  longitu.jinal  sections,  or  to  tensions  and  com- 
pressions on  helices  at  4.j'  to  the  axis. 

77.  A  cylindrical  envelope  is  filled  with  a  plastic  substr.nce  of  which  the 
F  is  very  small,  and  is  subjected  in  the  dn.-ction  of  the  axis  of  the  cylinder  to 
a  |)-"ssurc  of  2(K)AVcni.-.     Find   the  pressure  on   the  sides  of  the  envelope. 


l^xplain  the  result  when  a  ='2, 


i         200  ,  , 

Ans.  zk/cm. 

o-l 


78.  .\  bar  is  «*  vtched  under  a  force  of  intensity  p.     If  the  bar  is  prevented 
from  contracting,  find  the  lateral  stress;  also  find  the  extension. 

Ans.  —'—■  i 

a-l'    E   n(„-\)' 
70.  For  mild  -steel,   A'-I.j,000  and  G=GOOO  t(ms/sq.   in.     Find   K  and 
roi.s.son's  ratio.  i„,,    io,ooO;  4. 

80.  For  wrought  iron,  /v- 10,000  and  G  ^-ylnO  tons/sq.  in.     Find  E  and 
I'oisson's  ratio.  ,l„.,.  1 4,500  tons/sq.  in. ;    Vr". 

81.  For  copper,  K  =  1 2,(KK)  and  E  =  7800  tons/sq.  in.     Find  (!  and  Poisson  's 
'"''"•  -ln.t.  2S02  tonsAsq.  in.;    V;'- 

82.  Taking  the  value  of  the  cx-fTicient  of  ela.sticity  (E)  and  the  coeffic'ient 
■  if  rigidity  {G)  U,  lx>  l."),(KK)  and  TuTM  tons  for  ste<>l,  i3,9.')0  and  .'J4.J0  tons  for 
wrought  iron,  and  't.-jOO  and  37:)0  tons  for  cast  iron,  find  the  coefficient  of 
elasticity  of  volume  (A'),  and  al.so  the  values  of  .1  and  the  coefficient  of  dilata 
tion  (x),  assuming  the  metals  to  Iw  isotrojMc. 


Ans.  Steel 3f 


Wrought  iron. 


.1,11 


Cast  iron \\\ 


K 
12777} 
105r,!)j| 

f.7.s:)| 


A 
WG 


V.r, 


83.  .V  copp<'r  culx>  3"  '3"y;j"i.s  subjected  to  a  livdrostatic  pressure  of 
4.4.S  tons/sq.  in.;  hnd  the  volume  of  the  .strained  solid,  A'  Ix-ing  12,(H)0 
•""'''''"I-  in-  .1  „...  2().09  cubic  inches. 

84.  Ten  cubic  inches  of  wrought  iron  nni\  10  ciil)!,.  inches  of  water  arc 
subjected  to  lluid  pressure  of  :;  tons  |,;r  s(|ii;ir(<  inch;  lind  ih,.  n<'W  volumes. 
If  the  iron  is  spherical,  what  an>  the  old  and  new  diameters? 

85.  .\  body  is  distorted  withoiil  .•ompression  or  expansion;  (ind  the  work 
•I'w-  .1".v.   j'j  l.V,'  +  .Y,=  -t.V,,»4.2(7','  +  7V  +  7V)l'/N. 

86.  Find  the  work  recpiired  to  tuist  a  hollow  .yiin.ler  of  external  radius 
A'l,  internal  radius  li,,  and  length  /  through  an  a.igl(<  .». 

Ans.  .«^(««-W). 
4/ 


ill 


EXAMPLES. 


373 


87.  The  tension  on  a  soft-steel  test-piece  is  lOOOfc/cm.'.  If  £-22X10* 
and  "  =3§,  find  G  and  show  that  the  maximum  distortion  is  2'  2". 

Am.  846,154Jfc/cm.'. 

88.  Show  that  a  simple  elongation  is  equivalent  to  a  cubical  dilatation  and 
a  p.'iir  of  shearing  or  distorting  stresses. 

8q.  I'"in(l  the  resultant  shearing  stress  at  any  point  in  the  surface  of  the 
transverse  section  of  an  elliptic  cylinder.     (.Art.  lU,  Case  h.) 

a    h'c' 

.\its.  20  — — J,  p  Ix^ing  the  perpendicular  from  the  centre  upon  the  tan- 
put  to  the  ellipse  at  the  given  point,  and  26,  2r  the  major  and  minor  axes. 

go.  A  cylinder  uticlcrgcx-:,  torsion  round  its  axis.  Show  that  the  curveit 
I'l  I II  triution  are  concentric  circles. 

Qi.  Find  the  least  .sectional  area  of  a  rope  which  will  transmit  5(K)  11.1*. 
:ii  KH)  f/s,  the  max.  working  stress  being  10,000  Ibs./sq.  in. 

Ans.  .55  ins. 

q2.  Find  the  min.  diam.  of  a  shaft  whicii  will  transmit  500  H.P.  at  ;; 
siirf:ice  velocity  of  5    f/s,  the  safe  working  stress  being  10,000    Ibs./sf].  in. 

Ans.  3.75  in. 

03-  A  granite  cube  of  6  cm.  side  is  subjected  to  a  compression  of  24  tons. 
FituI  the  maximum  tangential  intensity  of  .-hear  and  the  maximum  distor- 
tiiiii,  /•;  Ix'ing  3()0,000A/cm.'  and  » «4.  .4ns.  333ik/cm.  ;  ^  =  9'  33". 

Q4.  A  closed  cylindrical  tube  is  subecte<l  to  an  internal  pleasure  which 
ilivi  lopes  ill  the  surface  of  the  pipe  a  stress  of  400ifc/cm.'  parallel  to  the  axis 
ami  a  hoop  stress  of  8(X)AVcm.-.     Determine  the  radiul  stress,  a  Ixiiig  3§. 

AriB.  680A-/cm.'. 

OS.  Show  that  the  hoop  (/)  or  radial  (q)  stresses  at  any  radius  r  in  a  thick 
liolluu  cylinder  or  sphere  are  connected  by  the  relation 


III 


O'l. 


ii7. 


ii.-. 


■'\ 


if 


''  d 

j-{qr)  -/  for  the  cylinder,     and     ^  (qr)  '-2jr  for  the  sphere. 

A  tulH'  of  wTought  iron,  of  2  ins.  internal  and  4  ins.  external  radius, 
■.■i(  i|  to  an  iiiieriuil  pressure  <.f  .■.(),0()0  ll>s./s(|.  in.,  the  pres.sun'  on  .lie 

iH'ing  iiU.      Find  the  tensile  stress  in  (he  nmtcrial  at  any  radius  r. 

•J(Mi(Mi7 
r' 

Find  the  max.  and  min.  stress  in  the  w.tlls  of  a  thick  cylinder  of  S  ins. 
I   iiiid    11   in.    external  diam.,   the    inside    fluid    pressiin«    In-itig  L'tHK) 

ill-  An.x.  3»39  and  1939  Ibs.M].  in. 

Till'  r;isl-iron  cylinder  of  an  hydraulic  pnw  has  an  e\i  •rnal  (ii:inielcr 
111  Miternnl,  and  is  subjected  to  an  internal  pressure  of  /)  tons  jter  s(|uarf 
Fitid   the   princiiMi!   s»n>!«eK   ;,?    tbo   i.ii'rr  sml   um-r   tirtumfrmires. 

the  pressure  is  A  toue  |ier  square  inch,  and  if  the  internal  diameter  is 


374 


THEORY  OF  STRUCTURES. 


10  in.,  find  the  work  done  in  stretching  the  cylinder  circumferentiallv,  E  beiryt 
8000  tons. 

Arm.  At  inner  circumference,  q=^p,  a  thrust,  and  / |p,  a  tension. 

At  outer  circumference,  </  =0,  and  /  =  - Ip,  a  tension. 

Wuriv  =126  ft.-lb.s.  jx-r  square  foot  of  surface. 

99.  The  chamlKT  of  a  27-ton  breech-loader  has  an  external  diameter  of 
40  ins.  and  an  internal  diameter  of  14  ins.  Under  a  powder  pressure  of  IS  tons 
jier  square  inch,  find  the  principal  stresses  at  the  outer  and  inner  circum- 
ferences, and  also  the  work  done;   E  being  13,000  tons. 

Ans.  At  inner,  7  =  IS  tons,  compression;  at  outer,  <7=0. 

At  inner.  /=  -23,'s  tons,  tension;  at  outer,  /=  -5,V  tons,  tension. 
Work  =1 J  ft  .-tons  per  sq.  ft.  of  surface. 

100.  What  should  be  the  thickness  of  a  9-in.  cylinder  (a)  '  linh  has  to 
withstand  a  pressure  of  SOOO  lbs.  j)er  square  inch,  the  maximum  allowable  ten- 
silo  stress  l)eing  24,000  lbs.  per  square  inch ;  (/))  which  has  to  withstand  a  pres- 
sure of  (i0(K)  lbs.  jHT  .square  inch;  the  maximum  allowable  tensile  stre.ss  being 
lO.lMX)  lbs.  per  square  inch?  Ans.  (a)   1.86  in.;  (b)  4i  in. 

loi.  A  tul)e  of  wrought  iron,  inside  radius  3  ins.,  outside  4  Ins.,  outside 
I>ressure  0.  What  is  the  insi<Ie  j)ressure  required  to  produce  a  maximum 
tensile  stress  of  l.">,000  lbs.  jx-r  square  inch?  Find  the  fractional  increase 
in  size  of  the  inside  radius.  Am.  4200  Ibs./sq.  in.;  .00161  in. 

102.  A  tul>e  of  wrought  iron,  inside  radius  2  ins.,  outside  3  ins.,  no  pres- 
sure inside;  outside  pressure  =  4200  lbs.  [kt  .square  inch  out.side.  Find  the 
I  ire  ular  compressive  stress  at  any  point,  and  also  the  diminution  of  the  outer 
r.dius.  30_).|0 

Ans.  7560+       ,     ;  .00009  in. 
r 

103.  .\  metal  <ylih.ler  of  ititemal  radius  r  and  external  radius  nr  is  sub- 
jected to  an  internal  jircssure  of  p  tons  \)vr  square  inch.     Show  that  the  total 

work  done  in  stretching  the  cylinder  circumfereiitially  is   *'V-      -  ft.-tons 

A    /( '  - 1 

[)er  s(|uare  foot  of  surface,  E  iK'ing  tiie  metal's  coefficient  of  elasticity. 

104.  The  metal  of  a  cast-iron  hydraulic  [jress  of  .WS  mm.  diameter  is  222  mm. 
thiik,  and  has  to  l)ear  an  internal  pressure  of  402  atmospheres.  Find  thu 
maximum  hoop  stress.  .\ns.  632</cni. 

105.  What  must  In-  the  thickness  of  the  metal  in  a  cast-iron  hydruulic 
press  01'  :m)  nun.  diameter  which  has  to  In-ar  an  internal  pressure  of  2S0  atnos- 
pheres  together  with  a  maximum  tension  of  iV./mm.?  Ans.  18.66cm. 

106  An  hydraulic  press  ha.s  to  rai.se  a  weight  of  1,162,400/,-.  The  iiitcr- 
n;d  |>r(  ssure  is  ."i73  atmospheres  and  the  diameter  of  tlM>  cylinder  is  .'»60  mm. 
The  tension  in  the  metal  is  not  to  exceed  700A/cm'.  Find  the  thickness  of 
•  he  nietiil.  Ann.  .10.85  cm. 

107.  The  nitual  thickness  of  the  metal  in  the  preceding  example  waa 
25.4  cm.  \\  hat  was  the  maximum  tensile  sfres.s  indueeii  in  lifting  the  given 
w^'Rhl?  Atui.SHk/cm. 

108.  The  thickness  of  the  metal  in  a  cast-iron  cylinder  of  108  mm.  diam- 


fill 
h 


EXAMPLES. 


375 


ptor  is  3o  mm.  Find  the  greatest  hoop  stress  in  the  metal  due  to  a  maximum 
internal  pressure  of  2iiOk/cm.  Ans.  486Jfc/cm.". 

log.  In  a  thick  cylinder  the  initial  tensile  stress  on  the  outer  skin  and  the 
initial  compressive  stress  on  the  inner  skin  are  each  3  tons/sq.  in.  Find  the 
resultant  strcss«>s  on  the  outside  and  inside  when  the  cylinder  is  subjected  to 
an  internal  fluid  pressure  of  4J  tons/.sq.  in.  Aim.  4.2  and  4.5  tons/sq.  in. 

no.  A  gun  of  12  ins.  internal  and  24  ins.  external  diameter  is  subjected 
to  a  maximum  internal  pressure  of  40,000  lbs.  per  square  inch.  Find  the 
stress  produced  at  r  =  «,  9,  and  12  ins.     Am.  66,666;  37,037;  26,667  Ibs./sq.  in. 

111.  Pipes  of  a  water-pressure  supply  company  are  to  withstand  a  possible 
pressure  of  1000  lbs.  per  square  inch;  they  are  of  6  in.  internal  diameter. 
Wiiat  is  the  outside  diameter,  the  safe  tensile  stress  of  the  metal  being  3000 
lbs.  i)er  square  inch?  Ans.  8.485  ins. 

112.  Assuming  that  the  annulus  forming  the  section  of  a  cylindrical  boiler 
in  composed  of  a  number  of  infinitely  thin  rings,  show  that  the  tension  at 

the  circumference  of  a  ring  of  radius  r  is  ■  i_^_^  per  unit  of  surface,  and  that 

R         A 

tiio  pressure  is  — I ^^  ,  A  and  B  denoting  arbitrary  constants,  and  m  being 

tlie  cocfficent  of  lateral  contraction.  Find  the  values  of  A  and  B,  p,  and  pi 
I'l  iiig  respectively  the  internal  and  external  pressures. 


113.  Show  that  in  the  case  of  a  spherical  boiler  the  circumferential  stress 
Hiul  pressure  are  respectively  -r— —  and  -j  +  /^_i  wm-3  •     ^'"*^  ^  ^^^  ^• 

114.  An  hydrauUc  cylinder  having  an  internal  diameter  of  6  ins.  and  a 
ihiekiiess  of  3i  ins.  is  subjected  to  an  internal  pressure  of  2  tons  per  square 
nil  h.     Draw  a  curve  of  hoop  stress  for  this  cylinder  to  scale. 

Ans.  217/r'-72(4r'+361). 

115.  A  cast-iron  water-main  is  30  ins.  internal  diameter  knd  I  in.  thick. 
U  hat  is  the  greatest  head  of  water  which  it  can  safely  stand?  If  the  pi|K>8 
are  wrought  iron,  what  ought  its  thickness  to  be  if  the  working  stress  is  10,000 
iiis.  per  square  inch,  and  if  the  longitudinal  seams  are  60  per  cent  of  the  strength 
<it  the  unhurt  plate?  Ans.  461  ft.,  /  being  3000  Ibs./sq.  in.;  .5  in. 

116.  The  3-in.  i)lungrr  for  the  cast-steel  cylinder  of  un  iiitcnsificr  is  con- 
iie.ie.l  with  a  piston  whi<h  works  in  a  4S-in.  cylinder  under  a  pressure  equiv- 
il.  Ill  (o  a  120-ft.  head  of  water.  Find  the  proixT  thiikness  of  the  metal 
(I  tlic  iutcnsifier,  allowing  a  maximum  stress  of  20,000  lbs.  jwr  square  inch. 

Alls.  2.46  in. 

117.  The  barrel  of  a  gun  consists  of  two  rings  A  and  B,  the  bore  of  the  gun 
;.n.l  the  diameters  of  the  inner  and  outer  rings  being  in  the  ratio  of  1:2:4. 
i  lie  ring  .1  is  shrunk  upon  B,  pro<lucing  a  pressure  of  5  tons  per  square  inch 
1 1  tlH<  surface  of  contact.  If  the  firing  of  the  gun  produces  an  internal  pres- 
-iire  of  20  luiut/sq.  in.,  find  the  stresses  induced  in  the  gun. 

Ans.  -19§,  +  1J  tons;    -15,-6  toM. 


CHAPTER  VI. 


I   i  : 
I  ; 


FRICTIOX. 

I.  Sliding  Friction.— Friction  is  the  resistance  to  motion  which 
is  always  (l('V(>lope(l  when  two  substances,  whether  solid,  liquid, 
or  gaseous,  are  pressed  together  and  are  compelled  to  move  the 
one  o\(T  the  other.  If  /'  is  the  mutual  pressure,  and  if  F  is  the 
force  which  must  act  tangent iaily  at  the  point  of  contact  to  pro- 
duce motion,  the  ratio  of  F  to  P  is  called  the  coefTicient  of  friction 
and  may  be  denoted  by  /.  The  value  of  /  does  not  depend  upon 
the  nature  of  any  ^inrjh  substance,  but  uix)n  the  nature  and  con- 
dition of  the  surfaces  of  contact  of  a  pair  of  substances.  It  is  not 
the  .same.  e.g..  for  iron  upon  iron  as  for  iron  upon  bronze  or  upon 
wood;  neitlKT  is  it  the  .'*ame  when  the  surfaces  are  dry  as  when 
lubricated. 

The  laws  of  friction  as  enunciated  by  Coulomb  are: 

(1)  That  /  is  ind(>pen(lcnt  of  tli(>  velocity  of  rubbing;  (2)  that 
/  is  independent  of  the  extent  of  surfaces  in  contact;  (.3)  that  /  de- 
I^ends  only  on  the  nature  of  the  surfaces  in  contact. 

Th(>  friction  b.'twe(.n  two  surfaces  at  rest  is  greater  than  when 
they  ar<>  in  motion,  but  a  slight  vibration  is  often  .sufficient  to  change 
the  friction  of  rest  to  tiiat  of  motion. 

.Morin's  ela))orate  friction  exi)eriments  completely  verified  the.se 

jx>r 


laws  within  certain  limits  of  pressure  (from  J   lb.  to  128  lb.« 


S(iuare  indi)  and  velocity  (the  maximum  velocitv  1 
second),  and  under  the  conditions  in  which  th(>v 


)eing  10  ft.  per 


\  U' 


ew  of  his  mot 


were  made. 


tab!'. 


•e  important  results  are  given  in  the  foil 


lowmg 


ipparatus    employed    in    carrying   out    these    exiMTimc 


con,Mt"d  of  a  box  which  could  be  loaded  at   |)1( 


wa-:   nuide    to   slide   along  a   1 


asure,  and  w 


iits 
Inch 


lorizontai   Iwd  by  means  of  a  cord 


:57(i 


SLIDIXG  FRICTIOS— INCUSED  PLANE. 


377 


Material. 


WcHid  on  u"(K)d. 
Mii.il  i>ii  wood. 


M(  i;d  on  metal. 


M(i:il  and  wood  on  1 
cicli  other  or  oack  \ 
(ill  it.self I 


State  of  Surfaces. 


Dry 

Dry 

Wet 

Dry 

Wet ; ; ; 

Slightly  oily 

()ccii.sionaIly  lubricated  as  usual. 
Constantly  lubricated 


Coefficient  of  Friction. 


.25  to  .5 
.2    "   .6 
.22  "   .26 
.15  "   .2 

.3 

.15 
.07  to  .08 

.05 


j.assiiijr  over  a  pulley  and  carrying  a  weight  at  the  end.  The 
(•(iiituct-surfaces  of  the  bed  and  box  were  formed  of  the  materials 
to  lie  experimented  upon.  The  pull  was  measured  and  recorded 
liy  ;i  spring  dynamometer. 

.More  recent  experiments,  however,  have  shown  that  Coulomb's 
laws  cannot  be  regarded  as  universally  applicable,  but  that  /  de- 
jMiids  upon  the  velocity,  the  pressure,  and  the  temperature.  At 
\irv  low  velocities  Morin's  results  have  been  verified  (Fleeming 
.I'likiii).  At  high  velocities  /  rapidly  diminishes  as  the  velocity 
iiiinascs.  Franke,  having  carefully  examined  the  results  of  various 
srrics  of  experiments,  especially  those  of  Poiree,  Bochet,  and  Gal  ton, 
has  suggested  the  formula 


/=/( 


,  -av 
0         > 


(•  licing  the  velocity  and  /o,  a,  coefficients  depending  upon  the  nature 
arnl  condition  of  the  I'ubbing  surfaces. 

For  example, 

/(I  -.20  and  a  =  .04  for  cast  iron  on  steel  with  dry  surfaces. 

/ii-.lM)  and  a  =  .02  for  wrought  iron  on  wrought  iron  with  dry 
surfaces. 

1    .24  and  «  =  .028r)  for  wrought  iron  on  wrought  iron  with 
-liu'litly  (lamp  surfaces. 

I'all  lias  shown  that  at  very  low  pressures  /  increases  as  the 
|'!i<-^uri'  diminishes,  while  Ronnie's  experiments  indicate  that  at 
V'  i\  iiish  pressures  /  rapidly  increases  with  the  pressure,  and  this 
1-  1"  iliaps  partly  due  to  a  depression,  or  to  an  abrasion  of  the  rubbing 

-'irl':ices. 

2.  Inclined  Plane. — I.et  a  body  of  weight  Y*  slide  uniformly  up 
'■'    inclined    plane   under   a  force   Q  inclined   at   an    angle  ^    to 

t'i"  plane. 


378 


THEORY  OF  STRUCTURES. 


Lot  F  be  the  friction  resisting  the  motion,  R  the  pressure  on 
the  plane,  and  a  the  plane's  inclination. 

R         «  The  two  equations  of  equilibrium  are 


F  =  Q  cos/3-Psina 
R=  —Q  sin  /?+P  cos  a. 


rru      t  P      Ocos^-Psma  ^    r  r  •  x- 

1  nereiore  -5  =  — jz—. — ^  ,  „ =  coefficient  of  f nction  =/. 

R     —Qsmjj  +  Pcosa  ' 

Let  the  resultant  of  F  and  ft  make  an  angle  <j>  with  the  normal 
to  the  plane.    Then 


tan  ^="5  = 


F      Qcos^-P  sin  a 


■Q  sin  li  +  P  cos  a' 


or    -p 


Q     sin(a  +  ^) 
cos(/3-^)' 


(f>  is  called  the  anj?^  of  friction.  It  has  also  been  called  the  angle 
of  repose,  since  a  body  will  remain  at  rest  on  an  inclined  plane  so 
long  as  its  inclination  does  not  exceed  the  angle  of  friction. 

If  a=0=^,  then -p  =  tan  ^=/. 

The  work  done  in  traversing  a  distance  x=Q  cos  fix.  If  Q 
is  variable,  the  work  done  =  iQ  cos  3dx. 

Jo 

3.  Wedge.— The  wedge,  or  key,  is  often  employed  to  connect 
members  of  a  structure,  and  in  generally  driven  into  position  by  the 
blow  of  a  hammer.  It  is  also  employed  to 
force  out  moisture  from  materials  by  induc- 
ing a  pressure  thereon. 

The  iigure  represents  a  wedge  descend- 
ing vertically   under  a  continuous  pressure 
P,  tims  producing  a  lateral  motion  in  the    j 
horizontal  member  C,  which  nmst  therefore 
exert  a  pressure  Q  upon  the  vertical  face  AB. 

The  member  //  is  fixed,  and  it  is    as- 
sumed  that   the  motion  of  the  machine  is 
uniform,  so    that    the    wedge    and  C    are    in  a  state  of  relative 
equilibrium. 

Let  Ri,  R2  be  the  reactions  at  the  faces  DE,  DF,  respectively. 


WEDGE. 


379 


thoir  directions  making  an  angle  ^,  equal  to  the  angle  of  friction 
with  tlie  normals  to  the  corresponding  faces. 

]jot  a  be  the  angle  between  DE  and  the  vertical,  a'  the  angle 
l)ot\veen  DF  and  the  vertical. 

Consider  the  wedge,  and  neglect  its  weight,  which  is  usually 
inappreciable  as  compared  with  P. 

Resolving  vertically. 


Ri  cos(90° -a  +  <j>)+R2  cos(90° -a'  +  <t>) 

=P=Risin(a  +  <^)+fl2sin(a'  +  ^).     (1) 
Resolving  horizontally, 


Ri  sin  {9(f -a +  <f>)-R2  sin  (90°-a'  +  ^)=0, 
or      ffi  cos  (a  +  <^)=fl2C0S  (a'  +  ^) 


(2) 


Consider  the  member  C,  and  neglect  its  weight. 
Resolving  horizontally, 


fti  cos  (a  +  ^)=Q  =  fl2  cos  (a' +  <^) (3) 

Assuming  the  wedge  isosceles,  as  is  usually  the  case,   a=a', 
ami  therefore, 

hyeii.  (2),    Ri=R2,    andbyeq.  (1),    2Rism{a  +  <f>)=P.     (4) 
Hence,  by  eqs.  (3)  and  (4), 


Q     cot  (a  +  (ji)     external  resistance  overcome 
~  effort  exerted 


(5) 


P  2 

f.V.B.— This  ratio  of  resistance  to  effort  is  termed  the  mechan- 
ical advantage,  or  purcha.se,  of  a  machine.) 

Suppose  the  motion  of  the  machine  reversed,  so  that  Q  becomes 
the  elTort  and  P  the  resistance. 

The  reactions  Ri,  R2  now  fall  helow  the  normals,  and  the  equa- 
tions of  relative  equilibrium  are  the  same  as  the  above,  with  -^ 
Mil)stituted  for  4>-    Thus 


|^icot(«-.^). 


ifi) 


;i;n 

111 


3^0  THEORY  OF  STRUCTURES. 

The  two  cases  may  be  inclucled  in  the  expression 


Q 


■p  =  \  cot  (n±(l>). 


(7) 


For  a  given  value  of  P,  Q  increases  with  «. 

If  there  were  no  friction,  <^  would  be  zero,  and  eq.  (7)  would 
become 

Q  _  cot  or 

Thus  the  efTect  of   friction  may  be  allowed  for   by   assuming 
the  wedge  frictionless,  but  with  an  angle  increased  by  2^  in  the 
first  case,  and  diminished  by  lV  •»  the  xecond  case. 
Again,  when  P  is  the  effort  and  Q  the  resistance,  eq.  (5)  shows 

that  if  «  +  9i>90°,   the  ratio  ^  is  negative,  which   is  impossible, 

while  if  a  +  4>  =  90°,  j,  is  zero,  and  in  order  to  overcome  Q,  how- 
ever small  it  might  be,  P  woukl  require  to  be  infinitely  great.     Hence 

o  +  ^  must  be  <  90°, 

and  below  this  limit  j^  diminishes  as  ({>  increases. 

Similarly,  it  may  be  shown  from  eq.  (7)  that  when  Q  is  the  effort  . 
antl  P  the  resistance, 

^must  hv  <rt, 

and  that  below  this  limit  ^  increases  with  ^. 

Elfiriennj.—  Wmng  the  uniform  motion  of  the  machine  let 
any  point  a  desren.l  \ertically  to  the  point  h.  The  corresponding 
horizontal  displacement  is  evidentlv  'Ibc. 

Tlie  motive  work=Pa,.; 
"  useful  work    =(?-26c. 


SCREWS. 


381 


Q ■ 2br    Q 
Honce  the  efficiency  =  -p-'",-  =  p  2  tan  a  =  tan  a  cot  (a +  ({>),  by 

0(1.  (n). 

This  is  a  maximum  for  a  given  value  of  <f>  when 

aiK  I  the  max.  efficiency  =  tan  (45°  -  -^)  cot  ( 45°  +  -^  ) 

l-tan-:r\      ,      .     , 
2  \      1-sm  (f> 

,     ^      ^  /  ~l+sin  <j> 
^1  +  tan^/  ^ 

For  the  reverse  motion  the  efficiency 
Pab 


'Q-2bc 


=  cot  a  tan  (a  —  <f>). 


Tins  is  a  maximum  when  a  =  45°+-:5  •     Thus  the 


max.  efficiency  =  cot  (45  )  tan  (45° — -^A 


—\     1— sin  <!> 
1+sin  4>' 


4.  Screws. — A  screw  is  usually  designed  to  produce  a  linear 
motion  or  to  overcome  a  resistance  in  the  direction  of  its  length. 
It  i-;  set  in  motion  by  means  of  a  couple  acting  in  a  plane  perpen- 
iliiular  to  its  axis.  A  reaction  is  produced  between  the  screw  and 
nut  wliich  must  necessarily  be  equivalent  to  the  couple  and  resist- 
ance, the  motion  heinq  steady. 

Take  the  case  of  a  sgj/are-threailed  *  screw.  It  may  be  assumed 
tliit  the  reaction  is  concentrated  along  a  helical  line  whose  diam- 
(  N  r,  r/,  is  a  moan  between  the  external  and  internal  diameters  of 
ill'  thread,  and  that  its  distribution  along  this  line  is  uniform.     It 

*  Siiuarp-thrpadwl  screws  work  more  accurately  than  thase  with  a  V  thread 
!■  I'  llit>  ('irk'ioncv  of  the  latter  has  been  shown  to  be  very  little  less  than  that  of 
ill'  f.irnipr  (Poncelet).  On  the  other  hand,  the  V  thread  is  the  stronger,  much 
i  "  iiiflal  Ix'iiiK  removed  in  cutting  it  than  is  the  case  with  a  square  thread.  Again 
^wtli  a  \'  thread  there  is  a  tendency  to  burst  the  nut,  which  does  not  obtain  in  a 
•rtw  with  a  square  thread. 


.i !  r.i'J 
ye,    «,  ^ 


....'if 


382 


THEORY  OF  STRUCTURES. 


^ 


will  also  be  supposed  that  the  axes  of  the  couple  and  screw  are 
coincident,  so  that  there  \vill  be  no  lateral  pressure  on  the  nut. 
Let  M  be  the  driving  couple; 

Q   "    "  axial  resistance  to  be  overcome; 
r    "    "  reaction  at  any  point  a  of  the 
helical  line,  and 
"   "  angle  between  its  direction  and 
the   normal  at  a;   ^   is  the 
angle  of  friction. 
Let  a  '*    "    angle  between  the  tangent  at  a 
and  the  horizontal;   a  is  called  the  pitch-angle. 
Since  the  reaction  between  the  screw  and 
nut  must  be  equivalent  to  M  and  Q,  then 
<?= algebraic  sum  of  vertical  components  of  the  reactions  at  all 
points  of  the  line  of  contact. 


Fig.   387. 


=  I[r  cos  fa  +  <f))]  =  cos  (a  +  <f>)I{r), 


(1) 


and  .1/  =  algebraic  sum  of  the  moments  with  respect  to  the  axis  of  the 
horizontal  components  of  the  reactions  at  all  points  of  the  line  of 
contact, 


=  2|^rsin(a  +  ^)-J=-  sin  («  +  ^)2'(r). 


(2) 


Lot  the  couple  consist  of  two  equal  and  opposite  forces  P,  acting 
at  the  ends  of  a  lever  of  length  p,  so  that  M  =  Pp. 
Then,  by  eqs.  (1;  and  (2), 

Q      Q     2 


and  the  mechanical  advantage 


Q    2p 


(3) 


Q    2p 
If  ^=0,  -p^^  cot  n  and  the  effect  of  friction  may  be  allowed  for 

by  iussuniing  the  screw  frictionless,  but  with  a  pitch-angle  equal  to 


\ 


SCREWS. 


383 


Again,  let  the  figure  represent  one  complete  turn  of  the  thread 

developed  in  the  plane  of  the  paper.     CD  o 

is  the  corresponding  length  of  the  thread; 

l)E  the  circumference  i:d;  CE,  parallel  to 

tlie  axis,  the  pitch  h:  and  CDE  the  pitch- 

,  Fio.  388. 

angle  a. 

The  motive  work  in  one  revolution        =Af2T=Pp27r. 
The  useful  work  done  in  one  revolution =QA. 


Qh       2p       ,      js    h 
Hence  the  efficiency  =  p  ^^  ^"d"^°^  ^"■^*^^^2Jr 

—j cot  (a 4-^)  =tan  a  cot  (a  +  <f>). 


nd 


(4) 


<i> 


This  is  a  maximum  when  a  =45"  -^,  its  value  then  being 

1— sin^ 
1+sin^' 

In  practice,  however,  a  is  generally  much  smaller,  efficiency 
being  sacificed  to  secure  a  large  mechanical  advantage,  which,  accord- 
ing to  eq.  (3),  increases  as  a  diminishes. 

If  „  +  ,^  =90°,  ^=0,  so  that  to  overcome  Q,  however  small  it  may 
be.  would  require  an  infinite  effort  P. 


Therefore 


a  +  ^<90°. 


Suppose  the  pitch-angle  sufficiently  coarse  to  allow  of  the  screw 
being  reversed.  Q  now  becomes  the  effort  and  P  the  resistance.  The 
(iiicetion  of  r  falls  on  the  other  side  of  the  normal,  and  the  relation 
between  P  and  Q  is  the  same  as  above, -^  being  substituted  for  <(>. 

Thus 

Q     2p 

p  =  ^cot(a-<^), 

ami  therefore  the  mechanical  advantage 


:i;.f-. 


384 


THEORY  OF  STRUCTURES. 


I' 

If  «  =  ^,  n"^'^'  '"^"^  ^*'  overcome  P,  however  small  it  may  be,  Q 


would  require  to  he  infinite. 


Therefore 


rt  ><^. 


If  'i<V>,  reversal  of  motion  is  impossible,  and  the  screw  then 
possesses  the  property,  so  important  in  j)ractice,  of  serving  to  fasten 
seeurely  togi^ther  ditilereiit  structu'".!  parts,  or  of  looking  machines. 

Again,  it  may  be  necessary  to  take  into  accoiuit  the  friction 
between  the  nut  and  its  seat,  as  well  as  the  friction  at  the  end  of  the 
screw.  The  corresponding  moments  of  friction  with  respect  to  the 
axis  are  (Art.  8) 


Ahh^-tli" 


.Q 


hd^-d^  ""'•  ^¥'' 


f  bemg  the  cw^fhcient  of  friction,  di,  d-z  the  external  and  internal 

diameters  of  the  .seat,  and  d'  tlie  diameter  of  the  end  of  the  .screw. 

5.  Endless   Screws  (Fig.  ;},s<)j.— A  screw  is  often  made  to  work 

i  witii  a  toothed  wheel,  tus,  for  example,  in  raising 

sluice-gates,  when  the  .screw  is  also  made  sutii- 

cieiitly  hue  to  prevent,  by  friction  alone,  the 

gates  from  falling  back  under  their  own  weight. 

The  theory  is  very  similar  to  the  preceding.     Let 

(he  .screw  drive.     A  tooth  ri.s<'s  on  the  tliread, 

and  tile  wiieel  turns  against  a  tangential  resi.«t- 

ane(>  (),  wiiicli  is  approximately  parallel  to  llie 

axis  of  th(>  screw. 

Let  I""ig.  .'{(K)  represent  one  complete  turn  of 
the  thread  develoiK>d  in  the  plane  of  (lie  \Ky\m,  a  being  the  pitcii- 
angle  as  Ix'fore. 

Consider  a  tooth.  It  is  acted  upon  by  Q 
in  a  direction  i)aral!el  to  the  axis,  and  by  the 
reaction  R  iM-tween  the  tiireail  and  tooth, 
making  an  iiiigle  </.  (the  angle  of  friction) 
with  til'-  normal  to  the  thread  CI). 

Therefore  Q     R  cos  (n +r>>).  Fio,  390. 

Again,  the  horizontal  compotu'iit  of  It  viz.,  ff  sjn  Cn+d^^   ha-^  a 


Flo.  3S«. 


BOILING  FRICTION. 


385 


iiioinent  R  sin  (a +  <!>)—  with  respect  to  the  "axis  of  the  screw,  and  this 

must  be  equivalent  to  the  moment  of  the  driving  couple,  viz.,  Pp 
(Art.  4).    Therefore 

Pp  =  R-^sm{a  +  ^). 

Thiif  the  relation  between  P  and  Q  is  the  same  as  in  the  preceding 
article. 

Similarly,  if  the  wheel  acts  as  the  driver, 


P     d 


Q 


6.  Rolling  Friction. — The  friction  between  a  rolling  body  and  the 
surtacc  over  which  it  rolls  is  called  rolling  friction.  Prof.  Osborne 
Kcyiiolds  has  given  the  true  explanation  of  the  resistance  to  rolling 
II  till'  case  of  elastic  bodies.  The  roller  produces  a  deformation  of  the 
smtaccH  in  contact,  so  that  the  distance  rolled  over  is  greater  than 
till  actual  distance  between  the  terminal  points.  This  he  verified 
li\  I  Npcriment,  and  concluded  that  the  resistance  to  rolling  was  due 
In  !  he  sliding  of  one  .surface  over  the  other,  and  that  it  would  raturally 
iiiiicasc  or  diinini.sh  with  the  deformation.  In  proof  of  this  he  found, 
for  example,  that  the  resistance  to  an  iron  roller  on  India-iublH>r  is 
/."  t  iriies  ius  great  as  the  resistance  when  the  roller  is  on  an  iron  surface. 
Ilnicc  the  harder  and  smoother  the  surfaces,  the  less  is  the  rolling 
liiHioii.  The  resistance  is  not  sensibly  affected  by  the  use  of  lubri- 
(Miits.  as  the  advantage  of  a  smaller  coeHicient  of  friction  is  largely 
<  iiinlcn-cted  by  the  increa.sed  tenden'-y  to  .slip.  Other  experiments 
in    Mt  required  to  show  how  far  the  resistance  is  modified  by  the 

<|iri'(|. 

<  !<'ii('ral!y,  as  in  the  ca.se  of  ordinary  roadways,  the  resistance  is 
(lnrll\  governed  by  the  artiount  of  the  deformation  of  the  surface 
iiii'l  I'v  tlie  extent  to  which  its  material  is  crashed.  I^-t  a  roller  of 
"■  L'tii  11'  (Fijf,  ;{ni)  i„.  ,„,  tho  |K)iiit  of  motion  utider  the  action  of  a 
1  iii/'iilal  pull  U. 

Ilic  resultant  reaction  Ix'tween  the  surfaces  in  contact  must 
I'll--  through  the  point  of  intersection  of  R  and  IF.  I^t  it  aLso  cut 
till'  siirf.nci'  in  the  {Hiint  R. 


•Hp!i 


3S6 


THEORY  OF  STRUCTURES. 


Let  d  be  the  horizontal  distance  between  B  and  W; 
p  "    "   vertical  "  "       B   "    R. 

Taking  moments  about  B, 


R* 


or 


Rp  =  Wd. 

R  —  the  resistance  =  If  - . 
P 


Fio.  391. 


ma^^  Coulomb  and  Morin  inferred,  as 
the  result  of  a  series  of  experiments, 
that  d  is  independent  of  the  load  upon 
the  roller  as  well  as  of  its  diameter,*  but  is  dependent  upon  the 
nature  of  the  surfaces  in  contact. 

7.  Journal  Friction.  —  Experiments  indicate  that  /  is  not  the 
same  for  curved  as  for  plane  surfaces,  and  in  the  ordinary  cases 
of  journals  turning  in  well-lubricated  bearings  the  value  of  /  is 
probably  governed  by  a  combination  of  the  laws  of  fluid  friction 
and  of  the  sliding  friction  of  solids. 


•  Pupiiit'a  experiments  led  him  to  the  conclusion  that  d  is  proportional  to  the 
square  root  of  the  diameter,  but  this  requires  further  verification. 
Let  /I  be  the  coefficient  of  slidinK  friction. 
The  resistance  of  the  roller  to  sliding  is  /iW,  and  "rolling"  will  be  insured  if 

R<li\V,  i.e.,  if  — <tan  <f>,  which  is  generally  the  case  so  long  as  the  direction  of  R 

does  not  full  below  the  centre  of  the  i„ller. 

Afsunie  that  K  is  applied  at  the  centre.  The  radius  r  may  be  substituted  for  p, 
since  d  is  very  small,  and  hence 

R-W± 

An  pr|iiation  of  the  same  form  applin.s  to  a  wheel  rolling  on  .  hard  roadway 
over  (ilLstiirlcx  cf  smull  hritf.  I,  and  iilw>  when  mlling  on  soft  ground.  In  the  latter 
cu-^c  ihr  ri'.«istiilK'i'  i.s  pr<iportii<nal  (o  the  pnKlurl  of  the  weight  upon  the  whwl 
into  tlir  iltpth  of  the  rut,  und  the  depth  for  a  small  arc  is  inversely  proportional 
to  the  niiliiis. 

l".\|M'rirmrits  on  the  tract iinml  resistance  to  vehicles  on  ordinary  roads  are  few 
in  iiMrnliiT  and  ineoniplcie,  so  that  it  is  imf><)Msiblc  to  draw  therefrom  any  general 
eiPticlii^ion 

1  roll!  the  rxiMriinents  carried  out  by  Kaston  and  .Viiderson  it  would  appear 
thai  tlir  value  <if  il  iti  inches  varies  from  1.(1  to  2  ('.  fur  wagons  on  soft  gnnind,  and 
that  the  rcsistiincc  is  not  .sensihlv  alTi'cted  bv  the  use  if  springs.  I'pon  ik  lianl 
rojui  in  f.iir  condition  the  resistance  was  found  to  be  from  J  to  }  of  that  im 
soft  grounil,  tin-  average  value  ol  d  being  )  inch,  and  waa  very  seiwibly  diminished 
by  the  use  of  springs. 


JOURNAL  FRICTION. 


387 


The  bearing  part  of   the  journal  is  generally  truly  cylindrical, 
and    is  terminated  by  shoulders  resting 
against  the  ends  of  the  step   in  which 
llio  journal  turns. 

Consider  a  journal  in  a  semicircular 
hearing  with  the  cap  removed.  When 
the  ''ap  is  screwed  on,  the  load  upon 
the  journal  will  be  increasec'  by  an 
amount  approximately  equal  to  the 
tension  of  the  bolts.  Let  P  be  the 
load. 

Assume  that  the  line  of  action  of  the  load  is  vertical  and  that 
it  intersects  the  axis  of  the  shaft.  This  load  is  balanced  by  the 
reaction  at  the  surface  of  contact,  but  much  uncertainty  exists 
as  to  the  manner  in  which  this  reaction  is  distributed.  There 
are  two  extremes,  the  one  corresponding  to  a  normal  pressure  of 
constant  intensity  at  every  point  of  contact,  the  other  to  a  normal 
pressure  of  an  intensity  varying  from  a  maximum  at  the  lowest 

I  mint  A  to  a  minimum  at  the  edge  of  the  bearing  B. 

Let  I  be  the  length  of  the  bearing,  a.id  consider  a  small  ele- 
ment JS  at  any  point  C,  the  radius  OC  ("T)  making  an  angle  0  with 
the  vertical  0-4. 

First.     Ix't  p  be  the  constant  normal  intensity  of  pressure. 

P  =-  lipJS  cos  01)  =  plKDD')  =  2plr. 
Factional  re.si.>i/nnrc  =  Jf/pJ.S/)  =/;>/J(J.S)  ^jplra-^JP^. 

riio  frictional  resistance  probably  api)roximate8  to  this  limit 
when  the  journal  is  new. 

Srronil.  liCt  p-  pn  cos  0,  so  that  the  intensity  is  now  pro|)or- 
ii'itial  to  the  depth  CI)  and  varies  from  a  maximum  po  at  .4  to  nil 

I I  li.     This,  jXThaps,   represents  more  accurately  the  pressure  at 
dili'Tent  iM)ints  when  the  journal  is  worn.     Therefore 

P - l^(pJS cos 01)  =  i(/(„J.S'  eosii  01) 
"'Ipolrfj  co.s2  OdO^iJr'-, 

."id  the  frirtional  rcvKtance^ I(fpJSI)  ^^fpolr^fP-. 


388  THEORY  OF  STRUCTURES. 

Hence  the  frictional  resistance  lies  between  /P:r  and  fP-. 

It  may  be  represented  by  [iP,  ft  being  a  coefficient  of  friction 
to  be  determined  in  each  case  by  experiment. 

The  total  mnmenl  of  frictional  resistance  must  necessarily  be 
equal  and  op{)osite  to  the  moment  M  of  the  couple  twisting  the 
shaft;  i.e., 

M  =  /iPr. 

Thus  the  total  reaction  at  the  surface  of  contact  is  equivalent 
to  a  single  force  P  tangential  to  a  circle  of  radius  ;ir  having  its  centre 
at  0  and  called  the  friction  circle. 

The  .work  absorbed  by  a.xle  friction  per  revolution 

=  M-2z  =  2finPr. 

The  work  absorbed  by  axle  friction  per  minute 

=  2/<-PrA'=-  f^-, 

.^'  being  tiie  number  of  revolutions  and  r  the  velocity  per  minute. 

.  Tiic  work  ab.sorbe(l  by  frictional  resistance  produces  an  equiv- 
ali'iit  amount  of  heat,  which  should  bo  dissipated  at  once  in  order 
t  )  i>reveiit  the  journal  from  becoming  too  hot.  This  may  be  done 
liy  giving  the  journal  sulli'-ient  hcarin<f  tiitrf<u-e  (an  area  ecjual  to 
till'  product  of  the  diameter  and  the  l(>ngth  of  the  bearing),  and  by 
the  employment  of  a  suitable  unguent. 

Sup|)os('  that  //  units  of  heat  per  s(iuare  inch  of  In-aring  .surface 
{hi)  are  dissipated  |ht  minute. 

Let  /  inches  Iw  the  length  and  d  inches  the  diam<>ter  of  the  journal. 
Tlieii  //f//  -heat-units  dis.sipated  =heat-unit.s  equivalent  to  frictional 
resistance 

jt-P,i\    itPv 
12./        V2J' 

J  b  irig  Joule's  equivalent,  or  778  ft. -lbs., 


ax 


12.///      P.V  ,      Vllh      Pr 

,"•"        /  fi       Id 


JOURNAL  FRICTION. 


389 


p 
IjCI  ^5  =  ?  =  pressure  per  square  inch  of  bearing  surface.    Then 

12Jh 


pv-- 


=  a  constant. 


In  Morin's  experiments  d  varied  from  2  to  4  in.,  P  from  330 
ll)s.  to  2  tons,  and  v  did  not  exceed  30  ft.  per  minute;  so  that  pv 
was  <5000,  and  the  coefficient  of  friction  for  the  given  limits  was 
found  to  be  the  same  for  sliding  friction. 

Much  greater  values  of  pv  occur  in  modern  practice. 

Rankine  gives  p(i'  + 20)  =44,800  as  applicable  to  locomotives. 

Thurston  gives  pr  =  60,000  as  applicable  to  marine  engines  and 

to  stationary  steam-engines. 

P>ictional  wear  prevents  the  diminution  of  I  below  a  certain 
limit  at  which  the  pressure  per  unit  of  bearing  surface  exceeds  a  value 
/(  given  by  the  formula 

P='pld  =  pkcP, 


where 


In  practice  A;  =  i  for  slow-moving  journals  (e.g.,  joint-pins),  and 
varies  from  IJ  to  3  for  journals  in  continuous  motion.  The  best 
practice  makes  the  length  of  the  journal  equal  to  four  diameters 
lie.  A- =  4)  for  mill-shafting. 

.\gain,  if  the  journal  is  considered  a  beam  supported  at  the  ends, 

q  Ix'inp  the  maximum  permissible  stress  per  square  inch,  and  C  a 
I'ocllicieiit  deiHMiding  upon  the  method  of  support  and  upon  the 
iiiaiirier  of  the  loading.    Therefore 

1 

For  a  given  value  of  P,  d  diiniiiishes  a.s  7  increases.  Also,  it  has 
1m(  II  shown  that  the  work  absorlxnl  by  friction  is  directly  proportional 


390 


THEORY  OF  STRUCTURES. 


Ilonce.  for  both  roa'oiis.  d  should  bo  a  iniiumum  and  the  shaft 
should  be  made  of  tlie  strongest  and  most  durable  material.  In 
praetice  the  pressure  i>er  sfjuare  ii.eh  of  bearing  surface  may  be  taken 
at  about  2  tons  per  s(juare  inch  for  cast  iron,  3J  tons  jX'r  square  inch 
lor  wrought  iron,  and  (]\  tons  per  square  inch  for  cast  steel. 

It  would  apiM'ar,  however,  from  the  recent  exixTiments  of  Tower 
and  otliers.  that  the  nature  of  the  material  mu/ht  k'come  of  minor 
inij)ortance,  wliile  that  of  a  suitable  lubricant  would  be  of  paramount 
importance.  They  show  that  ihe  friction  of  properly  lubricated 
journals  follows  the  laws  of  fluid  friction  nmch  more  closely  than 
tiiose  of  solid  friction,  and  that  the  lubrication  might  lie  made  so 
perfect  as  to  prevent  any  absolute  contact  between  the  journal  and 
its  bearing.  The  journal  would  therefore  float  in  the  lubricant,  .so 
that  there  would  be  no  metallic  friction.  The  loss  of  power  due  to 
frictional  resistance,  as  well  as  the  consequent  wear  and  tear,  would 
be  very  considerably  diminished,  while  the  load  uixjn  the  journal 
might  be  increased  to  almost  any  extent. 

Again,  Tower's  exi)eriments  iuilicate  that  the  friction  diminishes  as 
tlie  tcmiwrature  rises,  a  result  which  had  already  Ix^en  exiK-rimentally 
determined  by  Hirn.  It  was  also  inferred  by  Him  that  if  the  tem- 
perature were  kept_  uniform,  the  friction  would  be  approximately 
proportional  to  v^r,  and  Thurston  has  enunciated  the  law  that, 
wiUi  a  cool  bearing,  the  friction  is  approximately  proportional  to 
i'v  for  all  sjx^eds  exceeding  1(X)  ft.  jxt  minute. 

With  a  speed  of  I'rO  ft.  [x-r  minute  and  with  pre.ssures  varying 
from  KM)  to  750  ll>s.  jxt  .stiuan-  inch.  Thurston  found  exix-rimentally 
that  /  varied  inversely  as  the  sijuare  root  of  the  inlensity  of  the 
pressure.  The  sam(>  law.  but  without  any  limitations  as  to  s|x>e(l  or 
pressure,  had  Ix^en  previously  stated  by  Hirn. 

8.  Pivots.— Pivots  are  usually  cylindrical,  with  the  circular  edgo 
of  tlie  base  re?noved  and  .sometimes  with  the  whole  of  the  base  rounded. 
{\)nieal  pivots  are  employed  in  special  ma(  hines  in  which,  e.g.,  it  is 
important  to  keep  the  axis  of  the  .shaft  in  an  invariable  po.sition. 
Spherical  pivots  are  often  used  for  shafts  subject  to  sutlden  shocks 
or  to  a  lateral  movement. 

(n)  Cijlituiriral  I'irnls.~U  the  shafts  are  to  be  run  .slowly,  the 
iiitfiisity  of  pressure  ip)  on  the  .step  should  not  be  .so  great  as  to 
squeeze  out  the  lubricant.     Ueuleaux  gives  the  following  rules: 


PIVOTS. 


391 


Tlic  maximum  value  of  p  in  pounds  per  square  inch  should  be 
7!)0  for  wrought  iron  on  gun-metal,  470  for  cast  iron  on  gun-metal, 
and  1400  for  wrought  iron  on  lignum-vitJE. 

For  rapidly  moving  shafts 

n  Ix'ing  the  number  of  revolutions  per  minute,  c  a  coefficient  to  be 
ilclcrininod  by  experiment  (  =  .0045),  and  P 
the  load  upon  the  pivot. 

Sup{)ose  the  surface  of  the  step  to  be 
divided  into  rings,  and  let  one  of  these  rings 
1)('  bounded  b\  the  radii  x,  x  +  dx. 

Ill  one  revolution  the  work  absorbed  by 
the  friction  of  this  ring 

=  /ip-2?rJ-dx-2rj-. 
Hence  the  total  work  absorbe  '  in  one  revolution 


Fia.  393. 


fipn'^ 


Jx'^-di^ 


mid  dx,  di  are  the  external  and  internal  diameters  of  the  surface  in 

COlltilCt. 

If  the  ichole  of  the  surface  is  in  contact,  ^2=0,  and  the  work 

Again,  the  moment  of  friction  for  the  ring 

=  ftp  ■2nxdx-x  =  2u7:px^dx, 

and  tlie  total  moment 

A  ,,2        dx^-di' 

"   L  '^!^^P^'dx  =  ^!ir.p—^ 


392 


THEORY  OF  STRUCTURES. 


uP 
If  ^2  =  0,  the  moment  =  ~d\. 

o 

Thus,  in  both  cases,  the  work  absorbed  by  friction  =  2;:  times  the 
moment  of  friction. 

Let  D  be  the  mean  diameter  of  the  surface  in  contact  = -——?. 

Ix't  2y  be  the  width  of  the  surface  in  contact =di—d2. 


Then 


work  absorbed  =  fi7:P  (d  +  ~j-^ . 


Sometimes  shafts  have  to  run  at  high  speeds  and  "to  bear  heavy- 
pressures,  as,  e.g.,  in  screw-propellers  and  turbines.  In  order  that 
there  may  be  bS  little  vibration  as  possible,  p  must  be  as  small  as 
practicable,  and  this  is  to  some  extent  insured  by  using  a  collar- 
journal. 

Let  N  be  the  number  of  collars,  and  hi  di,  d2  be  the  external 
and   nternal  diameters  of  a  collar. 

Then  work  absorbed  by  friction  per  revolutio.    per  collar 

=  ~g-  (di^-dj^)  =  iftn  -y   .  2_fi  2  =  2t X moment  of  friction. 
According  to  Reuleaux,  the  mean  diameter  of  a  collar 


=Z)  =  . 


.V2' 


n  being  the  number  of  revolutions  per  minute. 

Also,  the  width  of  surface  in  contact -rfi -d2  =  .48\/D]  and  the 
maxinmm  allowable  pressure  per  square  inch 


p^ 


46940 
n 


I  i' "  ti' 


(h)  Wear. — The  wear  at  any  point  of  the  elementary  ring  must 
no^'pssarily  l>o  i)roportional  to  the  friction  up,  and  ahn  to  the  amount 
of  rubbing  surface  wliich  passes  over  the  point  in  a  unit  of  time, 
i.e.,  to  the  velocity  ojx.  w  being  tlio  angular  velocity  of  the  shaft. 


PIVOTS. 


393 


Hence  the  wear  at  any  point  is  proportional  to  fipwz. 

((■)  Conical   Phots.— As   before,    suppose    the 
surface  of  the  step  to  be  divided  into  a  number  V— — f- 
of   elementary    rings.     Two    cases    will    be    dis-     ^?-^^*4-— 
cussod : 

First.    Assume  that  the  normal  intensity  of 
prcissure  p  at  the  surface  of  contact  is  constant. 

Let  X,  x+dx  be  the  distances  of  D  and  E, 
r('si)ectively,  from  the  axis. 

The  total  moment  of  friction 


-X,-}— 

I 
I 


Fio.  394. 


^  r\j>DE.2nx-x=^  f\^dx 
Jx,  ^'^  sm  aj*% 


2  iijm 

3  sin  a 


(xi^-xz^), 


ii,  J2  being  the  radii  of  the  top  and  bottom  sections  of  the  step. 
Also,  P,  the  total  load  on  the  pivot, 

=  ppDE  sin  a2nx  =  2npJ"xdx 

=i:p{xi^-x^). 

,,  .     .  2   flP   Xi^-X2^ 

I lence  total  moment  of  friction  =  ^  -r—    2_-r  a- 

Second.    Assume  that  the  wear  is  of  such  a  nature  that  every 
]mn*,  e.g.  D,  descends  vertically  through  the  same  distance. 
Thus  the  normal  wear  a  sin  a, 

or    fjqwx  a  sin  a, 

or    •     px  <x  sin  a. 

In  the  present  case  a  is  constant,  and  hence  px  =  a  constants 
Thus  total  moment  of  friction 

=  l\pDE'2rax  =  t^~        zdx 
Jx,  ^'  amaJxt 


sin  a 


sin  a{x^~x^). 


394 


i*. 


Also, 


Hence  total  moment  of  friction 


THEORY  OF  STRUCTURES. 
P  =  I'pfyE  sin  a  ■  2r.x 

=  2r/)j  £'dx  =  2r.px{xi  -  Xz). 
liP 


:(.ri+X2). 


T 


2sin«* 
id)  Schieles   Pivots. — The  object  aimed  at  in  these  pivots  is 
to  give  the  step  such  a  form  that  the  wear  and 
the  pressure  are  the  same  at  all  points. 

Let  0  be  the  angle  made  by  the  tangent  at 
■'^yo       any  point  of  the  step  with  the  axis. 

Let  y  be  the  distance  of  the  point  from  the 
axis.    Then 

jyy  <x  sin  0; 

and  hence,  if  p  is  constant, 
Fig.  395. 

y  oc  sin  0    or    y  cosec  d  =  SL  const. 

is  the  equation  of  the  generating  line  of  the  step.  This  line  is  knowTi 
as  the  tradrix,  and  also  as  the  anti-fridion  curve.  If  the  tangent 
at  D  intersects  the  axis  in  T, 

DT=y  cosec  fl  =  a  const. 

The  curve  may  be  traced  l)y  passing  from  one  point  to  another 
and  keeping  the  tangent  DT  of  constant  length. 
The  above  equation  may  be  written 

2/^  =  a  const.  =0, 


or 


dx    ydx     ^^^\dx)' 


which  may  be  easily  integrated,  the  result  being  the  analytical  equa- 
tion to  the  curve,  viz., 

x  =  a  \og,\- -^ — ^ j  +Va2-^  +  aconst. 


BELTS  AND  ROPES. 


395 


>^./ 


Scliiclo  or  anti-friction  pivots  are  suitable  for  high  speeds,  but  have 
not  been  very  generally  adopted. 

Q.  Belts  and  Ropes. — Let  the  figure  represent  a  pulley  movable 
about  a  journal  at  0,  and  let  a  belt  (or  rope),  acted  upon  by  forces 
r,.  T-2  at  the  ends,  embrace  a  portion  ABC  of  the  circumference 
sui)tcii(ling  an  angle  a  at  the  centre. 

In  order  that  there  may  be  motion  in  the  direction  of  the  arrow, 
T\  must  exceed  T2  by  an  amount  sufficient  to  overcome  the  frictional 
n'si^tinirc  along  the  arc  of  contact  and  the  resistance  to  bending  due 
to  till-  stiffness  of  the  belt.  r.j., 

Coiisidor  first  the  frictional  resistance, 
ami  suppose  the  belt  to  be  on  the  point  of 
.v/('/)/)(«;/. 

Any  small  element  BB'  (=ds)  of  the 
belt  is  acted  upon  by  a  pull  T  tangential 
to  the  pulley  at  B,  a  pull  T-dT  tangential 
to  tlic  ])ulley  at  B',  by  a  reaction  equiva- 
htit  to  a  normal  force  Rds  at  the  middle 
point  of  BB',  and  by  a  tangential  force, 
or  frictional  resistance,  ftRds. 

Let  the  angle  COB  =  6,  and  the  angle 

zjo/r  ^do. 

Resolving  normally, 


Fia.  396. 


iT  +  T-dT)sin-^-Rds=0. 

Resolving  tangentially, 

dd 


(1) 


{T-T-dT)  cos -^-{iRds^O, (2; 

ft  iuiii!!  the  coefficient  of  friction. 

„     .     dO  .  .      ^  ,    dd  dd   . 

Now  (10  being  very  small,  sin  -^  is  approximately  y>  cos  -^  i.s 

ajii.ruxiinately  unity,  and  small  quantities  of  the  second  order  may 
1m'  ilisrofrarded. 

Ibiice  eqs.  (1)  and  (2)  may  be  written 


ami 


Tdd-Rds=0, 
dT-iiRds='0. 


(3) 
(4) 


396  THEORY  OF  STRUCTURES. 

dT 
Therefore  dT^uTdd,    or    -~r  =  luUd. 

Integrating,  log.T  =  n0+C, 

C  being  a  constant  of  integration. 

Wlientf=0,     T^Tj,      and  hence    log,7'2=C. 


(5) 


Tlierefore 


T 
log,  yT  =/*<?, 


or 


»"» 


When  d  =  a,     T  =  Ti,    and  hence 


Ti 


(6) 


(7) 


^   ij 


e  being  the  number  2.71828,  i.e.,  the  base  of  the  Naperian  system  of 
logarithms.    The  angle  a  may  be  called  the  angle  of  lap. 

If  IX  is  increiused  by  ;9,  the  new  raiio  of  tensions  will  be  e"^  times 
the  oKl  ratio;  so  that  if  a  increases  in  arithmetical  progression,  >he 
ratio  of  tensions  will  increase  in  geometrical  progression.  This 
rapid  incroa.se  in  the  ratio  of  the  tensions,  corresponding  to  a  com- 
paratively small  increase  in  the  arc  of  contact,  is  utilized  in  "  brakes  " 

for  the  purpose  of  absorbing  sur- 
plus energy.     For  example: 

A  flexible  brake  consisting  of 
an  iron  or  steel  strap,  or,  again, 
of  a  chain,  or  of  a  series  of  iron 
bars  faced  with  wood  and  jointed 
together,  embraces  about  three 
fourths  of  the  circumference  of 
an  iron  or  wooden  drum.  One 
end  of  the  brake  is  secured  to  a 
fixed  iwint  0,  and  the  other  to  the 
ond  n  of  a  l<>vfT  AOB  turning  about  a  fulcrum  at  0.  A  force 
applied  at  .1  will  cause  the  brake  to  clasp  the  drum  and  so  produce 
friction  which  will  gradually  bring  the  drum  to  rest. 


Fio.  397. 


BELTS  AND  ROPES.  397 

Let  w  be  the  angular  velocity  of  the  drum  beforp  the  brake  is 

apjilit't-l.    , 

I,ct  /  be  the  moment  of  inertia  of  the  drum  with  respect  to  its 

axis. 

'Iho  kinetic  energy  of  the  (lrum=-;^. 

\\licn  the  brake  is  applied,  the  motion  being  in  the  direction  of 
the  iirrow,  let  the  greater  and  lesser  tensions  at  its  ends  be  Ti  and  Ta 
nviM'ctivcly. 

Let  n  be  the  number  of  revolutions  in  which  the  drum  is  brought 
to  rest.    Then 

^Ia/^=^{Ti-T2)r:dn (8) 

d  liciiiK  the  diameter  of  the  drum. 

Also,  if  P  is  the  force  applied  at  A,  and  if  p  and  q  are  the  per- 
pciiflicular  distances  of  0  from  the  directions  of  P  and  T2  respectively, 

Pp  =  T2q (9) 

A-:ain,  Ti  =  T2e'",        (10) 

(V  li(  iiig  the  angle  subtended  at  the  centre  by  the  arc  of  contact. 
Ilruce,  by  eqs.  (8),  (9),  (10), 

n  =  ..,.   ,  7-— r (11) 

1 1'  the  motion  of  the  drum  were  in  the  opposite  direction,  q  would 
i"  I  he  pcrpondicuiui  distance  of  0  from  the  direction  of  Ti,  and 
then  I'll  =Tiq. 

I'locceding  as  before, 

,  qluPe"' 

"  ~2Pp{e'"'-l)7:d' 

;iiii|  tliorpfore  the  number  of  turns  in  the  second  ca.se,  before  the 
liiiini  coinos  to  rest,  is  e""  times  the  number  in  the  first,  which  is 
<i';iM<lU('iitly  the  preferable  arrangement. 

"ilu'  coctrieient  of  friction  /<  varies  from  .12  for  greasy  .shop-belts 
I  :i  mill  pulleys  to  .5  for  new  belts  and  hempen  ropes  on  wooden 
iliums.     In  ordinary  practice  an  average  value  of  fx  for  dry  belts 


398 


THEORY  OF  STRUCTURES. 


on  iron  pulloys  is  .2S,  and  for  -.virc  roj)os  .24;  if  tho  bolts  arc  wet 
/!  is  about  .38.  ' 

Formidcc  (6)  mid  (7)  are  also  true  for  non-circuJnr  pulleys. 

10.  Effective  Tension.  -Tlio  pull  available  for  the  transmission 
..f  power  7-,- 7-2  =  . S'.  L.-t  II.P.  In-  the  horse-power  transmitted, 
(the  sjieed  of  tiaiismission  in  feet  [x-r  second,  a  the  sectional  area 
of  the  rope  or  belt,  and  .v  the  stre.ss  jht  S(iuare  inch  in  the  advancing 
portion  of  the  belt. 

Then  if  Ti  and  T2  are  in  pounds, 

IIP      (T'i-T2)r      Sr 

The  working  tensile  stress  jxr  square  inch  usually  adopted  for 
leather  lx>lts  varies  from  2So  lbs.  (Morin)  to  355  lbs.  (Claudel), 
an  average  value  being  3()0  lbs.  In  wire  rop's  sr)()0  lbs.  jjer  square 
inch  may  lie  considered  an  aAerage  working  tension. 

Hem|M>n  roix-s  for  the  transmi.ssion  of  power  generally  vary 
from  4h   to  M  ins.  in  circumference. 

II.  Efifect  of  High  Speed.- When  the  six'ed  of  transmission  is 
great  the  effect  (.f  centrifugal  force  nmst  be  taken  into  account. 

The  centrifugal  force  of  tlu>  eli-ments  ^/.v  =  -— -.  „.  |H,i„g  (},e 

.sjx'cific  weight  of  the  '-It  or  rop.',  and  r  the  radius  of  the  pulley. 
Eq.  (3j  above  now  becomes 

TdO  -  lid.^ -=0 

'/      r 


or 


'I  ' 


and  hence,  Iiy  e(|.  (4;, 
Integrating, 


since  T^T.  when^-0. 


'  fulO. 


log. 


SLIP  OF  BELTS. 
.\lso,  T  =  Ti  when  0  =  a,  and  therefore 


Ti 1-2 

g_ 


wa 


=  6'-. 


or 


Ti=T2e'"' v2(e"'-i). 

g 


the  v.ork  transmitted  per  second 

(Wll     \ 
TaV-YvajfC""-!), 

wliich  is  a  maximmn  and  equal  to  |7'2(e'"'  — l)r  when  v«    ,,_.. 

aiiil  the  two  tensions  are  then  in  the  ratio  of  '2ei"  +  l  to  3. 

Tlic  sjx'ed  for  which  no  work  is  transmitted,  i.e.,  the  limiting 
siticil,  is  given  by 


Tag 


wa 


T21 f'  =  0,    or    v^ 


g 

wa" 


12.  Slip  of  Belts. — A    length  /  of  the   belt   (or  rorvO  becomes 
/(!--,^|  on  the  advatiritui  xide  and  nl+Trj    on  the  ftlark  side 

—    arid    ;»2  = 

n  '        (t 

rinw  the  advancing  pulley  draws  on  a  greater  length  than  is  given 
iH  i((  the  driven  pulley,  and  its  s{)eed  must  therefort>  exceed  that  of 
ill'   latter  by  an  amount  given  by  the  ecjuation 


T 

will  If   /)i=-—    and   ;)2  =  — ,   /?  being  the    coefficient  of  elasticity. 

*  n  '  /I 


reduction  of  speed,  or  slip 
<l)c('d  of  driving-pulley 


'('-^)-'("-f)  „- 


'(•■^^) 


Pi- Pa 


Tlie  slip  or  creep  of  the  Mt  mea-sures  the  loss  of  work.     In 
!!'!:!rv  prnrticp  th»  loss  with  leather  belting  does  not  exceed  2  per 


400 


THEORY  OF  STRUCTURES. 


f-i'iit,  while   with    wire  rops   it  is  so  .small  that  it  may  be  disre- 
pardod. 

13.  Prony's  Dynamometer.— This  ilynanioineter  is  one  of  the 
coinmonest  forms  of  fri-tioii-biake.  The  motor  whose  power  is  to 
1k'  measured  turns  a  wlieel  E  which  revolves  between  the  wood  block 
B  and  a  band  of  wood  blocks  .1.    To  the  lower  block  is  attache    a, 


O^ 


A-j^ 


Fin.  398. 


lexer  of  radius  p  carrying  a  weight  /'  at  the  free  end.  By  means  of 
the  .^icrews  (',  I)  the  blocks  may  be  tightened  around  the  circumference 
until  tlif  unknown  moment  of  fri.-tional  resistance  FR  is  ecjual  to  the 
known  moment  /'/),  A'  being  the  radius  of  the  wheel. 

The  weight  /*,  which  rests  ujton  the  ground  when  the  screws  are 
.«l.'i(  k,  is  now  just  balanced. 

TIu'  work  absorbed  b>  friction  per  minute  =  2?:ffF«  =  2rr/m, 
n  JH'ing  the  number  of  revolutions  [ht  mimile. 

14.  SLffnecs  of  Belts  and  Ropes.— The  belt  on  reaching  the 
pulley  is  bent  to  the  curvature  of  the  jx-riphery,  and  is  straightened 
again  when  it  leaves  the  pulley.  Thus  an  amount  of  work,  in- 
crea-iing  with  the  slilTne.ss  of  the  belt,  must  be  expended  to  overcome 
the  resistance  to  bending.     As  the  result  of  ex|)eriment,  this  resistance 

has  liieii  expre.s.xed  in  the  form  ^       T  Iwing  the  tension  of  the  l)elt, 

n  its  sectiotiiil  ;irea,  /("  the  radius  of  the  pulley,  and  !>  a  ctM'tticient  to 

l)e  determined. 

According  to  Hedtenba<'her.    h    ^.'M  ins.  for  hem|HMi  rofies. 

h     ].(]7 ' 

••  "   Ueuleuux,  h    SA     "     "    leather  k-lts. 


STIFFS  ESS  OF  BELTS  AND  ROPES. 


401 


Lot  the  figure  represent  a  sheave  in  a  pulley-block  tvirning  in  the 
(liiictioii  of  the  arrow  about   a  journal 
of  radius  r. 

Let  Ti  be  the  effort,  ^2  the  resistance. 

Tlie  resistance  due  to  the  stiffness  of 
till'  Ix'it  may  be    allowed    for  by  adding 

■  to  the  force  T2.    The  frictional  resist 

aiicc  at  the  journal  surface  is  P  sin  ^  or 
/■/',  /'  b(>ing  the  resultant  of  Ti    T-^. 

The  motion  being  steady,  and  taking  moments  about  the  centre, 

nTo 


Fio.  399. 


TiR=[T2  +  \~)R+fPr, 


or 


'■■-■^»+i^'+'^''- 


If  Ti  and  T3  are  parallel,  P  =  l\  +  T2,  and  the  last  equation 

lii'ciiiiies 

7'i-T2  +  y^-+      'T1  +  T2). 


l.rt  the  pulley  turn  through  a  small  angle  0. 
The  counler-ejfiriennj  of  the  sheave 

motive  work     TiO     Ti     ,       2lr      a      I 


useful  work      T^f)    T2     ^li    frbR-ff 

III  the  rase  of  an  endless  Iwlt  connecting  a  pair  of   pullej-s  of 
ri'lii  A'l./^j.  the  resistance  due  to  .stiffness  may  be  taken  equal  to 

1  '/'   *  /•  7  '       'X'">S  "1^  niean  tension  (  ==—  .^      ). 

Till'  resistance  due  to  jotirnal  friction  =^/r/' (77- +  „-). 

Till"  useful  resistance  -  T1-T2    S. 
Hence  the  counter-efTiciencv 


MhwM-^'i)- 


402 


THEORY  OF  STRUCTURES. 


In  wire  ropes  the  stress  due  to  bending  may  be  calculate  as 
follows : 

Let  X  be  the  radius  of  a  wire.  The  radius  of  its  :  vis  is  sensibly 
the  same  as  the  radius  R  of  the  t'llley. 

The  outer  layers  of  the  f  will  be  stretched,  and  the  inner 
shortened,  while  the  axis  wi.    loiiuiin  unchanged  in  length.     Hence 

X     change  of  length  of  outer  or  inner  strands  _  unit  stress 
R  ~  length  of  a.\is  E        ' 

and  the  unit  stress  due  to  bending  =  fig. 

15.  Wheel  and  Axle.— Let  the  figure  represent  a  wheel  of  radius 
/( turning  on  an  axle  of  radius  r,  under  the  action  of  the  two  tangential 

forces  P  and  Q,  inclined  to  each  other  at 
an  angle  d. 

The  resultant  R  of  P  and  Q  mu.«t 
equilibrate  the  resultant  reaction  between 
the  wheel  and  axle  at  the  surface  of 
contact. 

I^t  the  directions  of  P  and  Q  meet 
inT. 

If  there  were  no  .'riction,  the  resultant 
reaction  and  the  resultant  R  would  neces- 
sarily pass  through  0  and  T. 

Taking  friction  into  account,  the 
direction  of  R  will  be  inclined  to  TO. 
Ix-t  its  direction  intersect  the  circum- 
ference of  the  axle  in  the  point  .4.  The  angle  Ix'tween  T.\  and  the 
normal  AO  at  .4.  the  motion  being  steady,  is  equal  to  the  angle  of 
friction;   call  it  <f). 

Taking  moments  about  O, 

n) 


Fro.  400. 


Also, 


Pl>~Ql>-Rr>^\n  </.  =  0. 

7?2  =  /'2+(^--.i^2/V^COSfl. 


(2) 


^•t  /-sin  6 -    ,  '-* ,  fi  biinc  the  cof  fficient  of  friction. 

'  ^     \  l+/i^ 


WHEEL  AND  AXLE. 
Eq.  (1)  may  now  be  written 

Pp-Qp-fRr=0.    . 


•    •    «    • 


403 


.    .    (3) 


If  P  and  Q  are  parallel  in  direction, 

<?  =  0    and    R  =  P+Q. 


Let  the  figure  represent  a  wheel  and  axle. 


I/'t  P  be  the  effort  and  Q  the  weight  lifted,  the  directions  of 
P  and  Q  being  parallel. 

"    W  be  the  weight  of  the  "  wheel  and  axle." 

'    Ri  and  /?2  be  the  vertical  reactions  at  the  bearings. 

"    p  be  the  rudius  of  the  wheel. 

■<    q      "  "         "      axle. 

"    r      "  "         "      bearings. 

Take  moments  al)out  the  axis.    Then 


Pp-Qq- RiTnin  <f>-R2r  sin  (ffO.     .    :    j    .    (4) 


I'.tit 


Ri+Ra-W  +  P  +  Q. 


(5) 


Ilriire  Pp-(?g  =  (n'  +  P  +  ^)rsiri  ^'=(W  +  P+Q)fr, 


vv 


P(p-/r)=Q(g+/r)+/n'r. 


(6) 


404  THEORY  OF  STRUCTURES. 

Elpciency. — In  turning  througli  an  angle  d, 
niotivo  work^PjiO, 
useful  work  =QqO; 

tliorefore 


efficiency 


J'pO     Pp' 


xnd  tiie  ratio  yj  is  given  l)v  e(|.  ((5). 

i6.  Toothed  Gearing. — In  tootlied  gearing  the  friction  is  partly 
rolling  and  partly  slidiiig,  l)ut  the  former  will  be  disregarded,  as 
it  is  small  as  compared  with  the  latter. 


ov-c:- 


Fir..  \n2 

Let  the  pitch-circles  of  a  pair  of  teeth  m  contact  at  the  point 
li  touch  at  the  point  .1,  at>d  consider  the  action  hefnre  reachinfj 
tlie  line  of  centres  fh(),.  i.e.,  along  the  (trr  of  npprooch. 

Th(>  line  .1/^  is  normal  to  the  surfaces  in  contact  at  the  point  /?. 

Let  R  l>e  the  n>sultant  reaction  at  li.  Its  direction,  tiie  motion 
iM'ing  steady,  makes  an  angle  <',>,  equal  to  the  angle  of  friction,  with 
AB. 

Let  II  he  the  angle  lM>t\veen  0\()2  and  AH. 

Let  the  motive  force  and  force  of  resistance  l>p  res|x»ctivply 
equivalent  lo  a  force  /*  lanijential  to  ftie  pitch-circle  0\,  and  to 
a   force  Q  tangential   to  the   pitch-circle  Oj. 


TOOTHED  GEARING.  406 

T-ot  ri,  r2  be  the  radii  of  the  two  wheels. 

The   work  absorbed   by   friction   in   turning  through  the  small 
arc  (In 

=  (P-Q)ds (1) 

Consider  the  wheel  Oj,  and  take  moments  about  the  centre. 

*/Vi=/Mrisin(fl-</))+xsin0|,        ....     (2) 

where  AB=x. 

Similarly,  from  the  wheel  O2, 

Qr2  =  ftir2sin(/?-</.)-xsin^! (3) 

Hence 

9.  = ?:h — (4) 

sin  {d -<!>)+  —  sin  <j> 
ami  therefore 

I  — H — jjsm  <p 

sin  {0-tf))  -  — sin  ^ 
Hence  the  work  absorbed  by  friction  in  the  arc  ds 

(— +— )  J  sin  dxh 
sin  iO-tj)) sin  0 

Tn  [jreciselv  the  same  manner  it  can  lie  shown  that,  after  leamng 
tlic  line  of  centres,  i.e.,  in  the  arc  of  recess, 

T 

,.     sin  ("  +  vM  -~  ^"1  </» 

^1  = '> ...  (7) 

sin  ((?  +  0)+— sm  tj) 


406  THEORY  OF  STRUCTURES. 

and  the  work  absorbed  by  friction  in  the  arc  ds 


=Q 


(—  +  — )  J  sin  (fxis 


sin  {0  +  <p) sin  ^ 


(8) 


...  Q 


The  ratio  -7;  and  the  loss  of  work  given  by  eqs.  (4)  and  (6)  are 

rosiKJCtively  greater  than  the  ratio  77  and  the  loxs  of  work  given  by 

eqs.  (7)  and  (8).  and  therefore  it  is  advisable  to  make  the  arc  of 
approach  as  small  Jis  possible. 

Again,  by  e<i.  (4),  motion  will  bo  impossible  if 


sin  {0-(f>)  +— sin  ^  =  0; 


i.e.,  if 


cot  ^  =  cot  ^ 


ri  sin  0' 


and  this  can  only  be  true  if  the  direction  of  R  pa.sses  through  O2. 

Simple  appro.ximate  expressions  for  the  lost  u'ork  and  efficiency 
may  be  obtained  as  follows: 

0  differs  very  little  from  90°,  and  x  is  small  as  compared  with  Vi 
and  dilTers  little  from  tiie  corresponding  arc  .s-  mea.sured  from  A. 

Hence  the  work  ab.sorlx'il  by  friction  in  the  arc  ds 

^Qtan^{}-y-)sds  =  QiA  +  l-)sds, 


and  the  work  lost  in  arc  of  approach  .si 

1      1 


Jo         V-1     rJ  '•  Vn     rs'  2 


(9) 


The  useful  work  done  in  the  same  interval  =Qsi. 
The  counter-eijicietu-y  (reciprocal  of  efficiency) 


'1  1  \    '-2 


Vri     rj/  2 


(10) 


TOOTHED  GEARING. 
Similarly  for  the  arc  of  recess  sj, 


407 


/I  1  \  S'T 

the  lost  work  =Q/i(—+—)-2-»  •    •    •    (H) 


and  the  counter-efficiency  =  1  +  /« 


Vi     r2'  2 


.     (12) 


If  si  =S2  =  pitch  =  p  =  '^—^  =  ^^^^^,  ni,  712  being  the  number  of  teeth 

)li  M2 

ill  the  driver  and  the  follower  respectively,  the  expressions  for  the 
lost  work  given  by  eqs.  (9)  and  (11)  are  identical,  and  those  for 
tlu"  counter-efficiency  given  by  eqs.  (10)  and  (12)  are  also  identical. 
Thus  the  whole  work  lost  during  the  action  of  a  pair  of  teeth 

=  0;z(i+i)p2, (13) 


and  the  counter-efficiency 


■MhDl 


(14) 


I 


'^   \ni     112' 


(15) 


This  last  equation  shows  that  the  efficiency  increases  with  the 
miiiihor  of  teeth. 

If  the  follower  is  an  annular  wheel nmst   be  substituted 

ri     r2 

f,„.  _  (  __  i,i  the  above  equations.      Thus  with  an  annular  wheel 

'1     ra 
the  counter-efficiency  is  diminished  and   the  efficiency,   therefore, 

iiurcasc'd. 

It  has  been  assumed  that  R  and  Q  are  constant,  as  their  variation 
tiuiii  ii  constant  value  is  probably  small.  It  has  also  been  assumed 
that  only  one  pair  of  teeth  are  in  contact.  The  theory,  however, 
!mi!<N  good  when  more  than  one  pair  are  in  contact,  an  effort  and 
resistance  corresponding  to  P  and  Q  being  supposed  to  act  for  each 
pair. 


408 


TIILORY  OF  STRUCTURES. 


17.  Bevel- wheels.— Let   /.I,  IH  rcprc^'iit    the   (lovclopmonts  of 
the   axes    of    tlic    pitch-circles  //i,  II 2  of  a  pair   of   bevel-wheels 

when  the  pitch-cones  are  spread  out 
fiat,  0\,  (>2  being  the  corresponding 
centres. 

The  preceding  formula'  will  apply 
to  bevel-wheels,  the  radii  being  OJ, 
O2I,  and  tiie  j)itch  being  measured  on 
the  circumferences  7.1,  IB. 

18.  Efficiency  of  Mechanisms. — 
Generally  speaking,  the  ratio  of  the 
effort  I*  to  the  resistance  Q  in  a  mechan- 
ism may  be  expressed  as  a  function  of 
the  coefficient  of  friction  ju.  Thus 
P 


Ki- 


•to:<. 


Q 


=  F{n) 


If,  now,  the  mechanism  is  moved  so  tliat  the  points  of  application 

of  /'  and  Q  traverse  small  distances  Ji,  Jy  in  the  directions  of  the 

forces, 

.,       ...  .  QJy        1    Jm 

the  efhciencv  = -r. .-  =Tr— ;  -r . 

•      PJx     I'{,n)Jx 

Table  of  (.'oefficikxts  of  Axlk  Fkktion. 


.  I   Onl 


Itry. 


(Jreiisy 


Wcl. 


nil 


Hell-iiiotal  on  licll-iiietal 

Urass  (in  lirnsx 

Hrii.ss  (111  cast  iron 

Cast  iron  on  lifll-nictal 

Cast  iron  on  lirass 

Cast  iron  on  rust  iron 

('a.-^l  iron  on  liKinini-,  ila' 

l,ictiMtn-vit:r'  on  cmsi  iron j ..  .  . 

I.i)."iii'ii-vi!;i'  on  lii;Miirn-\  it;i'  ..,.!... 

\\r'iu:^lil  iron  on  Ijcll-rnclal '|,2.">1 

\\'rou:;lit  iron  on  rasl  iron 

Uroiijtlii  iron  on  lii;nutn-vita'.   .  .  .;.  1.S7 
i 


IS.-, 


.Hit 

]:\7 


.ISO 


CUtluIl. 

(I<t7 
07!» 
(172 

.(175 
(17.') 
()7r> 
1 

.110 


()75 
.07.-. 
.  12.-. 


Contin- 

Pure 

Lard 

I.illiri- 

Car- 
riage- 

ami 
Plum- 

Grease. 

rution. 

urease. 

bago. 

.019 

.(I.-.4 

.005 



.10 

.(Wl 

.().->» 
.()i»2 

14 

.11 

14 

,17 

.15 

.(J7 

.().-,4 

.09 

11 

.054 

Rut    the  ratio     '    depend.-^  only  ujion  tlic  geonietricid  relations 
b«'twcen   the  dilT(  '('lit   parts  of  the  mcclijinism,  and  will   therei'ore 


Mr 


EXAMPLES. 


409 


rcniaiii  the  sanip  if  it  is  assumed  that  n  is  zero.  In  such  a  case  the 
.liriciicy  woulil  ho  pcrfpct.  or  tho  motive  work  (PJx)  would  be 
iMpial  to  the  useful  work  (QJy),  and  therefore 

1    -iy 
F(0} 


1 


llciii'c  tlie  efficiency 


'F(^y 


EXAMPLES. 

I.  In  a  pair  of  four-shoaved  blocks  it  is  found  that  it  requires  a  force  P"  to 
r.iHi'  a  weight  HP',  and  a  force  of"  to  raise  a  weight  15P'.  Show  that  the 
plural  relation  between  the  force  P  and  the  weight  \7  to  be  raised  is  given  by 


p 


l-illd 

2. 

3. 

V.Tlii 

l!i.'  il 

4. 

li;,.! 

r:ii<.' 


the  cfliciencj- when  raising  the  weights  5/"  and  loP'.  Ans.  f;  f. 

Fintl  the  mechanical  advantage  when  an  inch  bolt  is  screwed  up  by  a 
.  spanner,  the  effective  diameter  of  the  nut  being  If  ins.,  the  diameter  at 
ISC  (if  the  thread  .84  in.,  and  .15  being  the  coefficient  of  friction.    An-i.  80. 

rind  the  turning  moment  necessary  to  raise  a  weight  of  1000  lbs.  by  a 
al  s(juare-threaded  screw  having  a  i)itch  of  6  ins.,  the  mean  diameter  of 
Incad  iK'ing  4  ins.  and  the  coefficient  of  friction  J.         -l/i«.  1930  in.-lbs. 

The  radii  of  the  pullej-s  of  a  differential  pullej'-block  are  0  ins.  and  53  ins. 
the  efficiency  when  a  pull  of  200  lbs.  in  the  hauling-chain  is  required  to 
a  weight  of  1  ton. 


5.  l'"in(l  the  mechanical  efficiency  of  a  screw-jack  in  which  the  load  rotates 
Willi  till'  head  of  the  jack  in  order  to  eliminate  collar  friction.  Threads  per 
ill  h,  :>:  mean  diameter  cf  threads,  IJ  ins.;  coefficient  of  friction,  0.14.  .\lso 
fill. I  ilircffii-icncy  when  the  load  iloes  not  rotate.  Ans.  30',';  17.1'j. 

f).  In  a  Hi  ting-machine,  an  effort  of  "ifi.O  lbs.  just  raises  a  load  of  2260  lbs. ; 
uli.it  is  the  ineciianical  advantage?  If  the  efficiency  is  .755,  what  is  the  velocity 
I':,'  M.? 

I!  nil  the  same  machine  an  effort  of  11.8  lbs.  raises  a  load  of  580  lbs.,  what 
i-  !.M\v  the  efficiency? 

7.  rill'  mean  diameter  of  the  threads  of  a  i-in.  bolt  is  0.45  in.,  the  slope  of 
'1  '  ilniaij  0.07,  and  tiie  cDcfriiient  of  friction  O.lfi.  Find  the  tension  of  the 
1'  It  whiii  pulled  U|)  bv  a  force  of  20  lbs.  on  the  end  of  a  spanner  12  ins.  l.inp. 

Ans.   1<)20  lbs. 

>>.  if  in  a  Weston  pulley-block  only  40  l»r  cent  of  the  energy  expended  is 
■•■  ii/i.l  111  lifting  tile  load,  what  should  Ix-  the  dianu'ter  of  the  smaller  pari  of 
111  >iinp(iuii(l  ))uiley  wlien  the  largest  diameter  is  8  ins.  in  order  that  a  j)ull 
'     'n  ili<.  nil  the  chain  may  raise  a  loud  of  550  lbs.?  Ans.  7.42  ins. 


I 


410 


THEORY  OF  STRUCTURES. 


9.  The  pitch  of  tho  screw  of  a  Hfting-jack  is  i  in.  Disregarding  friction, 
what  force  must  be  applied  at  the  end  of  a  24-in.  handle  to  raise  a  weight  of 
2()()4  lbs.?  Ans.  6 J  lbs. 

10.  The  law  connecting  a  force  P  at  the  handle  of  a  screw-jack  with  the 
weight  W  to  be  overcome  is  of  the  form 

P~bW+c. 

When  ir='J3nO  lbs.,  P=.30  lbs.,  and  when  TF=500  lbs.,  P-10  lbs.;   find  the 
values  of  the  coefficients  b  and  c. 

Also,  if  the  h:imlltMlcs(Til)esacircleof  19in3.,andif  the  pitch  of  the  scrow  is 
I  in.,  find  the  velocity  ratio.  Ans.  6=tii,  <'=4};   318.47. 

11.  In  the  preceding  example,  find  the  efficiency  in  each  case  and  also  find 
the  weight  which  must  be  lifted  so  that  the  efficiency  may  be  25  per  cent. 


I  J.  A  belt  laps  one  half  of  a  l4-in.  pulley  which  makes  2000  revolutions  per 
minute.    Tho  maximum  ten.sion  is  not  to  exceed  40  lbs.  per  inch  of  width. 
/<=.2.S,  find  the  width  of  the  Iwlt  and  the  power  transmitted,  the  weight 


If 

of  the  belt  beitig  l..j  lbs.  per  square  foot. 


13.  A  belt  embracing  one  half  the  circumference  of  a  pulley  transmits  10 
H.P. ;  the  i)ulley  makes  ,30  revolutions  per  minute  and  is  7  ft.  in  diameter. 
NcgkM'ting  slip,  find  T,  and  T,,  /« Ix-ing  .12,5.  Ans.  15415  lbs.;  1041§  lbs. 

14.  How  many  roj^^s  4  ins.  in  circumference  are  required  to  transmit  200 
H.P.  f rom  a  pullev  10  ft.  in  diameter  and  making  90  revolutions  per  minute? 

Ans.  10. 

15.  .\  J-in.  ropo  pusses  over  a  fi-in.  pulley,  the  diameter  of  the  axis  being 
J  in.;  tho  load  upon  the  axis=2Xthe  rojx'  tension.  Find  the  efficiency  of  the 
j)ulley,  tho  coefficient  of  axle  friction  l)eing  .OH  and  the  coefficient  for  stiffness 
.47.     Hence  also  deduce  tho  efficiency  of  a  ]iair  of  three-sheaved  blocks. 

Ans.  .807  ;  .427. 

16.  A  iM'lt  laps  l.")0°  round  a  pulley  of  .'i  ft.  diameter  making  130  revolutions 
per  minute;  tiic  ci)officiorit  of  friction  is  0.35.  What  is  the  maximum  ])ull 
on  the  Im'U  when  20  H.P.  is  Ix'ing  transmitted  and  the  belt  is  just  on  the  jxHnt 
of  .xlippin-i?  Ans.  .898  lb. 

17.  If  tho  pulleys  are  .")0  ft.  centre  to  centre,  and  if  the  tight  is  three  times 
tho  slack  tonsidM,  find  tho  loofith  of  the  belt,  the  coefficient  of  friction  l)eing 
i  and  tho  diainotor  of  one  of  tho  pulleys  12  ins.  Ans.  185.287  ft. 

18.  .'•  (l-in.  leather  Ix-lt  }  in.  thick  and  weighing  0.4  lb.  per  lineal  foot 
connects  two  pulleys,  each  3  ft.  in  diameter,  on  parallel  shafts,  and  is  found  to 
conitnoiico  to  slip  when  tho  moment  of  resistance  is  4(X)  ft. -lbs.  and  the  revo- 
lutions iire  ."lOO  |H'r  minute  Taking  tho  ctx'fficiont  of  friction  between  the 
I)uilovs  and  h-lt  toljc  .24,  estimate  the  greatest  and  least  tensions  when  on  the 
])oiiit  of  slipping. 

19.  .\  strap  is  hung  over  a  fixed  pulley,  and  is  in  contact  over  an  arc  of 
length  o(|ual  to  two  thirds  <if  t4ie  total  circumference.     Under  these  circum- 


EXAMPLES. 


411 


stances  a  pull  of  475  lbs.  is  found  to  be  necessary  in  order  to  raise  a  load  of 
l.")()  ll)s.  Determine  the  coefficient  of  friction  between  the  strap  and  the  pulley- 
riiii.  Ans.  .275. 

JO.  The  tight  tension  on  *  20-in.  bolt  tmbraeing  one  half  the  circumference 
(if  the  pulley  is  1200  lbs.  Find  the  maximum  work  the  belt  will  transmit, 
tile  tiiickiiess  of  the  belt  being  .2  in.  and  its  weight  .0325  lb.  per  cubic  inch. 
(Cotlficieiit  of  friction  =.28.)  Ans.  68.75  H.P. 

21.  \n  endless  belt  weighing  J  lb.  per  lineal  foot  connects  two  35-in.  pul- 
li  v^  anil  traiLsmits  5  H.P.,  each  pulley  making  300  revolutions  per  minute. 
till'  liiiht  and  slack  tensions,  /t  being  .28. 

22.  Find  the  width  of  belt  necessary  to  transiniit  10  H.P.  to  a  pulley  12 
iii«.  ill  iliiuncter,  so  that  the  greatest  tension  may  not  exceed  40  lbs.  per  inch  of 
tlir  wiiith  when  the  pulley  makes  1500  revolutions  ner  minute,  the  weight  of 
the  liilt  iwr  square  foot  being  1.5  lbs.,  the  angle  of  wrapping  180  degrees, and 
takiiij;  the  coefficient  of  friction  as  0.25.  Ans.  8  ins. 

23.  A  roix;  is  run  three  times  round  a  post,  one  end  being  held  tight  by  a 
fun  c  of  10  lbs.    Find  the  pull  on  the  other  end  which  will  produce  slip.    (/i=.25.) 

Ans.  1000  lbs. 

24.  A  rope-pulley  carrying  20  ropes  is  16  ft.  in  diameter  and  tra'^ismits 
(MMi  11.!'.  when  running  at  90  revolutions  per  minute.  Taking  /i  =  .7  and  the 
aniili'  of  contact  =180°,  find  the  tension  on  the  tight  and  slack  sides. 

25.  In  a  rope-drive  the  rope  weighs  0.8  lb.  per  foot.  The  tension  being 
J'«i  11k.  (in  the  tight  side  and  80  lbs.  on  the  slack  side,  find  approximately  the 
sis;  ill  the  two  cases,  supposing  the  shafts  to  be  50  ft.  apart.     Obtain  the  formula 

youciiiploy. 

26.  .\  pulley  3  ft.  6  ins.  in  diameter  and  making  1.50  revolutions  per  minute, 
ihivi*;,  l>v  means  of  a  l)elt,  a  machine  which  absorbs  7  H.P.  What  must  be  the 
\Miltli  (if  Inlt  so  that  its  greatest  tension  shall  be  70  lbs.  per  inch  of  width,  it 
li  lie  assumed  that  the  tension  in  the  driving  side  is  twice  that  on  the  slack 

s'.j. ;' 

27.  The  efficiency  of  a  single-rope  pulley  is  found  to  be  94  per  cent.  Over 
h'.v  m.iny  of  such  pulleys  mast  the  rope  pass  in  order  to  make  it  self-sustaining, 
i  I ..  t.i  have  an  efficiency  of  under  .")0  per  cent?  An.f.  12. 

28.  A  cable  from  a  shij)  is  wound  three  times  round  a  post  and  a  force 
ei|ii.il  to  the  weight  of  100  lbs.  is  applied  at  the  other  end;  how  much  energy 
i-  Ir-irnyeil  when  the  ship  is  brought  to  rest  after  dragging  the  cable  10  ft. 
tilt  r,„.|fl(ieiit  of  friction  being  .2? 

2Q.  Power  is  transmitted  from  a  pulley  5  ft.  in  diameter,  running  at  110 
rcvciliitioiis  ixT  minute,  to  a  pulley  8  ins.  in  diameter.  Thickness  of  belt  =0.24 
in.:  innil'ilus  of  elasticity  of  belt,  9000  lbs.  jier  .square  inch;  tension  on  tight 
siili  [(-r  inch  of  width  =60  lbs.;  ratio  of  tensions,  2.3  to  1.  Find  the  revolu- 
tini  <  per  minute  of  the  small  pulley.  Ans.  792. 


i 


I 


4i: 


THEORY  OF  STkUCTlRtS. 


30.  A  belt  \vt'ighing  J  lb.  ikt  lineal  foot  onriccts  two  42-in.  pulleys,  on 
making  '-'40  revolutions  [kt  minute.  Fiiui  t.V  limiting  tension  for  which  work 
will  Im'  transmitteil.  Also  find  the  tight  and  slack  tensions  and  the  efticiency 
when  the  Ik4i  transmits  .')  H.P.    (Diameter  of  a.xle  =2  in.;  coefficient  of  friclioti 

-  JS.)  Ans.  m\  lbs.;    1(M>.S2  lbs.;   44.32  lbs. 

31.  .\  circular  saw  makes  1000  revolutions  jht  minute  and  is  driven  by  a 
Ix'lt  3  ins.  wide  and  \  in.  thick,  its  weight  ix>r  <'ubic  inch  iK-ing  .0325  lb.  The 
ix'lt  i)asses  over  a  lO-in.  puliey  embracing  one  half  the  circumference  and 
transmits  li  H.I*.  Find  the  tight  and  slack  tensions,  the  coefficient  of  friction 
being  .2S.  -Uw.   130.16  lbs.;  .J4.r>fi  lbs. 

32.  The  most  efficient  sjH'ed  of  a  10-in.X}-in.  belt  w 'ighing  .032.')  lb.  |)pr 
cubic  inch  is  SO  ft.  jht  second,  the  corresponding  tight  and  slack  tensions 
Ix'iiig  in  the  ratio  of  7  to  3.  The  coefficient  of  friction  is  J.  Find  the  angle 
subtended  at  tin  centre  of  the  [)ulley  by  the  arc  of  contact.  Also  find  the 
tight  and  slack  ten.sioiis  and  the  work  transmitted. 

33.  .\  ('otton  rope  \\  ins.  in  diameter  weighs  0.72  lb.  \wt  lineal  foot  and 
may  In-  worked  with  430  lbs.  total  tension.  Find  the  horse-power  transmitted 
at  (K),  HO,  and  UX>  fe<'t  [xt  second  velocity  of  rope,  the  tight  tension  being  4J 
tinu's  the  slack  tension. 


34.  In  .-i  travelling-crane  the  driving-rope  runs  at  5000  ft.  per  minute. 
Find  the  tension  due  to  centrifugal  action,  having  given  that  a  rojie  1  in.  in 
diameter  weighs  0.2H  lb.  (x-r  foot  of  length.  Ans.  60.4  lbs. 

35.  In  an  endless  Ix'lt  |)assing  over  two  pulleys,  the  least  tension  is  1,")0  li)s., 
the  ((H'Hicient  of  friction  2S.  and  the  angle  subtended  by  the  arc  of  contact 
14S°.     Find  the  greatest  tension.     The  diameter  of  the  larger  wheel  is  7S  ins., 

of    till 


laller  10  ins.,  of  the  iM'ariiigs.  3  ins.     Find  the  efficiency.     A  tightci  - 
ng-pulley  is  made  to  press  oti  the  slack  side  of  the  Ix'lt.     .\ssuming  that  the 
to  the  ('(K'fficient  of  elaslicitv  in  the  ratio  of  1  to  80,  find 


working  tension  is 

the  iiicremejil  of  the  arc  of  contact  on  the  i)elt-i)ull<'V,  the  tension  of  tlic  slack 


■<iili'.  and  the  force  of  tlii'  tiiihlening-pulley. 


.1/ 


30!>  lb: 


36.  Two  pullcy-i  3  ft.  11  ins.   in  di.imeter.  running  at    |."»tl  revolutions    i<>r 
miiiiiti'.  iirc  connected  i)y  a  leather  1k-II   weighing  0.6  lii.   [n-r  foot   in  Jeii!;lli. 


Taking  ," 
37    A 


3,  (iiid  the  greatest  ten-ion  in  the  Ix'lt  when  transmitting  7J  11. P. 


.1/ 


.360  lb: 


band   einbr.Mciiig  tliree   foilrths   of   tile    ciniiiiifcieiice  of  a 


bnike-pulley    keyed 


revolving  shaft   has  one   e\tremity  .'ittjiched   to  the 


etid  ,1  of  the  lever  .ic/f,  and  the  other  to  the  ftxnl  |ioinl  <>  (Ix-lween  .1  and  H\ 

ssiire  iK'lween  the  band  and  pulley 


ilto'lt    ullicli   the  lever  o- 


Tlle 


illcctetl  by  a  joiie  applieil  at  right  ;(iigli 


Ihi'  lever  at  the  end  /*.     Show 


th.'it    the  time  ill  uhicli  the  a\le  is  brought   to  re^t    is  about    2J   times  a-"  gK'at 


yh 


■volviiig  III  (iiic  direiiiun  ;ii  in  ;lie  opiwisitt 


1/ 


38.  The  powei  of  an  engine  making  n  revolution.i  |ier  liiinilti  is  tested  liy 
;!  I'limv  brake  having  it-  ariii  of  li^ngth  /  c(inne<ted  with  a  spriii'r  biilimr 
which  register-  a  force  /'.      The  arm  IS  vertical  and  the  weight    IT  of  the  brake 


EXAMl-LES. 


413 


H  -M,|,<,rtcd  by  a  stiff  spring  fixed  vertically  below  the  centre  of  the  wheel. 
Wh.ii  Vrror  in  H.H.r.  would  he  introduced  by  placing  the  spring  x  ft.  away 
[nmi  I  lie  central  [wsitioa?  ■        l^^^  ^ 


Alls. 


Pr 


\  B  being  the  B.II.P. 


30  \  string  of  wood  blocks  embraces  the  24-in.  pulley  of  nn  engine,  one 
,,,„l  „f  tlu^  string  b<-ing  attached  to  a  load  of  112  lbs.  and  the  other  to  a  spring 
1,,1  ,n.  c  which  indicates  12  lbs.  when  the  pulley  is  making  (50  revolutions  |)cr 
,„i„„ir  Find  the  work  given  out  at  the  brake  and  the  coefficient  of  friction 
U,,uru  the  blo..ks  and  pulley.  Ans.  1.148  H.P.;    .3.^.. 

40  III  a  Pionv-brake  test  of  a  Weatinghouse  engine  the  blocks  weie  hxed 
,„  ,  M-in.  (ly-whe«-l  with  a  (>-in.  face,  and  the  balance  reading  was  4S  lbs.; 
ill,.  ,li<tance  from  centre  of  .shaft  to  centre  of  balance,  measured  horizt.ntally, 
w,~  id  ins.,  and  the  numlKT  of  revolutions  jxr  minute  was  024.     Find  the 

Ans.  14.3. 

'  '!li     \ii  (-nginc  makes  l.W  revolutions  i)er  minute.     If  the  diameter  of  the 
hr.l  r-puUe,-  is  4.5  ins.  and  the  pull  on  the  brake  is  50  lbs.,  find  the  HHP._ 

42  \  small  water-motor  is  tested  by  a  tail  dynamometer.  The  pulley  is 
ls,„-  ill  diameter;  the  weight  is  (JO  lbs.;  the  spring  registers  a  pull  of  .V)  lbs.; 
,1„   nuiniur  of  revolutions  iK-r  minute  =.'J00.     Find  the  H.H.P.         Anx.  i. 

43  \  Ueynol.is  water-brake  has  reces.ses  of  9  ins.  internal  and  IS  ins.  ex- 
l,,,n  ,1  aiamcter.  Find  the  velocity  in  feet  lier  second  with  which  the  water 
„,:m  ,  Ir.ulate  so  as  to  absorb  l.-J  H.P.  at  200  revolutions  iicr  minute,  the  axis 
oith.  ,v,vssesl)eingat4.-)°totheplane(tf  thedisk.  Atis.  <.13ft. 

44  \  Irou.le  water-brake  has  recesses  of  ()  ins.  internal  and  18  ins.  external 
,l,„n, ',..-  the  axis  of  the  reees.ses  l)eing  at  4.')°  to  the  piftne  of  the  .hsk.  The 
.|w|  n,  ikcs  ,S()  revolutions  (K-r  minute.  The  resi.stance  to  motion  is  balanced 
Ik  -h  1I,s  at  the  end  of  a  4S-in.  lever.  Find  the  hor8e-iH)wer  develo|)ed 
uhI  ;h.   xdocitv  of  the  water  in  fi>et  (x-r  secon.l  in  the  direction  of  the  recess 

Anx.  3.0.-.  ll.P.;   114  ft. 
'4s    \  horizontal  axle.10  ins.  in  diameter  has  a  vertical  load  m\mi  it  of  20  tons 
an.l  .  hoii/.ontal  pull  of  4  tons.     The  r.H-ffi.ient  of  fricti...,  is  0.02.     Fm.l  the 
h,:,   .v,„.r,Ml.d   iK-r  minute,   and   the   hor.si-iH.wer  wasle.l   in   frution     when 
nn  :  ,  :,()  revolutions  jK-r  minute.  Anx.   l.V,  units;    3.03  H     . 

4'.  Ilic  vertical  pn-ssure  U|H.n  a  steel  pivot  of  100  mm.  diameter  is  -lOOfr 
■  ni  i„  |,ivui  makes  100  revolutions  iK«r  minute.  What  is  the  work  absorU-d 
I,.,     ■  -iini,   iKT   second.   /.  iK.ing  .07?  ^^»x-  -'^-^^  •'"'• 

17    \  shaft  makes  20  revolutions  ikt  minute  in  a  Ix-aring  of  0.2..  m.  (hani- 
,  ,        I,  ,1„.  load  on  the  iM-aring  is  Sm)k;  find  the  work  consume.!  i^-r  second, 
;„    ,,  ,r  •'"•''•  '^'  ^"'• 
"\  t'l-fi    plank  AH.  hinged  at  li,  has  its  middle  iH.int  sup|K.rte<l  on  a 
Ji        MiiMlstonc  and  carries  a  weight  of  100  lbs.  at   the  en.l  A.     The  grind- 
er  .,„hs  m)  lbs.    iiid  makes  1?:.  revolution!.  l>er  minute  on  :.  l-m.  ii\le. 

1  iIh.  r.K-fhcient   ..f  plank  fri.-tion  an<l  rolling  friction  to  W  .:!  and  .().. 

r.  -       u,  Iv,  lind  ill  how  manv  turns  the  grindstone  will  come  t..  rest  when  the 

.        .     ■  Aiix.  2.04  turns. 

Mi.'        pdwer  cetts»'s  to  act.  •  »  •  u_oi 

: .    \  l-in.  axle  makes  400  1,  v.iiuii.....-,  |>er  mirmtt-  on  antj-frufrm  wneels 


■■     !| 


I 


l<      I 


i#h<ii 


414 


THEORY  OF  STRUCTURES. 


30  ins.  in  diameter,  which  are  mounted  on  3-in.  axlea.    The  load  on  the  axle 
is  5  tons.    Find  the  horse-power  ab?r.-bed.     (/i-O.l;   0-30°;   d -0.01  in.) 

Ans.  1.7. 

50.  Find  the  work  absorbed  by  friction  per  revolution  by  a  pivot  3  ins.  long 
and  carrying  (}  tons,  its  upper  face  being  6  ins.  in  diameter,  coefficient  of  friction 
.04,  and  2rt  being  90°.  Ans.  .33936  in.-ton. 

51.  Calculate  the  horse-power  absorbed  by  a  footstep-bearing  8  ins.  in 
diameter  when  supporting  a  load  of  4000  lbs.  and  making  100  revolutions  per 
minute,  (a)  with  a  flat  end,  (6)  with  a  conical  pivot  a -30°,  (c)  with  a  Schick 
pivot.     (Take    /i=.03.) 

.\ns.  (a)  .51;  (6)  1.02;  (c)  .76,  if  upper  radius -length  of  tangent -2X 
lower  radius. 

S3  The  diameter  of  a  solid  cylindrical  cast-steel  pivot  is  2}  ins.  Find  the 
diameter  of  an  equally  efficient  conical  pivot. 

53.  The  pressure  upon  a  4-in.  journal  making  50  revolutions  per  minute  is 
6  tons,  the  coefficient  of  friction  being  .05.  Find  the  number  of  units  of  heat 
generated  per  second.  Joule's  mechanical  equivalent  of  heat  being  778  ft.-lbs. 

54.  A  water-wheel  of  20  ft.  diameter  and  weighing  20,000  lbs.  makes  10 
revolutions  per  minute;  the  gudgeons  are  6  ins.  in  diameter  and  the  coefficient 
of  friction  is  .1.  Find  the  loss  of  mechanical  effect  due  to  friction.  If  the 
motive  power  is  suddenly  cut  off,  how  many  revolutions  will  the  wheel  make 
iaefore  coming  to  rest?  .4ns.     JfH.P.;   10.9. 

SS  A  wedge  with  a  taper  of  1  in  8  is  driven  into  a  cottered  joint  with  an 
estimateil  pressure  of  600  lbs.  Find  the  force  with  which  the  two  parts  of  the 
joint  are  drawn  together  and  the  force  required  to  withdraw  the  wedge,  the 
coefficient  of  friction  being  .2.  Ans.  1128  lbs.;   307  lbs. 

56.  .\  grindstone  with  a  radius  of  gyration  -12  ins.  and  making  120  revo- 
iutioits  |H>r  minute  is  suddenly  left  to  the  influenra  of  gravity  and  axle  friction 
and  comes  to  rest  in  160  revolutions.  Find  the  coefficient  of  axle  friction^ 
the  diameter  of  the  axle  being  1 }  ins. 

57.  A  fly-wheel  weighing  8000  lbs.  and  having  a  radius  of  gyration  of  10  ft. 
is  disconnected  from  the  engine  at  the  moment  it  is  making  27  revolutions  per 
minute,  it  sto|)8  after  making  17  revolutions.  Find  the  coefficient  of  friction, 
the  a.vle  U-ing  12  ins.  in  diameter.  An».  .2325. 

58.  .V  niiiwiiy  truck  weighing  12  tons  is  carried  on  wheels  3  ft.  in  diameVr; 
the  jdumiilM  are  4  ins.  in  diameter,  the  coefficient  of  friction  f\.  Find  the  ret  st- 
ance of  the  truck  HO  far  as  it  arises  from  the  friction  of  the  journals. 

.inn.  37 J  lbs. 

59.  A  tramcar  whp<'l  is  30  ins.  in  diameter,  the  axle  2J  ins.;  the  coefficient 
of  a.\le  friction  .OS,  of  rolling  friction  .09.    Find  the  resistance  per  ton. 

Ann.  28.37  lbs. 

60.  .\  bearing  16  ins.  in  diameter  is  acted  upo.'i  by  ahoriBontal  force  of  .50 
tons  iiiid  a  vertical  force  of  10  tons;  the  ((x-fficient  of  friction  is  ,'b.  Find  the 
horse-jMiwcr  absorlxMl  by  friction  jier  revolution.  Ant.  .906  H.P. 


EXAMPLES. 


415 


6i.  A  steel  pivot  3  ins.  in  diameter  and  under  a  pressure  of  5  tons  makes 
60  revolutions  per  minute  in  a  cast-iron  step-well  lubricated  with  oil.  How- 
much  work  is  absorbed  by  friction,  the  coefficient  of  friction  being  .087 

Ans.  .85J  H.P. 

62.  A  pair  of  spur-wheels  are  4  ins.  and  2  ins.  in  diameter;  the  flanks  of  the 
tetth  are  radial;  the  larger  wheel  has  16  teeth;  the  arc  of  approach -arc  of 
recess  =  i  of  the  pitch.  Show  how  to  form  the  teeth,  and  End  their  efficiency. 
I  Coefficient  of  friction -.11.)  ^^-  •97- 

63.  Find  the  work  lost  by  the  friction  of  a  pair  of  teeth,  the  number  of 
teeth  in  the  wheels  being  32  and  16,  and  the  diameter  of  the  larger  wheel, 
wliich  transmits  3  H.P.  at  50  revolutions  per  minute,  3  ft.    Ana.  3.646  ft.-lbs. 

64.  The  driver  of  a  pair  of  wheeb  has  120  teeth,  and  each  wheel  has  an 
addendum  equal  to  .28  time  the  pitch;  the  arcs  of  approach  and  recess  are 
ea.h  equal  to  the  pitch;    the  tooth-flanks  are  radial.     Find  the  efficiency. 

Coefficient  of  friction -.106.)  Ana.  .9m. 


CHAPTER  VII. 


OX  THE  TRANSVERSE  STRENCiTH  OF  BEAMS. 

1.  The  Moment  of  Resistance.— Let  the  piano  of  the  paper  lie 
a  plane  of  .symmetry  with  resjx'ct  to  the  l)eain  PQRS.  If  the  beam 
is  subjected  to  the  action  of  external  forces  in  this  plane,  PQRS  is 


D  c 


A  B 


Fig.  404. 


aT- 


Fio.  40.5. 

bent  and  a.«isunies  a  curved  form  P'Q'R'S'.  The  upper  layer  of 
fibres  Q'R'  is  extended,  the  lower  layer  P'S'  \>f  compressed,  while 
of  the  layers  within  the  lM>am,  tho.se  nearer  P'S'  are  compressed 
and  those  nearer  Q'R'  are  extended.  Henc(>  there  nuist  be  a  layer 
M'y  iM'tween  P'S'  and  Q'R'  which  is  neither  compressed  nor  ex- 
tended. It  is  called  the  neutral  surfnrc  (or  cylinder),  and  its  axi.s 
is  ))er|><'ndieular  to  the  plane  of  flexunv  In  the  present  treatise 
it  is  pro(,v<.se(l  to  deal  with  flexure  in  one  plane  only,  and,  in  general, 
it  will  be  found  more  convenient  to  refer  to  M'\'  as  the  ne-itral 
line  (or  luis).  a  term  only  vised  in  reference  to  a  tranyvcrse  section. 

If  a  force  act  upon  the  beam  in  the  direction  of  its  length,  the 
lower  !av(>r  P'S',  instead  of  being  compressed,  may  b«'  stretclie<i. 
In  such  a  ca.se  there  is  no  ni-titral  surface  irilbin  the  lM>ain,  but  theo- 
reticallv  it   still  exists  somewhere  without  the  beatu. 

("ciisider  an  indefinitely   thin    sHee    of    the    Iwam  AliCD  at  a 

discance  x  from  an  origin  in  th<'  neutral  axis  and  of  thickness  di. 

If  A'}"("iy  i.s  tills  (lenient  in  the  bent  beam,  the  following  asFunip- 

tions  are  made: 

416 


THE  ^fOMEST  OF  RESISTANCE. 


417 


(a)  That  the  flexuro  is  small ; 

(6)  That  the  material  of  the  beam  is  homogeneous; 

(r)  That  any  section  AD  '.vhich  is  plane  before  bending  remains 
plane  after  bending; 

id)  That  the  strains  of  the  several  layers  are  directly  propor- 
tional to  the  stresses  to  which  they  are  due,  so  that  the  layers  stretch 
mill  shorten  freely  under  the  action  of  tensile  and  compressive  forces, 
iKit  withstanding  the  connection  between  the  different  layers. 

Lot  Figs.  406  and  407  represent  enlarged  views  of  the  elements 
Ali(l)  and  A'B'C'D'  in  Figs.  404  and  405,  and  let  the  planes  A'ly 
mill  li'C  intersect  in  0.  The  point  0  is  the  centre  of  curvature  of 
the  l>ent  layers  between  A'B'  and  CD'. 


Oi lO 

p Q 

A" 'b 


Fio.  406. 


Fio.  407 


Let  R  be  the  raditi.'i  of  cunnture  of  the  layer  P'Q',  which  is  neither 
liiiiiihcned  nor  shortened,  and  which  is,  therefore,  subjected  to 
hi)  ^ircss  in  the  direction  of  its  length. 

Let  /  1)0  the  force  developed  along  the  layer  pY  of  sectional 
an  a  ihi  and  at  a  distance  y  from  P'Q'. 

Without  altering  the  conditions  of  equilibrium  it  may  be  as- 
Muiird  that  two  forces,  opposite  in  direction  but  each  equal  in  mag- 
iiiiilf  to  /,  act  along  FQ',  and  therefore  the  force  along  pY  is 
t'4ui\:ilont  to 

I'll  a  force  t  along  P'Q',  together  with 
'()  a  couple  of  moment  ly. 

The  fnrr(>  along  every  "thT  layer  nf  the  element  i«  also  equiv- 
alent to  a  similar  force  and  a  similar  couple. 


i 


W 


11 


I 


418 


THEORY  OF  STRUCTURES. 


Hence  the  forces  along  all  the  layers  are  equivalent  to 

(c)  a  force  It  along  P'Q',  together  with 

(d)  a  couple  of  moment  Ity, 

the  symbol  J  denoting  algebraic  sum,  as  the  forces  change  from 
tensions  to  compressions  on  passing  from  one  side  of  the  neutral 
surface  to  the  other. 


Again, 


R  +  tj    Op'      p'q' 
It    ~0I"     P'Q" 


or 


^  =^'^'r».r^'^  =the  strain  of  the  layer  p'q' 


E  da 


Therefore 


t  =  -^yda. 


Hence  the  total  force  along  P'Q' 


E 


=  It^j^Iyda=0, 


since  P'Q'  remains  unchanged  in  length. 

Tliercfore  lyda^O,  and  P'  is  the  C.  of  G.  of  the  face  A' I/. 
Thu/>  the  neutral  axis  is  the  locus  of  the  centres  of  gravity  of  the 
transrerse  sections  of  the  beams.  Also,  the  total  moment  of  the 
couple  acting  on  the  element 


E 


E.    E 


-■Hy-^r,Ifda=jJ-j.Ak'^, 


R 


k  Iwing  the  radius  of  gyration  and  /  the  moment  of  inertia  of  the 
tr;uisv('rs(>  section  of  the  beam  through  A'B'  with  respect  to  an  axis 
at  /''  porixMidicular  to  the  plane  of  llcxurc. 

The  inoincnt    j.l  is  gcnt'mlly  termed  the  moment  of  re-vstance, 
but  i.«!  sometimes  sijokea  of  tus  the  clastic  moment.     It  must  neces- 


'Itii 


THE  MOMENT  OF  RESISTANCE 


419 


sarily  balance  the  bending  moment  M  of  the  external  forces  which 
cause  the  flexure,  anil  therefore 


M.p. 


or         -r  = 


ME 
I  ~R- 


Let  fy  be  the  stress,  i.e.,  the  load  per  unit  of  area,  in  f'q'.    Then 

.  ,  .  E  jy        E 

t=fyda,  and  therefore  /j/  =  ^y  or  Z^'r- 

Again,  let  z  be  the  vertical  deviation  of  E',  Fig.  405,  from  MN, 
i.e.,  the  deflectim  of  E'  with  respect  to  the  neutral  axis.    Then 

<Pz 
1  '^dx'^ 


"i-(i)'r 


tlio  upf)er  or  lower  sign  being  taken  according  as  the  centre  of  curva- 
ture 0  falls  above  or  below  M'N'. 

Now,  -T-  is  the  tangent  of  the  angle  0  which  the  tangent  to  the 

neutral  axis  at  E'  makes  with  the  straight  line  M'N',  and  this  angle 
is  \(Ty  small.    Therefore,  approximately, 

--^  =  tanfl  =  », 
ax 

ami  may  be  taken  equal  to  zero  in  the  expression  for  the  curvature, 

s(i  that 


X 

R 


AH       id 


II 


ciice 


TE 


A^z    E     H    f 


dx-     R     I 


an  r( Illations  from  which  may  be  determined  the  deflection  (z), 
th.  sl()])e  (0),  the  curvature  (-^)  at  any  jxiint  of  the  neutral 
:i\:  .  and  the  stress  developed  at  any  distance  y  from  the  neutral 
Tlie  curvature  of  the  beam  is  Ji  =  ~Ui-     W    the    beam    has 


axi-. 


1 

'  {J 


420 


THEORY  OF  STRUCTURES. 


an    initial    curvature    d"-   the  change  of    curvature    is  evidently 

tin 


Ho 


R 


4*- 


El 


FiJ.  408. 
tlie  limit  of  elasticity 

y 


Curvature,  Fig.  408,  may  be  defined  as  the  anguiar 

change  (in  radians)  of  the  direction  of  the  curve  per 

change  in  angle      0 
unit  of  length  = jg =j^. 

The  lieain  is  strained   to   the  limit  of  safety  when 
either  of  the  extreme  layers  A'B',  D'C  is  strained  to 
In  such  a  case  the  least  of  the  values  of 


-  for  the   layers    in    question    is   the    greatest   value   consistent 

with  the  strength  of  the  beam.  If  fe  and  c  are  the  corresponding 
stress  and  distance  from  the  neutral  axis,  then 


Again,  Fig.  409  represents  on  an  exaggerated  scale  the  transverse 
section  of  the  beam  at  A'D',  the  upper  and 
lower  breadths  of  the  beam,  A'A"  and  D'D", 
iK'ing  respectively  contracted  and  stretched, 
and  iM'ing  also  arcs  of  circles  having  a  common 
centre  at  O'. 

Let  R'  l)e  the  radius  of  the  arc  P'P",  whose 
lct\gth  remains  unchanged. 

Let  mtJ  be  the  lateral  c(M>Hicient  of  elasticity, 
tn    being   a   numerical    coelticient.     As   before, 
for  anv  laver  at  a  distance  y  from  P'P", 
viE  _t  ^E 
R'   ~mj    W 
juid  therefore  R'  =  mR. 

Tluis,  within  the  limitx  of  elaslinty,  the  curvature  of  the  breadth  is  - 

that  of  the  length,  and  d(H's  not  sensibly  affect  the  resistance  of  the 
!»e:i!!i  to  iH'nding.  The  i!ifl\ience,  however,  upon  the  bending  may 
iH'come  sen.sibie  if  the  breadth  is  very  large  as  compared  with  the 
depth,  as,  e.g.,  in  the  case  of  iron  or  steel  plates. 


Fio.  409. 


1? 


MODULUS  OF  SECTION 
Modulus    of    Section.  — The  ratio  - 


421 

is  called  the  strength 


modulm  of  the  section  and  is  usually  denoted  by  the  symbol  Z. 

The  modulus  may  be  easily  determined  as  follows: 

Divide  any  section,  as,  e.g.,  that  showTi  by  Fig.  410,  into  a 
iiuniber  of  thin  layers  by  lines  parallel  to  the  neutral  axis. 

Let  ay  be  the  area  of  one  of  these  layers  at  y  from  the  neutral 
axis,  and  let  /„be  the  stress  developed  in  this  layer. 

If  the  width,  and  therefore  the  area,  of  this  layer 

are  diminished  in  the  ratio  of  y,  /  being  the  skin 
stress,  then 


/ 


and 


jOy  =  diminished  area  =  a/,  suppose, 

fyOy^fay'. 


Fio.  410. 


Treating  every  layer  in  a  similar  manner,  the  modulus  figure, 
.<li(t\vn  shaded,  is  obtained  and 

KJyOy)  =  total  force  on  one  side  of  the  neutral  axis 
~}I{ay')  =M', 

.1'  Ix'ing  the  area  of  one  side  of  the  modulus  figure. 

Hut  this  area  is  the  same  on  each  side  since  the  neutral  axis  is 
at  the  C.  of  G.  of  the  section. 

Hence  if  2 A  is  the  total  area  of  the  modulus  figure,  and  if  h  is  the 
fijirlive  depth,  i.e.,  the  distance  between  the  centres  of  gravity  of  the 
hkmIuIus  areas  above  and  below  the  neutral  axis, 

}Ah  =  moment  of  resistance  of  section =/—  -/Z, 


ami 


Ah  =  Z. 


The  table  on  page  422  gives  the  moments  of  inertia,  strength 
modulus,  and  the  square  of  the  radius  of  gyration  of  various  sections 
met  with  in  ordinary  practice. 


422 


THEORY  OF  STRUCTURES. 


■S2D 


i^^' 


32"      D 


Section  Modulus  -. 

of  Gj-ration  KK 

G 

D' 

12 

BD'-bd' 

BD'-M' 

HD 

\2{BD-bd) 

BD'-hd' 

BD'-bd' 

HD 

\2(BD-bd) 

Bd'  +  bD' 

Bd'  +  bD' 

0Z> 

l2(Bd  +  bD) 

12 

BD^ 

24 

D' 

18 

BD' 
24 

24 

32 

D' 
16 

rfD«-rf*^ 

■  '''  +  rf') 

16 


16 


l^BD'~hdl 
16  BD-bd 


WORK  OF  FLEXURE. 


423 


3.  The  Work  of  Flexure.-The  work  done  stretching  the  layer 

p'q'  (Fig.  407) 

1  1  u  ,      i  dj, 

^rytip'q'-P'Q')  "o'l'^'^o  W^^' 

and,  therefore,  the  work  done  in  distorting  the  element  ABCD 

Ilcnco  the   total  work  of  flexure  between  points  defitied  by  values 
xi  and  X2  of  x 


If  on  this  portion  of  the  beam  loads  are  concentrated  at  different 
,K.int.,  the  integration  must  te  taken  between  each  pair  of  consecu- 
UYv  loads  and  the  results  superposed.  ,     ,  ,.         ,        , 

llu.  expression  is  necessarily  equal  to  the  work  of  the  external 
for,...  between  the  same  limits,  and  is  also  the  energy  acquired  by  the 
1,..,,..  in  changing  from  its  natural  state  of  equilibrium 

If  the  proof  load  P  is  concentrated  at^one  point  of  a  beam,  and 

if  d  is  the  proof  deflection,  the  resilience  =  ^d. 

If  a  proof  load  of  intensity  w  is  uniformly  distributed  over  the 
l„,,m,  and  if  y  is  the  deflection  at  any  point,  the  resilience  =  2/ u-j/dx, 
,1„.  integration  extending  throughout  the  whole  length  of  the  beam. 

4  EquaUzation  of  Stress.-The  stress  at  any  pomt  of  a  beam 
nuL  a  transverse  load  is  proportional  to  its  distance  from  the  neutm 
,,!,„..  so  long  as  the  elastic  limit  is  not  exceeded.  At  this  limit 
;,;.,„.nals  which  have  no  ductility  give  way.  In  materials  possessing 
,l,;,,ilitv.  the  stress  may  go  on  increasing  for  some  distance  beyond 
,h,.  ,.!astic  limit  without  producing  rupture,  but  the  stress  is  no 
lo„.or  proportional  to  the  distance  from  the  neutral  plane,  its  varia- 
,1.„;  heing  much  slower.  This  is  due  to  the  fact  that  the  portion  in 
..:„n„-ssion  acquires  increased  rigidity  an<l  so  exerts  a  continually 
Mun'using  resistance  (Chapter  IV)  almost  if  not  quite  up  to  the  point 
of  m.pture,  while  in  the  stretched  portion  a  flow  of  metal  occurs  and 


424 


THEOkY  OF  STRICTIHES. 


an  approximately  constant  resistance  to  the  stress  is  developed. 
Thus  there  will  be  a  more  or  less  perfect  equalization  of  stress  through- 
out the  section,  accompanied  by  an  increase  of  the  elastic  limit  and 
of  the  appnrent  strennrth,  the  increase  depending  both  upon  the  form 
of  section  and  the  ductility. 

For  example,  if  the  tensile  elastic  limit  is  the  same  as  the  com- 
pressive, the  shaded  portion  of  Fig.  424  gives  a  graphic^!  represen- 


FiG.  424. 


Fio.  425. 


Fio.  426. 


:  'Ir 
(I   I 


tation  of  the  total  stress  in  a  beinv  of  rectangular  section  when  the 
straining  is  within  the  elastic  lir  t  Beyond  this  limit  it  may  be 
represented  a.s  in  Fig.  425,  and  will  \w  intermediate  between  Fig. 
424  and  the  shadetl  rectangle  of  Fig.  426,  which  corresponds  to  a 
state  of  perfect  equalization. 

By  means  of  a  specially  designed  extensometer,  the  author  has 
carried  out  a  number  of  experiments  with  a  view  to  determine  the 
changes  of  fibre  length,  within  the  limit  of  elasticity,  at  different 
ilepths  of  a  beam  loadeil  transversely.  For  a  full  account  of  these 
experiments  the  reader  is  referred  to  the  R.S.C.  Trans.,  Vols.  VII 
and  VIII. 

The  loading  w:us  of  two  kinds,  namely,  (1)  loads  of  increasing 
magnitutle  placetl  at  the  centre,  and  (2)  equal  loads  of  increasing 
magnitude  concentrated  at  two  points  eciuidistant  from  the  centre, 
the  maxinmm  B.M.  in  eacli  case  being  the  same  as  for  the  corre- 
sponding centrally  {)laced  load. 

In  all  mea.>*uren»'nts  tlic  In-ams  were  placed  on  supports  60  ins. 
apart  and  the  distance  between  the  extensometer  points  was  8  ins. 

Diagrams  were  plotted  showing  the  "  lengthenings  "  and  "  shorten- 
ings "  of  the  fibres  at  different  depths,  and  an  inspection  shows  that 
they  are  approximately  proportional  to  the  distance  from  the  line 
which  does  not  apparently  change  in  length.  These  experiments, 
therefore,  seenr.  to  verify  the  assumption  that  the  stress  developed  at 
any  point  in  the  beam  is  approximately  proportional  to  the  distance 
from  the  neutral  surface,  and  the  agreement  of  the  assumption  with 
fact  becomes  more  marked  as  the  homogeneousness  of  the  material 
increases. 


EQUALIZATION  OF  STRESS.  "*5 

The  results  of  the  experiments  justify   the   following   fe...    al 
inferences: 

With  timber  beams  (Figs.  427  and  428) : 

(1)  That  when  a  beam  is  loaded  at  the  centre,  the  position  oi 


Fio.  427. 


the  neutral  surface  under  increasing  loads  remains  practically  un- 

dianged  and  is  a  little  nearer  the  compression  than  the  .ension  side. 

(2)  That  when  loads  are  concentrated  at  points  equidistant  from 

the  centre,  the  neutral  surface  under  the  smaller  loads  is  at  some 


Fio.  428. 

considerable  distance  from  the  mid-depth  on  the  compression  side 
Tliis  distance  diminishes  as  the  load  increases,  and  under  the 
heaviest  loads  the  neutral  surface  seems  tc  have  gradually  returned 
to  nearly  the  same  position  as  when  the  beam  was  loaded  at  the 
centre. 

With  a  7.1"X3.33"  cast-iron  beam  (Fig.s,  4'29  and  Am): 
(1)  That  the  curves  for  the  loads  at  the  centre  show  less  varia- 
tion in  the  position  of  the  neutral  surface  than  when  the  loads  are 


V 


426 


THEORY  OF  STRUCTURES. 


concentrated  at  equi-^st-^nt  points.  The  axis  is  precisely  at  the 
mid-depth  of  th(  •  ,  .p  to  2400  lbs.,  but  l)eyond  this  load  there 
is  a  perceptible  movement  towards  the  compression  side. 


Fkj.  429. 


(2)  That  the  diapram  for  the  lK<am  under  loads  10  ins.  from 
the  centre  shows  a  slight  movement  of  the  neutral  surface  towards 


Fio.  430. 


the  tension  side  for  the  smaller  loads,  but  under  3fi00  lbs.  It  sud- 
ilenly  moves  to  the  compression  side  and  then  gradually  returns 
towards  the  centre  under  still  higher  loads. 


\w  ~ 


Via.  431. 


Fid.  43i. 


(3)  That  loads  at  20  ins.  from  the  centre  show  that  thi»  m^utra! 
aurlacc  is  much  nearer  the  tension  side  for  the  smaller  loada,  but 


EQUALIZATION  OF  STRESS. 


427 


prailually  moves  to  a  position  slightly  on  the  compression  side  for 
tiic  higher  loads. 

\Mth  an  8-in.  rolled  joist  (Figs.  431  to  434) : 

(1)  That  in  Fig.  433,  in  which  the  loads  are  concentrated  at 
8-in.  centres,  the  stress  in  the  material  is  almost  directly  propor- 
tional to  the  distance  from  the  neutral  axis,  which  seems  to  be  slightly 
alujve  the  centre  of  gravity. 

(2)  That  the  diagrams  for  the  30-in.,  20-in.,  and  15-in.  concen- 


IT    - 

^r 

- 

Via.  4.33. 


Fio.  134. 


tnitions  indicate  that  the  stress  in  the  material  increases  more  rapidly 
than  the  distance  from  the  neutral  axis,  while  the  increase  is  not 
.«(»  rapid  for  the  6-in.  (Fig.  434)  concentration  and  for  the  beam 
loadfd  at  the  centre.  In  the  last  case  the  neutral  axis  has  moved 
very  appreciably  above  the  centre  of  gravity. 


Fio.  43ft. 


Fio.  436. 


With  a  7.85"  X  3.425"  cast-steel  beam  (Figs.  435  to  438): 
3    rhat  in  all  cases  the  stresses  in  the  material  are  verj-  approxi- 


XXIV 

r^tto. 

Jt 

- 

XXVIII 


Fio.  437. 


Fio  438. 


\v\\Ay  proportional  to  the  diatanee  from  the  neutral  iurfftce,  and 
tint  this  neutral  surface  very  approximately  coincidos  with  the  centre 


428 


THEORY  OF  STRUCTURES. 


of  gravity,  indicating  that,  in  the  case  of  the  concentrated  loads, 
the  variation  of  stress  in  the  beam  in  question  is  closely  in  accord- 
ance with  theory. 

(2)  That  the  Figs.  435  to  438,  plotted  from  the  results  obtained 
for  a  lx»am  loaded  at  the  centre,  show  that  the  neutral  surface 
ha.s  very  appreciably  moved  towards  the  compression  side,  but  that  the 
stress  in  the  material  is  still  pro[X)rtional  to  the  distance  from  the 
neutral  surface  in  its  changed  po.><ition. 

Ex.  1 .  .1  limhrr  l>eam  6  in».  square  and  20  fl.  long  rests  upon  two  supports, 
and  is  uniformhj  loaded  with  a  trriijhl  oj  100  Ihs.  per  lineal  jnot.  Determine 
the  strc.tn  at  the  eentre  at  a  puinl  distant  2  ins.  from  the  neutral  line. 

Also  find  the  central  curvature,  E  being  1,200,000  lbs.  per  square  inch. 


I 

Hence 
Therefore 


■"if-. 


M 


10020-20 


<i(NN)0 
108 


ll 
2 


-5000  ft.-lbs.  -60,000  in.-lhi. 

1200000 
R       ■ 


/,-llllJ  Ibs./sq.  in.     and     /? -2160  ins. -180  ft: 


Ex.  2.  4  standpipe  .leetion  .33  //.  in  length  and  weighing  5720  lbs.  it  plated 
upon  two  supiHirls  in  the  sainr  horizontal  plane  .30  //.  apart.  The  internal 
diameter  of  the  pifte  is  30  ins.  and  its  thickness  J  in.  Determine  the  additional 
unijormlij  distributed  load  which  the  pipe  can  carry  bettceen  the  hearings,  so  that 
the  stress  in  the  metal  may  nowhere  exceed  2  ttmg  tons  per  square  inch. 

lA't  ir  U'  the  nvjuired  Iniwl  in  pounds. 

Thi-  weiitht  i.f  tlie  pij)e  Ix-tween  the  lM'arinfpi-JSX5720-.'>200  l'^-*. 

Thus  the  total  di^t/ibutiHl  weight  between  the  bearings -(IF +5200 
lbs.). 

The  .straining  i»  evidently  greatest  at  the  centre,  and  at  thin  |x)int 

,V-^'^~30X  12  in.-lb*.  -45(H'  +  .V200)  in.-lb«. 


.\lso, 
Therefore 


and 


/-Vl.VJ-V  i.v. 

4ft(iy-t-5200)     2X2240 
V15»      —      15 

ir -,30.000  111(1. 


Ex.  3.  An  iron  bar  is  hrni  into  the  arc  of  a  circle  of  500  ft.  rfia»Mfr;  the 
riirifirirnt  of  rlnsticitif  is  .'W),IMtf),(MI0  llm.  Find  the  moment  of  resistance  of  a 
aection  of  the  bar  and  the  maximum  intmsity  of  ttreu  in  the  metal,  (a)  vkm 


EXAMPLES. 


429 


the  bar  is  round  and  1  in.  in  diameter,  {h)  when  the  bar  is  sqwre,  having  a  aide 

>'l  1  I". 

//  the  metal  is  not  to  be  strained  above  10,000  lbs.  per  square  inch,  find  (c)  the 

diameter  of  the  smaUest  circle  into  which  the  bar  can  be  bent. 

Moment  of  resistance  -j^l  — 250X12  ^  "^^'^^^  in.-lba. 

For  caae  (o),  '  "  «4  '  *"**  ^^  moment  - 10,000  — 

625     .     ,. 
-  --tt  in.-lbe. 
4 

A'^"'  i"^^'    ""'^    /-^000lbs./9q.in. 

It  1 

For  case  (6) ,  /  -  r:^,  and  the  moment  -  lO.OOOX  — 

-833i  in. -lbs. 
A'*'-  i'l^^'    *"**    /-5000lb9./sq.in. 

,s     "WOO      30000000      _    .     „..,-.»♦ 
F'or  case  (c),  — r—  -     j^^^,    ,    and    H - 12.)  ft., 

or  I  ho  diameter -250  ft. 

i;x.  4.  7**  cut  out  of  an  elliptic  section,  with  its  major  axia,  2a,  vertical,  the 
Tiiiiviqular  section  which  han  the  greatest  moment  of  inertia, 

]a-x  2b  1)0  the  minor  axis  of  the  ellipse; 

2x  and  2y  bo   the  depth  and   width  respect  ivnly  of  th««    required 
T'l  taneular  section. 

Its  moment  of  inertia --yx'-'a  max., 


ami  therefore 

or 

Ajrain, 

sihl  ilifTorentiatliig, 

Thrfifom 


yx*-&  max., 
3yJjr  +  xdy-0. 


Kio.  4.!i. 


*•     V' 


wdi  ^W.V    n 
a"         Ir 

X 


'-U 


3a'    6« 


430 


THEORY  OF  STRUCTURES. 


Hence 


av3 


)  ~~r  *'''• 
X'    4 

Ex.  5.  The  driving-whcfls  o}  a  locomolivc  arc  d  jcet  in  diameter  and  the  length 
vf  the  crank-radius  in  r  feet.  Ac  a  tipccd  of  x  miles  per  hour  the  stress  developed 
in  the  connecting-nnl  I'.v  not  to  ixccvd  ]  fxxinds  per  si/uarc  imh.  The  connecting- 
rod  weighs  ]V  poundn,  is  I  jcit  long,  and  has  a  constant  moment  of  inertia  I  from 
ind  to  end. 

Assume  that  the  cormerting-rmi  may  be  consideref"   a  beam  supported 

at  the  two  end;.      In  addition  to  its  own  weight   W         arries  a  uniformly 

\V  i' 
distributed  load  of  —  —  developed  by  the  centrifugal  force,  v  being  the  linear 

velocity  of  the  crank-pin  in  feet  jx-r  second.    Thus 

.VJSO  2r    44  rx,  , 


and 


W'(l  +-)-£^=-mu.\.  H.M.  in  in.-lbs. 

V       gr!    S 

/ 


-/. 


1- 


c  being  the  distance  of  the  extreme  fibres  from  the  neutral  axis. 

Kx  6.  Thi  section  of  a  hook  taken  at  right  angles  to  the  directiem  of  the  puU 
is  ;j  ins.  deep  and  1  J  ins.  wide.  (  aiculate  the  maximum  and  minimum  stresses 
in  the  material  when  the  hmik  is  loatled  with  1 0  tons  and  the  distance  of  the  line 
of  pull  from  the  inner  edge  of  the  section  is  I  in. 

The  max.  stress/sq.  in.  ^stress  due  to  direct  pull 

±     Ix-nding 

10        lOvjJxlJ 
"axil^i'i  X1JX3' 

—  I.H  toriM  in  tension  and  SJ  ton.<f  in  rompreiwion. 

Ex.  7.  //  the  pin-hoirs  '..t  a  b'-idiir  rgr-ixir  vrre  drilled  nut  of  truth  si^lewttyi, 
and  the  main  Ixuh/  oj  tl«  hur  iirrc  •')  ins.  wiilc  and  '2  ins.  thick,  what  profxirtinn 
would  the  maximum  sirr.'ts  licar  to  the  mean  iu'er  niiif  cro.is-sietiiin  of  the  Ixir  at 
which  the  menu  line  of  force  was  |  in.  from  the  middle  of  tht  section,* 

I^'f  /'  tons  U-  the  pull  on  the  bar.     Then 

max.  strexs  at  section  -stress  due  to  direct  pull  4:  stniin  due  to  bending 


I' 


EXAMPLES. 


431 


'''AX2X5'     10^       20/ 


5X2^X2X5'     10 

p     r>o  P      17 

—  tons  /sq.  in.  in  tension  and  77;  —  tons/sq.  in.  m 
10  20  10  20 

compression, 

:iM(l  the  ratio  of  the  maximum  stress  to  the  mean  is  1.15  and  0.85  on  theten- 
Mdii  and  compression  sides  resixjctively. 

Ex.  8.  .4  .^tcd  boiler-plate  tube  3«  ft.  Umg  30  ins.  inside  diameter,  weigh* 
\im  lbs.  and  rests  upon  supjmrts  33  ft.  npa.c.  Find  the  maximum  intensity 
,.;  strvKs  in  the  metal.  What  additional  weight  may  be  suspended  from  the  centre, 
a^^uming  that  the  stress  is  nowhere  to  exceed  8000  lbs.  per  square  inch  f 

Let  ( =  thickness  of  tube.     Then 

4200-weightof  tube  in  pounds  =  — (15 +<'- 15  )36x490, 


or 


('+30/-'^     and     («..3.'>!)in. 


22/ 
/,  the  moment  of  inertia,  ■=—  ( 


22/15./j^^^[lx.^8.1. 


33 


3350 

8 


Hie  weight  of  the  unsupiwrted  length  "f  tulx-  =—X 4200 -3850  lbs. 
This  is  uniformly  distributed,  and  therefore 

*33Xl2-max.  B.M.  in  '"•-»«•-  153.51) Cl^^^'^O' 
I  lence,  /  -  "40.»  Hw./sq.  in. 

Next,  let  P  be  the  weight  required  at  the  rentre.     Then 

-X33X12+  ^^33X12 -max.  B.M.  in  in.-lbfl. 
4  8 

8000   ^H 
"15.359^  H^'^-^^-*' 


:uA  thcri'fore 


P-|S,S(i0lb8. 


Note.  If  r  if  the  Interior  radius  of  ■  hollow  tub«  of  thickneM  t,  tb«  Mctional 
area  of  the  tube -:T(rT"t'  r')-ar:rt,if  t  it  to  small  as  compared  with  r  that  t'  may 
be  diiregarded  without  appreciable  error.  Also,  the  moment  of  inertia  of  the  sec- 
tion   "(r      '     r')-»rr't,  approiimately.     Hence  the  moment  of   resistance  of  « 

4 
tiollow  tube  whose  thickness  is  small  as  compared  with  the  radius 


--Ki't-)rfr>t,  approaimatetx. 
r 


432 


THEORY  OF  STRUCTURES 


V.K.  9.  .1  nuft-iroti  nrtangular  girder  rests  upon  supports  12  ft.  apart  and 
carriin  a  irciglU  of  2000  Iha.  at  the  centre.  1}  the  breadth  is  one  half  the  depth, 
fiiui  the  sretionai  umi  n/  llir  girder  so  that  the  intensity  of  stress  may  nowhere 
exceed  4000  tbx.  per  .<(/«(;/<'  inch. 

•l/.vo,  find  the  depth  aj  n  ivnnight-iron  girder  3  ins.  wide  which  might  be  sub- 
stituted for  the  ca.'<t-iron  girder,  the  coeflicictit  of  strength  for  the  wrought-iron 
being  SOOO  lbs.  per  .square  inch. 

Ti'ke  (/  the  depth  ami  h(  =  Jr/)  the  breadth  of  the  cast-iron  girder.  Then, 
disregarding  the  weight  of  the  girder, 

"Tl-'Xl  •-'=">'*''•   »•^'•  '»  in.-lbs.  =4000^' =''^d'. 
4  o         12 

Therefore  </'  =2U>  and  d  =0  ins.,  so  that  the  sectional  area=-M  =  18  sq.  ins. 

IhI 
The  weight  of  the  girder  =  — ^  1 2  X  450  -  V  i'  lbs.    Taking  this  into  account, 

1  ''000 

-ix  Y<i'xl2xl2  +^— -12X12 -max.  B.M.  in  in-lbs. 
8  4 

4000^ 


or 


d'--ifi' -216=0    and    d -6.38  ins. 


6.3S 


Therefore  sectional  area  =   "—  X6.38  -20.3.53  sq.  ins. 


Next,  let   D  In-  the  required  depth  of  the  wrought-iron  girder.     Disre* 
garding  the  weights  of  the  girders, 


200(> 


3yf)> 


-mm 


w 


«        '^      '    6 
The  weight  of  the  wrought-iron  girder 


and    Z>'-9,    or    D-3in. 


■  ^12x48G-120I>lb«. 

144 


Taking  this  into  aeeount, 


-s — 12X12+— T-12xl2-max.  B.M.  m  m.-lbs. 
8  4 


3/>' 
-8000-jj-, 


or 


Z>«-.54/)-18-0    and     />-4.52in8. 


EXAMPLES. 


433 


Ex.  10.  Compare  the  stre,igth  modulus  of  a  rectangular  section  6  cm.  wide 
X  J4  cm.  deep  unth  that  of  a  double-tee  section  of  the  same  area,  the  flanges  being 
IS  cw.  X3  cm.  and  the  web  18  cm.X2  cm. 

For  the  rectangular  section, 

/       1    /6X24'\ 

-c-=i2i-nr-j""®' 

For  the  double-tee  section, 


U^ 


/       1   /18x24>-2X8xl8' 


c     12  V 


12 


) 


r 


■1080. 


Fia.440.  Fro.  441. 


,  ,.     1080     1.5        ..    ,  ,,       .     . ,    ^ 
Thus  the  ratio  of  the  two  moduh=-r^=-j^,  so  that  the  double-tee  sec- 
tion is  nearly  twice  as  strong  as  a  rectan- 
gular section  of  the  same  area. 

Ex.    11.   0.1,  Fig.  442,    is  the  neutral 
axis  of  a  cantilever  fixed  at  0  and  carrying 
a  uniformly  distributed  load  of  intensity  w 
together  uilh  a  uteight  W  at  the  free  end  A. 
\/pt  I  be  the  length  of  the  cantilever.    Then,  at  any  point  (x,  y)  in  the 
neutral  axis  with  respect  to  0, 


Fig.  442. 


-j^(/'-2te+x')+1F(/-x). 


Integrating, 


(,  iH'ing  a  constant  of  integration. 

If  the  cantilever  is  so  fixed  that  the  neutral  axis  at  0  is  horixontal,  then 

r    '*  at  O,  i.e.,  when  x-0,  and  therefore  c, -0.    Thus 

ax 

an  I'luation  giving  the  slope  "  of  tht>  axis  at  any  point.    Integrating  again, 

«»-T('*7-'7-n)-'^('M). 

an  I'luation  giving  the  deflection  of  any  point  in  the  axis. 

riicrc  is  no  constant  of  integration,  aa  x  and  y  vanish  together.  The 
[Kui  1  .1  is  evidently  the  most  deflected  point,  and  if  Y  ia  the  value  of 
y  ut  .1,  i.e.,  when  x-t, 


434 


THEORY  OF  STRUCTURES. 


or 


..r-e(f.f). 


if  P  is  the  uniformly  distributed  load  -  wl. 

Again,  let  JF  be  the  increment  of  Y  corresponding  to  an  increment  AW 
of  the  weight  W.    Then 

/P    ]V  +  JW\ 


and  therefore 


EIJY'^-JW, 


or 


E- 


1^  JW]^ 
3  AY  I* 


which  is  the  equation  commonly  employed  in  determining  the  value  of  the 
modulus  E. 


If  W  is  nil, 


.1^ 
*8  EI 


.1?L' 
'g  EI* 


and  if  Pv  —wl)  is  small  as  compared  with  W, 


V    1^ 
'"3  EI' 

The  deflection  may  be  approximately  found  by  assuming  that  the  neutral 

axis  0.1  is  bent  into  the  form  of  a  circular  arc 
of  radius  R.    Draw  the  horizontal  AN,  intcr- 
w'ctiiig    the    vertical    through   0   in  N.     The 
■^A         curvature  is  so  small  that  the  difference  in  the 
p.,Q   ^^  lengths  of  OA   and  AN  may  be  disregarded. 

Taking  ON  •»  F  -the  maximum  deflection,  thea 

AN'  ~0N(2R-0N)  -'RON,  approximately. 


Therefore 


Also, 


l'-2RY    and     l'-2S- 


I 


E   ^Ejr 
"r"   1' 


Ex.  12.    A  beam  xupported  at  O  and  A  in  the  same  horxMontal  plane  earmt 
a    uniformly  d'stihulrd  load  of  intensity  w 
together  xcith  a   ueight  W  concentrated  at  the         ot«H^^' 
middle  point. 

I/Tt   I  be  the  distance  between  the  sup- 
port h. 

First.     Let    the  beam   merely  rest  ujion 
the  sujifwrta   (Fig.  44-1).    Then,  Utween  0  and  B,  at  any  iwint  (x,  y)  of 


i" 


LXAMPLES. 


435 


the  neutral  axis,  with  reference  to  0, 


dh,    W+wl      wx* 
^""^'d?         2      ""      2' 

—~  being  the  reaction  at  0  (or  A).    Integrating, 

dx         i  o 

f,  b'ing  a  constant  of  integration. 

At  the  middle  point  the  neutral  axis  is  horizontal,  and  therefore 

$-0,    when    X-5-, 


so  that 
HtMice 


dx 

-EI^--EIt&ne 
dx 


W  +  wl  ,    ^^*J1(W    toh 
T"*        6      8\2'^3'/' 


an  equation  giving  the  slope  of  the  neutral  axis  at  any  point  between  0  and  B. 
Integrating  again, 


„,       W  +  wl  ,    wx*    l^  /W    wl\ 


in  whirh  there  is  no  constant  of  integration,  as  x  and  y  vanish  together. 

This  last  equation  defines  the  deflection  curve  and  gives  the  deflection  of 
any  I M  lint  of  the  neutral  axis  between  0  and  B. 

l-it  1'  be  the  inaxi""'ni  deflection,  i.e.,  the  deflection  at  the  middle  pointy 

wiuro  X  =-.    Then 

+EIY-^(|wl+W), 

EIY-^(|P+W), 
4» 


or 


if  /'  \*  the  uniformly  distributed  load  -wl. 

\i;;iin,  let  JF  be  the  increment  of  Y ,  corresponding  to  an  increment  iW 
.if  11 .    Then 

El{y  +  iY^-^\l^+W+AW), 


.:«!;;,  ill 


ami  therefore 


or 


EUY-  -^iW, 
1  iW  !• 


436 


THEORY  OF  STRUCTURES 


an  equation  also  commonly  employed  in  determining  the  value  of  the  mod- 
ulus E. 

5    PI' 


If  W  is  nil, 


384  EI     384  EI' 


and  if  P  is  small  aa  compared  with  11', 

48  EI  • 

The  deflection  may  be  approximately  found  by  assuming  that  the  neutral 

axis  OA  is  bent  into  the  arc  of  a  circle  of 
radius  R.     Draw   the    horizontal  OA,  inter- 
secting the  vertical,  through  the  middle  point 
B  B  iti  .V.     The  curvature  is  so  small  that  the 

Fig.  445  difference  in  the  lengths  of  ON  and  OB  may 

1)0  disregarded. 
Take  BN"}',  the  maximum  deflection.     Then 

Z)-V'  =  BA'(J/i'-«A')  =2ftB.V,  approximately. 


Therefore 

/'                                1  1' 
-^2Hy    and     r--t^ 

Also, 

M/,     U     E     8EY 
I       y      R       1'    ■ 

The  deflection  at  any  distance  x  from  0  is  Y  —y,y  being  given  by 

(i--x)'-2%. 

Second.    Let  bntk  ends  of  the  beam  be  fixed  so  that  the  neutral  axis  is  horizontal 
at  0  and  at  A.     Let  M,  be  the  moment  of  fixture  at  each  end.     The  moment  at 
O  evidently  tends  to  produce  rotation  from 
right  to  left.     Then,  at  any  jwint  (x,  y)  be- 
tween 0  and  B, 

,.         t^.^'y     /H'  +  icK        vx'     ., 

r''V. 

i/x" 

M,x, 


Integrating,  -El^ 
W  +  id 


■Elt&nO 


4     '■-  « 


Fio.  446. 


an  equation  giving  the  sloi*  0  at  any  (mint. 

Ay. 


There  is  no  constant  of  integration,  as  —  and  x  vanish  together. 


Also, 


dy 

dx 


0   when    /-/,    and  therefore 


W+wl       wP         I 


or 


EXAMPLES.  437 

8  12  ^8        13' 

W+wl  ,    wx*     ..X* 
Integrating  again,        -Ely=     ^    *      t;7~^'T» 

an  t  i|iiation  defining  the  deflection  curve. 

There  ia  no  constant  of  integration,  as  x  and  y  vanish  together. 

Ix>t  1'  be  the  maximum  deflection,  i  e.,  the  value  of  y  at  the  middle  point. 


where  x=-^.    Then 


If  IF  ism. , 


or 


384  EI  la     la 

If  P  (  -U'O  is  small  om  compared  with  U", 

I  Wl«  Wl 

y -r     and     M,  — - 

19a  EI  8 

Again,  the  B.M.  ia  nil  when  .V  -0  =  -^/yi, 

tW +ui\       vx'     ,, 


a  cimuiratic,  giving  two  values  of  x  and  defining  two  points  in  the  beam  at 
whiiii  the  B.M.  is  zero  and  at  which,  therefore,  pins  may  be  introduced  so 
that  the  beam  may  be  considered  as  consisting  of  two  cantikvera   OD,  AC, 


.^  D  C  ' 

Fio.  447. 


I;-^^H 


Fio.  448. 


siipiiorting  an  independent   intermediate  length  CD  (Pig.  447).     The  length 
ill)  •  ^AC)  is,  of  course,  the  least  of  the  values  of  x  gi\'en  by  the  last  equation. 

At  the  middle  point  of  the  beam,  i.e.,  when  x—--,  the 

»-"-(-2  )iF-2(2)  -'(s-^rS 

jW    wl\     ,tW    P\ 
Take      OK-'(t+^)  -^'^    a"d     the  vertical  BE-/ (^  +  ^). 


438 


THEORY  OF  STRUCTURES 


;■  ti 


Tlie  parabola  KDECL  ia  the  B.M.  diagram,  showing  that  the  bendin(f 
actions  on  the  central  span  and  on  the  two  side  lengths  are  opposite  in  char- 
acter. 

If  it  is  assumed  that  W  alone  acts,  then 


uul  l)etu'een  O  and  B, 


„    Wx     Wl 


.         r>..  „     .  ..  I       .  ,  Wl 

80  that  the  B.M.  at  B,  i.e.,  when  x--,  is  also  — . 

Thus  the  two  lines  K'D'E'  and  E'C'L',  Fig.  448,  are  the  B.M.  diagram, 
tho  value  of  Oiy{  -AC)  being  the  value  of  x  when 

„    Wx     Wl  I 

0-_-_    or    x-j. 

If  it  is  assumed  that  P(  -«*/)  alone  acts,  then 

wl' 


«nd 


^•-12 
,,    wl      wx'    wl* 


The  B.M.  at  B,  i.e.,  when  x  • 


I   .    wP       PI 
■J- "24°' 24- 


Thus  the  parabola  K"D"E"C"L",  Fig.  449,  is  the  B.M.  diagraiu,  tho 
leugth,s  OD",  OC"  being  the  values  of  x  in  the  quadratic 


„     wl      wx' 
0-JX-- 


wl' 
12' 


from  which 


so  that 


I 


(-^-). 


'"^ 


Fio.  449. 


OD"-. 21 U    and    OC"  =.789/. 


The  ordinate  of  the  B.M.  diagram  in  Fig.  446  is  evidently  the  cdgebraie 
sum  of  the  corresponding  ordinates  in  Figs.  448  and  449. 

.\Bain,  aiwuming  /  to  be  constant,  the  work  done  in  bending  the  beam 

in  the  first  case 

I 


1    ri  /W  +  p      wx\  • 

'Ei[       96  '     128         640r 


il' 


which  becomes 


EXAMPLES. 


h"^"^-"  "^  m^"^-^- 


439 


in  the  second  ease 


-.irrim'-f-r- 


PW 


wliich  becomes 


■£/V384     1440    384/' 


Third.    Let  the  beam  be  fixed  at  0  and  merely  rest  upon  the  support  at  A.    In 
the  first  place  consider  the  effect  of  the  uniformly  distributed  load  and  let 


Fio.  450. 

if,,  ff,  be  the  vertical  reactions  at  the  supports  0  and  A  respectively. 
Then,  at  any  point  (x,  y)  of  the  neutral  axis, 


,d'y 


wx' 


At  A,  M  -0,  and  therefore 

0-R./-^-Af„ 


or 


wl' 


»    •    •    •    •    \^) 


Integrating, 


.dy 


wx' 


-^^£-"■^-1  -'•"■ 


iiv.  iquation  giving  the  slope  of  the  neutral  axis  at  any  point.  There  is 
11, ,  .onstant  of  integration,  as  x and  ^vanish  together. 

„  X*  «'X*  -.    X*  fjy. 

Integrating  again,        -EIy~Kt—     24""      2 

•in  .  .juation  giving  the  delleilion  of  the  neutral  axis  at  aiiy  point.  There 
■;<,:<•,  .onstant  of  integration,  as  x  and  y  vanish  together.  The  deflection  at 
.1   I.e.,  when  x-/,  is  also  0.    Therefore 


440 


IHt.ORY  OF  STRUCTURES 


0  =  /^.-g-^-.U,- 


or 


/?./-3.V,-^. 


(0 


lence,  by  (A)  and  (C), 


5         .">  u7'     1 


Thus  the  fixture  of  one  cud  throws  /iiv  eighths  of  the  load  upon  that  end, 
while  Mrr'  eighlhx  of  the  load  is  borne  at  .4. 
The  li.M.  is  m7  at  |)uints  given  by 


0-R,x 


.^f  ---^ii-xHi-ix), 


I 


i.e.,  when  x  — —  and  when  x—l. 
4 

Take  OK  --'  and  Of  --^  (Fig.  450).    Then  the  parabola  KCEA,  with  its 

() 
vertex  £:  at  a  vertical  distance  DE  -  -^P/  above  the  middle  point  of  AC, 

is  the  H.M.  diagram,  and   the  liending  actions  on  the  portions  OC  and  AC 
an'  evidently  opposite  in  kind. 

The  maximum  dellection  is  no  longer  at  the  centre,  but  it«  position  may 

be  found  by  putting  -jt~^  '"  "'•!•  "*)•    Then 


Q-R, 


MX' 


-•»Ax, 


or 


0-A,, 

Mi 


S' 


/ 


from  which  x  -  —  (li*  -  ^  ;W)  -  .■>  ',  verv  nearly. 
H»  •  ' 

The   correttiMinding  value   « f   7.  j,,..,  the   maximum  deRection,  can  now 

In-  foiniti  liy  sul>Htituting  this  value  of  i  in 
J  ,'--  «'<|.  il<). 

c'-;^_^_     "''--..>  Ill  the  wrimd  ftlacf,  consider  the  effect  of 

•  r  ■■'  the   single  weight   H',  //,  and    R,  being  uKain 

the  n-aetions  at  0  and  il  respectively.    Then, 
Fio.  451.  In-tuitn  O  rtm/  W, 


: 


.1/ 


^>2^H,s-M^. 


•       •        • 


(D) 


Integrating. 


-/•/^-«|-«,., 


EXAMPLES. 


441 


ail  pquation  giving  the   slope  at  any  point  of  the  neutral  axis  between  0 
ami  B. 

Let  a  be  the  slope  at  B,  then 

I'      ..  I 


-FAUMa-Ri-^-Mi^. 


(E) 


I  ntcgrating  again,  -  Ely  -  R,-^  -  Mt-j, 


(F) 


an  I'ljuation  giving  the  deflection  of  any  point  of  the  neutral  axis  belwun  0 
wl  H. 

{jL-i  y,  be  the  value  of  y  at  B,  then 

I*  /' 

lirtween  B  and  A, 

.lf--B/g-«.x-Af.-»F(x-i). 
At  .1  the  B.M.  is  nil,  and  therefore 

0~RJ-M,-W^, 
I 


or 


Integrating,         ~^' ^-'^>^~'^*'~~2  {''j)   "♦"*» 
(-,  In  iiiK  a  ranstant  of  integration. 

Hut  't^  -tan  «  at  H,  i.e.,  when  x—x.    Theref<M« 
dx  '      '  i 

P  I 

-El  tan  a  -/?,-f  -  W,—  +C,, 

■till,  tiy  wj.  (E),  f,  -0. 

1 1,  ,„^  -  J?/^  -  ft,^'  -  .V.x  -  7  (x - 7) '. 

an  •"iniitioti  giving  the  slope  of  the  neutral  axis  at  any  point  between  B  and  A. 

r*         X*     W  /       l\' 
•  Itratiiig  again,      -*''»-''i^ -•"i^^'g  ('~  j)    "^^ 


Inl) 


r,  Ining  a  mnstant  of  integration. 
""t  yVg  **»*">  '"IT"     Therefore 


-&7y.-«4-*-'-i^'^' 


mi..  l.yc«|.  (F),  r,-0. 


..^=* 


•442 


Henre 


THEORY  OP  STRUCTURES. 
-£/y-/?,5-.V,|-^(x-^) (H) 


an  equation  ftiving  the  deflection  of  any  point  of  the  neutral  axis  betiDeen  B 
and  A. 

But  y  is  also  nil  at  .4,  i.e.,  when  j  "/.    Therefore 

/'       /'   ir 


or 


«,/-3.V.-^. 


(K) 


Hence,  by  eq».  (r?)  and(K),    /?i=[^»'    and     ^."^j"^^. 


so  that  the  ('ixtun>  throws  f/<iv»i  nixtiT'ilhH  of  the  weight  on  0  and  frt  nxteenlhM 
of  the  weight  on  .1. 

The  H.M.  is  nil  at  a  jM)int  Ix't  ween  O  and  B,  defin''d  by  the  value  of  x,  given 
by  making  .1/  =0  in  ci\.  (D).     Then 

O-ft.x-.V,    or    *"/f-'-Jx'- 

Take  OK'  -  ^  and  OC'  -  ,4'  (Fig.  451).    The  two  lines  KC'E'  an.l  E'A 

are  the  U.M.  diagram,  and 

DE'^lni. 
«5m 

The  maximum  lirfitction  is  evidently  at  some  point  Ix-fween  B  and  A,  and 
its  position  may  be  found  by  making  —  --O  in  o<|.  (H;.     Then 

0-//.--.W,x  --(x~-)  , 


which  bei-omes 
fr«)in  which 


0-  -5x'f  KWx    4/', 

X  ~l\\        ._  )      .."WV.  very  nearly. 


When  the  lienm  rnrrirs  the  two  lu.irl.,.  the  IVM.  ;it  miiv  puirit  \*  the  nlqt^wmf 
siini  of  the  (■oirc!<|i<inihii^  iirilitiMtt'-<  nf  the  K.M.  diugranis  obtaiiu-d  by  loit- 
sidcriiiK  the  1<muIs  .Me|itir:iti>ly. 

Si  iiliM*  the  xloiic-  anil  ihi-  ilcllcrtiiin  an-  the  alRcbniic  ^unis  of  the  corrp 
iii|Miiiitii)u  slo|ie4  and  ilclicrtiont  due  to  the  'M'iMirittc  jondx. 

If  then'  an-  a  niitnlH-r  of  weightx  conrentrated  nt  diffciciit  (Hiintjt  iilonf 
a  Ix-aiii.  each  weight  may  be  treati'd  inde|iendently  and  IIm*  several  n>!-uiu 
8U|i«-r|iosed. 


REINFORCED  CONCRETE  BEAMS. 


443 


Fio.  452. 


5.  Reinforced  Concrete  Beams. — In  concrete  beams,  reinforced 
liy  the  introduction  of  stwl  rods, 
ixpandeil  metal,  etc.,  on  the  tension 
^idc,  it  is  commonly  considered  good 
practice  to  assume  that  the  whole  j. 
of  the  tension  is  carried  by  the 
metal. 

Let  Ee,  E,  be  Young's  moduli    for  concrete   and  steel,   respec- 
tively, in  compreiifinn: 
ft,  ft     "    the  extreme  fibro  .stre-sses  in  the  concrete  and  steel; 
"     "   breath h  of  the  beam; 
''     ''  distance    lx»t\veen    the   upper    surface   of    the 

beam  and  the  centre  line  of  the  steel; 
"    "   distance  iK'twet'n  the   up|)er  surface  and   tie 

neutral  axis; 
"     "   distance  lx*t\veen   the  centre  line  of  the  steel 
and  the  Jjottoin  .surface  of  the  lK>am 
n  "    <<  area  of  the  .steel  reinforcement. 

For  equilibrium,  the  algebraic  sum  of  the  horizontal  forces  in 
th-  In-ani  must  bo  nil.    Therefore 

f!'t-ol.     0, 


b 
d 

xd 

h 


MP 


bxd    fj,    K,i\     r)d      1  -T 
2a  "fc^     'E^d  "rx    ' 


t. iking 
llii.-i. 


r== 


E.- 


M 


'^^rW       rW 
rbd  ' 


\\\\.i])  d('i«>rmines  thp  |H>sition  of  the  neutral  axte. 
1  111'  moment  0}  reKixlanre 

-^W)'+/.a(l-.^)rf 


, 


444 


THEORY  OF  STRUCTURES. 


4}M'x{l-'^), 


at 


=Mi-.t)- 


Ex.  13.  A  concrrle  beam  4  inx.  wiilr  hif  12  in*,  deep  i>  reinfnreed  by  a  /tUrl 
r«/  (if  J  in.  diametrr  itland  with  iV.s  ciiiirr  Ivk'  1  i'l.  alnnr  the  lension  jure  oj  the 
lirtim.  Yiiuny'K  miiduli  jor  ctmrntf  and  sUrl  in  comprrgsion  an-  3,000,000  (AV) 
and  :SU,(NN>,000  iK,)  lbs.  prr  .vi/u/irc  inch.  The  conrrele  may  be  subjected  to  a 
ctimprcssion  c/  (KM)  lbs.  /ht  square  inch  and  the  steel  to  a  tensiuti  of  18,000  lbs. 
per  square  inch. 

1     /I  \  '      Er       1 


Then 


*'"4''(>)    '•'"*'  *"••  '"•' 


E,      10' 


and 


N  2 X .UMi X  ,^_4 X  n  +  (.i9fi)'  - .\m 

Ax4xll 


Hpin-e  the  numtnil  oj  nsistann ,  so  far  a«  it  depends  upon  the  eompres»\ve 
strength  of  the  (•(ini'n't<'. 


.25C 


-ixfi00x4xll»Y.2.'i6(l  -■"'  ')  -34.005  in.-UjB., 


iind  the  moment  <>f  re.sintanc*',  .>«>  far  liM  it  depends  upon  the  tenitiU  strength 
of  the  steel, 


INOOO) 


.2.56 


X .  1 WJ  X  1 1  ( I  -  '"  -')  »  35,497  in.-Iba. 


The  smaller  of  the  two  n>sults  mu.it  U-  taken  i\n  the  actual  moment  of  riMtistanre . 

Kx.  II.  7'<i  design  <l  cunrrelr-sltel  slab  to  carrif  an  cxtirnal  liHui  oj  100  /'>•«. 
Iter  square  /(xi/,  the  slab  In  In  ftlaetil  iMtieeen  Inii  sleel  lieams  (it)  ifiji.  aparl.  Take 
the  same  moduli  i;.-  in  tin  /ninding  ixiinifilr,  and  let  t>2.'>  lbs.  and  IH.OOO  lbs.  lie 
the  utrixs^s  which  <iin  In   sajili/  itritln/Htl  in  the  inncriir  ni,d  steel  rcsfiertlivli^. 

Try  a  slab  5  iii>'.  ihifk,  n-iiiforred  hy  MitA  hmIs  of  j  in.  diameter,  .scaej-d  t) 
•as.  ii\mTX  Mild  pl.'Ked  4  ins.  ix-low  the  top  of  the  slali.     Consider  a  (x>rtiuii  of 

Vw.  4K<. 
the  slab  (\  in*,  wide,  ront.iipiiin  one  reinforcing  r<nl,  and  let  the  weijcht  of  the 

rfJIlcrele  ln'  l.'iO  U»s.  |MT  niliir  fcMit. 


FORMULA  Wl^Cbd'.  446 

6X5X12 
Weight  of  concrete  in  [)08ition  per  lineal  foot  =    —-.^  "  150  -31.25  lbs. 


1728 


"  steel 

"  external  lomi 


I  (  t  t  4  1  ( t  H 

t  t  it  it 


Totalioad  "       "       " 

i.r  ^ay  H2  lbs.  [xr  lineal  fixit. 

The  maximum  H.M.  on  the  portion  under  consideration 


-     .17  lb. 
-50.00  Ibe. 

-81.42  Ibe. 


82X5'X12 

H 


=.'1075  in. -lbs., 
ulii^h  must  not  exceed  the  Imxl  moment  ■/  rcaixtancf  if  the  slab  is  to  be  safe. 

Now, 


\'(.04!))'+2X.;)4»X  ,',X6x4-.049      ,oo 

,'»X0x4 


The  tnommi  of  resintanrr,  no  far  as  it  (iei)ends  ujwn  the  eompresnve  strength 
III  the  concrete, 

625X0X.1S.3X4'-       .183v 
-  11  -     — 1  =5156  in.-lbe., 

atiil  the  mnmmt  of  reirixlanrf,  so  far  as  it  depends  Upon  the  teruile  atrenglh  of 

I  111'  HlOt'l 

(I  QO 
l-~      \  -3313in..ibfc 

\s  the  smaller  of  these  moments  is  greater  than  the  max.  B.M.,  the  slab 

1-   iiii|ily  sjifc. 

\  more  exact  adjustment  might  lie  made  by  diminishing  the  thickness  of 
ihi'  loiicrclc  and  s|»acing  the  hmIm  a  little  further  a|>art.  The  properties  of 
ri'i.  ri'ic,   however,  are  so  variable   that  great   nicety  of  adjustnusnt  seema 

liMniM-i'ssiiry. 

6.  Fonntila  Wl-^Cbd^.-In  thi.s  fnrtimlu  b  is  a  transvjTse  ditnen- 

'inn  nf  ;,  JM-nir.  of  li'tiKth  /,  (If'i)tli  il,  and  carrying  a  load  W. 


Ii  iiiLs  IxK'ti  Hhowii  that 


lif  /  is  the  max.  stress, 


/       U-2 


it;  at 

r 


446 


THEORY  OF  STRUCTURES. 


But 

and 
Therefore 


A  is  proportional  to  bd, 

k '  d, 

c  "  "  "   d, 

M  '      "        '•  in. 

M.(P 


H7a 


ocIhP  ^CM^, 


C  U-ing  a  coefficient  which  defx^nds  upon  the  .lature  of  the  inaten: ., 
the  character  of  the  hiaiHiin,  and  tiie  in«'tiiod  of  sui)|M)rtinp  the  \rot\\n. 
Its  value  nuist  be  determined  In-  exixTiinent. 

This  formula  is  .sometimes  used  to  determine  the  breaking  weigiit 
of  a  iH'am,  and  ('  is  tlien  called  the  tnodulu.s  of  it(j>ture.  Values 
of  r  for  iron,  steel,  an<l  timU'r  Ix-ains  are  tabulated  at  the  end  of 
Chapter    IV. 

Kx.  l.'i.  .4  Ui-in.-iliipXCt-in.-u-iile  red  pnie  Uam  rrsling  upon  supparta 
20  It.  itfMirt  jiiil.i  iinilir  <i  Imiil  «/  14,J."i<)  llis.  ninrinlrnlnl  ut  the  centre.  Find  V. 
Dinrrgardiuy  the-  wcijIU  uj  tin   hmni, 

142.")OxL'OX  12 -*'••>•  10'. 


Tlicrefore 


(-5700. 


7.  Beams  of  Uniform  Strength.  A  Ix-am  of  uniform  strength 
i-;  a  iH'am  so  designed  that  tiie  greatest  stres,s  dr'velo|H'd  under  a 
given  load  is  the  same  at  every  section  from  end  to  end  of  the  Ix'ani. 
I^'t  (/  be  the  depth  and  z  the  greatest  width  of  a  section  of  .such 
a  iHam  at  any  distance  /  from  an  origin  in  the  neutral  axis.  Then 
/  is  the   max.    stie.ss  develoiMMJ   in   thin   secti«M>.     Its  moment  of 

2/ 

resi.stance  =  /  —  '^zif^.  since  /  is  to  lie  constant. 

Therefore  ^y^rx t he  B.M.  at  » he  section, 

a.V. 

which  may  Ik-  written  zy^^cM, 

c  Ix'ing  a  cfn-dicient  defM>nding  u|H)n  the  value  of  /  and  the  form  of 
the  section 

At  fioints  at  which  the  H..M  i^  ml.  qr^O,  and  therefore  either 
z  or  y  must  !»•  ml.  Throrrtnally,  tlien,  a  iM-ai"  n'«|uires  no  sectional 
area,  i.e.,  no  material,  at  such  jKants,  but  it  is  manifest  that  at  ei«ry 


ip™ 


BEAMS  OF  UNIFORM  STRENGTH. 


447 


jioint  the  beam  must  have  a  sufficient  sectional  area  to  take  up  the 
.'•liiaring  force. 

Case  a.  If  the  width  (i.e.,  z)  is  constant, 

:iii(l  the  ordinates  of  the  profile  in  elevation  are  proportional  to  the 
Ninan'  root  of  the  corresponding  ordinates  of  the  B.M.  curve. 
Cask  b.  If  the  depth  (i.e.,  y)  is  constant, 

zocM, . 

;iih1  the  ordinates  of  the  profile  in  plan  arc  directly  proportional  to 
thr  corresponding  ordinates  of  the  B.M.  curve. 

Case  c.  If  the  ratio  of  depth  to  width  (i.e.,  j)  is  constant, 

;iti(l  the  ordinates  of  the  profiles,  both  in  elevation  and  in  plan,  are 
jiMportional  to  the  cuIk«  root  of  the  corresponding  ordinates  of  the 
MM.  curve. 

If  the  load  on  the  Ix'am  consists  of  a  number  of  concentrated 
l.p.nls  Mtxi,  and   the  profiles  are  evidently  made  up  of  cubical 

|i:ir:tlM)ia8. 

(  'ase  d.  If  the  sectional  area  (yz)  is  con-'»tant, 

yccM, 

ami  the  ordinates  of  the  profile  in  elevation  are  again  proportional 
['<  tlic  com'sjxjnding  orditmtes  of  the  H.M.  curve. 

Anain.  when  M  "0,  y  =  0,  and,  therefore,  z  is  infinite.     Hence,  in 
a  Ixaiii  of  this  kind,  the  distribution  of  the  material  is  most  defective. 


^ 


^ 


Pin.  454. 


Fia.  455. 


HmmK   of  Approiimately  Uniform  Strenfith. — ^The   curved   line* 
ii    I  igs.  454  and  455  are   the  lh'>oretical  profiles  of  a  cantilever 


448 


THEORY  OF  STRUCTURES 


and  a  beam  of  uniform  strength.  It  is  sometimes  the  practice  to 
replace  the  curvetl  bouniiary  Unes  by  the  tangents  CD  at  the  section 
CC  of  greatest  depth;  a  depth  (or  width)  DD  is  thus  provided  at  .4 
which  is  sufficient  for  the  S.F.  at  that  point.  Such  a  beam  may  be 
said  to  be  of  approximately  uniform  strength. 

l^t  y  =  f[i)  be  the  equation  to  the  curved  profile.    Then 

is  the  tangent  of  the  angle  which  the  tangent  at  (x,  y)  makes  with  the 
neutral  axis. 

At  C,  i.e.,  when  x  =  0, 

}'  (0)  =  tan  CDE  =  j^j,  =  ~0A~' 


and  therefore 


1)D  =  2AD''20C  -20Af'iO) 
-CC  -20Af'm. 


Ex.  16.  .4  cantilever  OA  of  length  I  and  conxtant  width  t  earrie$  a  veight 
W,  at  the  free  end  A  and  n  unijitrmly  dintributed  load  fV,. 
At  any  point  distant  x  from  O, 


,y._c.W-f  j  W,U-z) +^ -^•(/-x)'(, 


which  may  be  written  in  the  form 


(Hr-'-^y 


(w:) 


•22  W, 


BO  that,  theoretically,  the  profile  of  ihe  cantilever  in  elevation  ia  an  hjrper- 
bola  with  ita  centre  at  a  diKtance      '...     */  fnim  O,  and  its  semi-axes  equal 

to  ip-/  and 

A  8uffi(i(>nt   area   ntu»t,  however,    be   pro- 
vide<i   at  .1    (i>   iiirry    the   S.V.   at    that  jmhiiI 
line  to  ir,.     This  I'lin  Im-  (1uiii>  by  Hulwtituiiiic 
for  the  i'urve<l  profile  the  taniceiitM  CD,  wtiii'h 
Fio.  4.'i6.  ^.jii  gjyp    ^    cantilever  of  aijproximately  uiii- 

ftjrm  fttrength. 


BEAMS  OP  USIFORM  STRENGTH. 
At  0,  i.e.,  when  x-0, 


449 


Also, 


^■-d(ii'.+f')-.^. 


ami  therefore  at  0,  i.e.,  when  x  -0, 

2tOC  X  -/'(O)  -  -c(W,  +  »rj 

nr  -nO) ^(^(T-. 

Ilonce  DD-CC-20Af'(0) 

The  dotted  lines  define  the  cantilever  of  uniform  strength  when  OA  is  the 
Icpttcr  I'arr. 

I!x.  17.  .4  beam  A  A  of  eonntanl  width  t  and  of  length  21  carries  a  rmiformly 
disinhiitfd  load  H',  and  a  weight  W,  concentrated  at  the  middle  point  0. 

At  any  point  distant  x  from  0, 


ry'-cAf~c["^'(l-x)  +  '-"'^\l^-x')\, 


wliii  h  may  be  written 


W^* 


-1, 


•<>  that,  theoretically,  the  profile  of  the  beam  in  elevation  consists  of  two  elliptic 
;r  -  r.i ,  the  elli|Mie8  having  their  centres  at  the  jiointB  O,,  0„  where 


CC* 


\t  O  i.e.,  when  *-0, 

^)r«.-  (r,  +  y')    -  ,--.  Fio.  457. 


\1^> 


!  tifri'fore  at  0 


-r;o)x2jOf;--'^, 


':•  Wtii 


;':r 


450 


aod 


THEORY  OF  STRUCTURES. 

fir,      fir, 

DD-CC--2OAxl'(0) 


~cc 


rlW, 


■.'"^'("•■^l 


-cc 


,w,  +  w. 


The  dotted  lines  show  the  beam  of  unilorm  strength  when  AA  is  the 

lower  face. 

Ex.  18.  Design  a  cantilever  of  approximately  uniform  strength  from 
the  following  data:  Length  - 1 1  ft.;  circular  section— load  at  free  end"  1960 
Um.;  working   stress  -  7680   Ibs./sq.  In.;  disregard  the  weight  of  the  beam. 

At  X  feet  from  the  fixed  end,  let  y  be  the  radius.    Then 

Z5??!l^-.V -1960(11 -x)r2,   or   y'-^^(ll-x) 
y       4  ss 

is  the  curve  of  the  theoretical  profile  and  is  a  cubical  parabola. 
At  the  fixed  end,  x  =  0,  and 


Again, 


ot 


l'M'.i 
the  radius  ==\|-Su  • '  "^i  ins. 


3%'--^ 


dX' 


88' 


ily     _343J_ 
dx'     264  y'' 

Hence,  if  the  tangent  at  C,  Fig  4.')4,  n-plaifs  the  curved  profile. 


CF 
—  at  C  -  -  ■'  ■'.'  ■  =  -  tun  CDE  ---rr 
,/r  -^lUCUi'  11 


dx 


34.1     1 


and 


CE-llX 


34.1    4  _7 
UM  49  "g" 


milFFXESS.  451 

Therefore  the  depth  DD  of  tht-  cantilever  of  approximately  uniform  strength 

-2.40-2(OC-r£:)-2(3i-li)-4§  ins. 

8.  Stiffness. — If  D  is  the  nmxiniuni  doflection  of  a  girder  of  span 

/.  -r  is  a  measure  of  the  stiffness  of  the  girtler. 

In  practice  the  deflection  of  an  iron  or  a  stee.  girder  under  the 

wtiiking  load  should  lie  between  ■r:r^.  and  ^^t;;:,  i.e.,  it  is  limited  to 

IJlX)  600 

1  or  2  ins.  per  100  ft.  of  span,  and  rarely  exceeds  y^^^,  or  1.2  ins.  per 
KM)  ft.  of  span. 

A  timber  beam  should  not  deflect  more  than  .r—   or  1  in.  per  30 

fi.  (if  span. 

The  proper  stiffness  of  a  girder  is  sometimes  secured  by  requirinj; 

tlic  central  depth  to  lie  between  -rj  and  -^,  its  value  depending  upon 

till'  material  of  which  the  girder  is  composed,  its  sectional  form,  and 

till'  work  to  be  done. 

In  all  the  cases  of  Kxs.  11  and  12,  the  maximum  deflection  Y 

P 
is  pr(>iK)rtional  to  ^,  and  therefore 


r  = 


EI 


I'  Ik  itig  a  coefficient  which  depends  upon  the  character  of  the  loading 
;hi(1  tlie  nature  of  the  supi)ort8.     Thus 

Y  pp 

T-  -» the  stiffness  =  -^j-, 

Al  n,  if  M„„  is  the  max.  B.M.  on  the  beam  and  /  the  corresponding 

-■Jr  stress  in  the  material. 


M 


""      c'      qd'' 


u 


Ml 


il  t"  inji  the  depth  of  the  beam  and  q  a  coefficient  depending  upon  its 
"Ml (Mai  form. 


MICROCOPY    RESOIUTION    TEST   CHART 

ANSI  and  ISO  TEST  CHART  No   2i 


1.0 


I.I 


1.25 


2.2 


^    1^     12.0 


1.4 


1.8 


1.6 


^     /APPLIED  IN/MGE     Inc 


J^ 


'M    *fl/    ■   0.^00   -   Phone 
■*■;   .'M      •,989      'o. 


452 


THEORY'  OF  cTRUCTURES. 


Therefore 


I  ~E 


Ex.  19.  A  cast-iron  beam  of  rectangular  section  and  oj  20  //.  span  carrier 
a  tmiformly  distributed  load  o{  20  tons;  the  coefficient  of  working  strength  is  2 
Urns  per  square  inch;  the  stiffness  is  1  in.  in  100  ft.;  E  is  8000  tons.  Find  the 
dimensions  oj  the  beam,  viz.,  b  the  breadth  and  d  the  depth. 


therefore 
Also, 

and  therefore 
Hence 


20X20  ,.,_/,     .,M'_W'. 


W'  =  1800. 

Y     J 5    20(240) '12_  45 

I  ''l'J00~;JS4     .SOOObci'       2bd*' 

W  =27,000. 
bd'    27000 


W-     18000 


15  ins.  =d 


and 


.     ISOO    ^. 
6=-—,-=  Sins, 
lu' 


Kx.  20.  Compare  the  strength  and  stiffness  of  two  similarly  loaded  beams  of 
tlie  same  material  <//  equal  lengths  and  equal  sectional  areas,  the  one  being  round 
and  the  other  square. 

F-ct  r  Ijc  the  radius  <if  the  round  Ix'am,  /r  the  intensity  of  the   surface 

si ress. 

Let  fj  be  !i  side  of  ihe  s(itmr<'  Ihmiiii,  /o  the  intensity  of  the  surfaee  stress. 

rrr*  a' 

Tljen  r.r-==a';    I,  for  round  har. -=-,-,  and  for  square  har  =  — . 

Also,  sinee  the  !>eanis  are  sitnilarly  loaded,  the  bending  moments  at  cor- 


resiKJiidiiiR  iK)ints  are  e(|Ual,  and  therefore 


7T    ■"    ..  1-" 


so  that 


/.  ":{  rr'':i\7  "\n:r 


Thus,  tmder  the  same  load,  the  round  Ix-ani  is  strained  to  a  Rreatfr  extent 
than  thv  square  l^eam,  and  the  Intfcr  is  the  sfniiiffcr  in  the  ratio  of  v  sx  to  v  «;). 

.\aBin,  the  stiffness  of  u  beam  is  ^,i.     In  the  presetit  earn",  p,  E,  and  /  are 


DISTRIBUTION  OF  SHEARING  STRESS. 


453 


the  same  for  the  two  beams  and  therefore  the  stiffness  is  inversely  proportional 
to  the  moment  of  inertia.     Hence 


the  stiffness  of  the  round  beam  _  12       1  22  _ 

the  stiffness  of  the  square  beam     »rr*  °°  3jt     '      21 


Ex.  21.  A  horizontal  beam  of  a  length  I  equal  to  tweiUy  times  the  depth  d 
renl.i  upon  supports  at  the  ends  and  carries  a  uniformly  distributed  load  P.  If 
Oil  tieulral  axis  is  at  mid-depth,  and  if  E  is  1 ,200,000  lbs.  per  square  inch,  determine 
the  stiffness  of  the  beam,  so  that  the  maximum  stress  may  nowhere  exceed  400  lbs. 
Alst)  find  the  uvrk  of  flexure.  • 

Let  }'  be  the  maximum  deflection.    Then 

PI  I       PP  / 1  \ 

^=800j^,  or  y—  6400  (-j)  -128,000; 

Y      5    PP      5      128000        1 
also,  tht sHffnesa ~j~^^.^ I200OOO"  " 720 ' 

Again,  by  Ex.  12, 

.     .^  ^    P*''     n,       128000  PI 

the  work  of  flexure-  240  £/—  '^^  240  X 1200000  "2250' 

9.  Distribution  of  Shearing  Strew.— I^t  Figs.  458  and  459  rep- 
rt'scnt  a  slice  of  a  beam  bounded  by  two  consecutive  sections  AB, 
A'li',  transverse  to  the  horizontal  neutral  axis  OC. 


0 


Fio.  488. 


C'li 


Pio.  4fiO. 


Let  the  absci.'wir  of  these  sections  with  respect  to  an  origin  in  the 
iKulial  axis  Im*  x  and  x  +  Ji.  so  that  the  thickness  of  the  slice  is  Jj. 

Ill  ti\('  limit,  since  Jx  iM'conics  indefinitely  small,  corresiwnding 
liiii  ar  dimensions  in  the  two  sections  an*  the  same. 


454 


THEORY  OF  STRUCTURES. 


lA't  I  be  the  moment  of  inertia  of  the  section  AB  (or  A'B'  in  the 
Hmit)  with  respect  to  tiie  neutral  axis. 

Let  c  be  the  distance  of  A  (or  .1'  in  tiie  limit)  from  the  nentral 
axis. 

Let  /i,  /2  be  the  unit  stres.ses  at  A  and  A'  respectively. 

Consider  the  portion  ACC'A'  of  the  slice,  CC  lK>ing  parallel  to 
and  at  a  distance  1'  from  the  neutral  axis.  Since  it  is  in  e(iuilibrium, 
the  algebraic  sum  of  the  horizontal  forces  acting  upon  it  mast  be 
nil. 

The  stress  at  P  distant  y  from  00'  =/i^. 

Therefore  the  horizontal  force  on  the  slice  PQ  of  thickness  dy 


=  h^zdy, 

and  the  total  horizontal  force  on  the  area  ACC 

-^j£yzdyJ-lAy, 

A  being  the  area  ACC  and  y  the  distance  of  its  C.  of  G.  from  the 
neutral  axis. 

Similarly,  the  total  horizontal  force  on  the  face  A'C'C 

and  the  resultant  horizontal  force  on  the  element  under  consideration 

A,.,   / 


'»-^'»=''Kl'41- 


But 
and 


Therefore 


3/i,  the  B.M.  at  yl,='-7, 


Ma,  "     "     "  / 


'     c. 


My-\fz^nfJ^—^t. 


J\f  being  the  change  of  B.M.  in  passing  from  A  to  A'. 


DISTRIBUTION  OF  SHEARING  STRESS. 


455 


In   the  limit  Jx  and  JM  become  indefinitely  small,  so   that 

J.U^rfA/^     the  S.F.  at  the  section  AB. 
Jx     dx 

Thus 


A-/2    ^M    S 
c     ~   I    ~l'^' 


and  the  resultant  horizontal  force  on  the  element  ACC'A' 

=Ay^=jAydx, 

This  must  be  balanced  by  the  shear  developed  over  the  surface 
CC'C'C,  when  the  thickness  becomes  indefinitely  small. 
Let  g  be  the  average  intensity  of  shear  over  this  surface. 


"    w  "    "   widthCC. 


Then 


qwdx= shear  developed 
=jAydx. 


Hence 


qwjAy. 


Ex.  22.  A  solid  rectangular  section  of  width  b  and  depth  d. 

Ama  between  CC  and  surface  "''  (9  ~"  ^')  • 

Distance  of  centre  of  gravity  of  this  area,  i.e.,  y   from  the  neutral  azia 


TherefoM 


and 


/— —     and    w—b. 

12 


05 /rf 


'    6d'U  / 


*— < 


^ 


Fio.  460. 


3  R 


This  is  evidently  ji^rcatert  and  -'-  ^  when  F-0,  i.e.,  at  the  neutral  surface. 


456 


THEORY  OF  STRUCTURES. 


The    intensity   of   shear  at  any  point  may  be  represented  by  the  hori- 
zontal distance  of  the  point   from  the  parabola  AVB  having   ita  vertex 

3  S 
at  the  point  V  where  OV  ='-  -—  =q^^ 

If  qmy  is  the  average  intensity  of  S,  the  S.F,  at  the  cross-section, 

3  _S 

gmax.      ^  Ixi      3 

bd 

Ex.  23.  A  solid  circular  section  of  diameter  d. 

Area  of  element  of  thickness  dy  at  distance  y  from 


the  neutral  axis=  i^iJl-r-y'/^y-    Therefore 

-       n        Id'  2  /<i'         \  * 


and 


The  width  u)-CC 


«d* 
64" 


-sif- 


H.n„  ^^^.y..±lC<l.y.)t 


64 


and 


64S/d'        A 


3:td*  V  4 
which  is  evidently  greatest  when  Y  =0. 


Thus 


16S 


and  the  intensity  of  shear  at  any  point  may  lie  represented  by  the  horizontal 
distance  of  the  point  from  the  parabola  .1 VB,  where  OK———. 


atd' 


Alao. 


9,v.  "3rt<i'"    »rrf'  '3' 
4 


DEFLECTION  DUE  TO  SHEAR. 


457 


Ex.  24.  A  triangular  section  of  depth  d  and  xvith  a  base  of  width  b, 

M 


Area  of  section  -■ 


2' 
bd  (jd-Yy    1  b   (2 


^      2         d*  2  d  V3  /  • 


Distance  of  centre  of  gravity  of  ACC  from  the  neutral  axia 
1/2.    ,A      2/1 


-'■-Kj'-^)-|(3-^-'')^ 


6  /2 


Therefore 


and 


which  is  a  maximum  when 


the  width  CC{  -u.)  — -  (|-d  -  y)  . 

^47(3^--)  (1^^-). 

o--dr(^d+F)+dF(|d-y) 


Fio.  462. 


or 


Y 

-^. 

am. 
9maz. 

'Y2l^'' 

9.V. 

-S-^\bd 

9«v. 

1    6d« 
24    / 

36 
24- 

3 

/ 

bd* 
"36" 

i.e.,  at  half  the  depth  of  the  beam. 
Thua 

Also 

Tliprefore 

sitii'o 

10.  Deflection  Due  to  Shear. — The  strain  energy  per  unit  of  vol- 

uini'  (lup  to  shear  =  ;jr;.    Therefore  the  strain  energy  of  the  elemen- 
tary volume  PQ  (Figs.  458  and  459) 


1  ASAvy  .  , 


458  THEORY  OF  STRUCTURES. 

and  the  total  strain  energy  due  to  shear 

the  integrations  extending  over  the  whole  length  and  depth  of  the 
beam. 

Ex.  25.  A  cantilever  of  length  I,  depth  d,  width  b,  and  loaded  with  W  at  tiu 
free  end.     Then 


12' 


■6,    and    ^i/=-(j-j/»). 


Therefore  the  total  shear  strain  energy 

d 


18IF'  p   /•2  /d'       A '     .      3 


Gbd 


and  if  9  is  the  corresponding  deflection, 

W      3WH 
2      5Ghd' 

.    6  Wl 
80  that  9'^^- 

Hence  the  total  deflection  due  to  bending  and  to  shear 

1  WP     6Wl 
"3  £/  ■*'5  GW 

If  y  is  the  deflection  due  to  bending, 
Y~\5  Gbd)  ■  V3 


\      3E/d\* 


Ell     lOG 


©• 


Taking  G~-E  and  y-rr,  this  ratio  becomes  -5^^.  ^nd  the  deflection 

due  to  shear  is  only  —  per  cent  of  that  due  to  bending. 
o 

It  appears  therefore,  that  the  deflection  due  to  shear,  in  the  case  of  solid 
cantilevers  and  beams,  is  a  very  small  quantity  and  may  be  disregarded  with- 
out sensible  error.  Its  magnitude,  however,  may,  and  often  does,  become 
appreriahlo  in  th(!  case  of  plate  girders,  built  beams,  rolled  joists,  etc.,  and 
should  not  be  neglected. 

II.  Curves    of   Maximum  Normal   and  Tangential  StreM.  — As 

already  shown  in  Chapter  V,  if  a  shear  stress  is  induced  on  any  plane 


CURVES  OF  NORMAL  AND  TANGENTIAL  STRESS 


459 


a t 


_^o6-».al) 


1. 
Fio.  463. 


in  a  strained  solid,  an  equal  shear  stress  is  developed  upon  a  second 
piano  at  right  angles  to  the  first.  Consider  an  indefinitely  small 
triaiijjular  element  abc  (Fig.  463)  of  a  horizontal  beam  bounded  by 
a  i>lane  be  inclined  at  6  to  the  vertical, 
tlic  horizontal  plane  ab,  and  the  ver- 
tical plane  ac. 

The  element  abc  is  kept  in  equi- 
lihrium  by  the  stress  p-ac  upon  ac, 
the  slioar  sab{=q-ab)  along  ab,  the  shear  qac  along  ac,  and  the 
stress  developed  in  the  plane  6c.  The  weight  of  the  element  is  neg- 
lected as  being  indefinitely  small  as  compared  with  the  forces  to 
which  it  is  subjected.  Let  the  stress  upon  6c  be  decomposed  into 
two  components,  the  one  X-6c  normal  and  the  other  F-6c  tangen- 
tial to  6c. 

Resolving  perpendicular  and  parallel  to  be, 

X  •bc'=p-ac  cos  6 +q-ab  COB  d+q-ac  sin  9 


and 
or 

and 


Y-bc=^  -p-ac sin d—q-ab sin 6  +q-accosd, 
X'^pcoa'd+q8in29 


„  sin  2d 

Y^-p— —  +  5  cos  2^ 


(1) 


(2) 


The  value  of  0  for  which  Z  is  a  maximum  is  given  by 


dX 


2q 


— =0=-p8in2fl+29cos2^,    or    tan2fl  =  +— .      .    (3) 
Substituting  this  value  of  d  in  eq.  (1), 


the  max.  value  of  X  •■ 


■iW^^- 


(4) 


The  upper  sign  gives  the  maximum  stress  of  the  same  kind  as  p, 
whili  tin*  maximum  stress  of  the  opposite  kind  is  given  by  the  lower 
s\f:i\,  if  the  total  stress  is  negative. 

The  value  of  6  for  which  F  is  a  maximum  is  given  by 


dY    ^  „  . 

—  =.0-  -p  cos  2^-2?  sm  2d, 


or    tan2«--^.      .    (6) 


460 


lUEOHY  OF  STRUCTURES. 


Substituting  this  value  of  0  in  eq.  (2), 

the  max.  value  of  Y  =  y^-j-  +  fi , 


(6) 


which  is  the  maximum  intensity  of  shear. 

The  values  of  X  and  1'  given  by  eqs.  (4)  and  (6)  have  been  obtained 
by  another  method  in  Chap.  V.  It  was  also  shown  in  the  same 
chajjter  that  the  planes  of  principal  stress  are  defined  by  the  relation 

2t 
tan  20  =  -. 

P 
Let  di,  O^he  the  values  of  d  for  which  X  and  Y  are  respectively 
greatest.    Then 

tan  20 1  tan  2^2=  -~i,=  -h 


and  therefore 


dr- 


q  2t 
.(?2  =  45"». 


Hence,  at  any  point,  the  angle  between  the  plane  upon  which 
the  normal  intensity  of  stress  is  a  maximum  and  the  plane  upon 
which  the  tangential  intensity  of  stress  is  a  maximum  is  equal 
to  45°. 

Again,  q  is  zero  when  Ox  =90°  or  0",  and  p  is  zero  when  di  =45°. 


-w 


thriiif- 


—¥ — 


y     \ 


un^'"  ,^^ 


l?(£_^'f_ 


-at\M 


Fks.  404. 


Fio.  4f).'>. 


Thus  the  curve  of  rprentest  normal  intensity  cuts  the  neutral  axis 
at  an  angle  of  45°,  one  skin  surface  at  90°  and  the  opposite  at  0°, 
while  tlie  curve  of  (jrcatest  tangential  intensity  cuts  the  skin  surfaces 
at  45°  anil  touches  the  neutral  axis. 

Fig.  404  serves  to  illustrate  the  curves  of  greatest  normal  inten- 
sity. There  are  evid(>ntly  two  set.s  of  these  curves,  referring  resjiect- 
ivcly  to  direct  thruxt  and  direct  tension. 

Fig.  405  illustrates  the  curve  of  greatest  tanrjential  intensity. 

Ex.  'J(i.  .\  |iil('li-|)iii(;  Ik'uiii,  14  ins.  w'uk',  1,'>  iiiH.  ilLt'p,  and  weighing  45 
lbs.  [XiT  culjif  foot,  is  placod  upon  8up|X)rts  10  ft.  9  ins.  apart,  and  earrie.s  a 


CURVES  OF  NORMAL  AND   TANGENTIAL  STRESS.        461 

loml  of  20  long  tons  at  the  centre.  Find  the  deflection  and  curvature,  E  being 
l.J7),(X)0  lbs.  Determine  the  stiffness  of  the  beam  and  the  uniformly  dis- 
tril.iiled  load  which  will  produce  the  same  deflection.  Also  find  the  maximum 
iiittnsitics  of  thrust,  tension,  and  shear  at  points  (a)  half-way  between  the 
iKUtnil  axis  and  the  outside  skin  in  the  central  transverse  section;  (b)  at  one 
third  of  the  depth  of  the  beam,  in  a  transverse  section  at  one  of  the  quarter- 
spans.  Also  find  the  inclinations  of  the  planes  of  principal  stress  at  these 
]ioints. 

If  y  is  the  maximum  deflection. 


ElY^mP.w). 


l4x  15' 
In  the  present  case  £  =  1,270,000  Ibs./sq.  in.;  I ^y-;    i-129  ins.; 

/'    wt.   of   beam  =ii^^0ix45=^  lbs.  =say     700  lbs.;    lF-20    tons 


144 
=  44,SOO  lbs.    Therefore 


32 


the  stiffness  4  =  f' (1^+44800) 


12 


1270000X14X15'     319' 


ami 


129 


the  deflection  "=  ^  =5T^  ""•'*  '"• 


Again,  (f  +^)  12  =  (f  +  w)  '-f  =max.  B.M.  in  in.-lbs. 


=  moment  of  resistance  ""E'' 


riierefore 


/700  ^448no^  ^-^  =1270000  14XJ5^ 
^__-+44800j--=      ^  12     , 


aiHl  '  -curvature  at  the  centre  -.z,.r,,so  that  the  radius  of  curvature  at  the 
/;  34.J4 


viitrc  is  8434  ins.  or  286  ft. 
Til  fiml  pnndq. 
Al  the  central  sectiort 


(f  .44™)'f -fi  '-^' 


12 


Therefore     p  =  138fi}lbs. 

Also, 


g=^.4y---^7",.;i4x3J(li+3f) -1680  lbs. 
^      '  14xlo' 


1 


4G2 


THEORY  OF  STRUCTURES. 


Hence  the  max.   shear    stress 


'\4 


P' 


^+f/'- 1817.5  lbs., 


and  the  max.   normal  st  esses  =  1386.75  ±1817.5 

=3204.25  lbs.  and  -430.75  lbs. 
At  the  quarler-span  B.M.- iwOO +44800) 'F-'FXM* 


Also, 
Therefore 


and 


Also, 


I-IMKX)  129  .     ,. 
=         "  —,  -  in. -lbs. 
8         4 

S.F.  =22400 -^F  -22,225  lbs. 

179900  129  ^£    14X15* 
8         4   ~2i       12      ' 

p=460i}lbs. 

,       22225X12,^     ^^^ 
^*«-14^-^^X^^S' 


and 


Hence 


max. 


5-l41ilb8. 
:.  shear  stress  =  J -^^^'  +  (141J)'  -370  lbs., 

and  max.  normal  stresses  =  460.5  ±  270  -  730.5  and  1 90.5.  lbs., 

For  the  principal  plane, 

2X1680 
in  the  1st  case,     tan  20  =  .^^5^^  =2.422    and    9-31°  30'; 


2d 


1386.75 

tan2e=?j^^J^=  .613    and    0-15°  45'. 
460jf 


12.  Moment  of  Inertia  Variable. — In  the  preceding  investigations 
the  moment  of  inertia  /  ha.s  been  assumed  to  be  constant. 

From  the  general  equations,  at  any  point  x,  y  in  the  neutral 
axis, 

dx^     c  ' 


and  therefore 


,j,<Py    f 


c  being  proportional  to  the  depth  of  the  girder  at  a  transverse  sec- 
tion distant  x  from  the  origin. 

Hence,  for.  beams  of  uniform  strength,  the  value  of  c  in  terms 
of  X  may  be  substituted  in  the  last  equation,  which  may  then  be 
integrated. 


MOMENT  OF  INERTIA   VARIABLE. 


463 


Ex.  27.  A  girder  of  uniform  strength,  of  length  I,  of  rectangular  section,  rests 
upon  two  supports  and  carries  a  uniformly  distributed  load  of  w  lbs.  per  unit  of 
knijth,  which  produces  a  maximum  stress  of  f  Ibn.  at  every  section  of  the  beam. 

Show  that  the  central  deflection  is  ^^-  ^  (^^  I  when  the  breadth  (6)  is  constant 

and  the  depth  variable.     Find  the  deflection  when  the  depth  (d)  is  constant  and 

the  breadth  variable. 

Let  2z  be  the  depth  of  the  beam  at  x  from  the  middle  point,  and  let  y  be  the 
detlection  of  the  neutral  axis  at  the  same  point.    Then 

-(li)'(^'■)'• 
-g-ii4©'(^'■)"-"(^•■)"*. 


and 

Hence 

where 


Integrating, 
Integrating  again, 


-y-a  /  8ip-"-fdi— |(»8ine+«M9)+fc, 


2x 
I:  l)eing  a  constant  of  integration  and  sin  ''  -"T- 


When 


j.  =  '      y_o    and    e-^.    Therefore 


0--  --^k' 
2  2 


When  x=0,  y,  i.e.,  the  deflection,  is  greatest  and  fl-0.    Therefore 


—  Viou.  ""2 


la(r.       \      >r-2/l/6\» 


464 


THEORY  OF  STRUCTURES 


Again,  let  Fig.  466  represent  a  cantilcvt  r  of  length  I,  sperific  weight  w, 

circular  section,  and  with  a  parabolic  profile, 
the  vertex  of  the  parabola  being  at  .4. 

Let  21)  be  the  depth  of  the  cantilever  at  the 
fixed  end. 

Let  the  cantilever  also  carry  a  uniformly 
distributed  load  of  intensity  p. 

Consider  a  transverse  section  of  radius  z 
at  a  distance  j  from  the  fixed  end. 

Let  X,  y  he  the  co-ordinates  of  the  neutral 
axis  at  the  same  section.    Then 


Fio.  466. 


El 


f/x'      6    /  2  4  dx* 


Butz'=-ril-x). 


Therefore  ^  ^  (' "  ')  ^/^^  =  g"  7 '  ^  '^  ^  ^  ^  ^  *^  ' 


or 


itEh*,l'u     XKb\,  p 


Integrating, 


4   l',lx     6    n        -2  1      2 ^' 


There  is  no  constant  of  integration,  as   -  =0  when  x— 0.     Integrating  again. 


nEf^ily     "•rfe'//x'_x'\      px^ 
4    /»dx""  (i   /   ^2      6''       4 ^  ' 


There  is  no  constant  of  integration,  as  x  and  7  vanish  together.     Thtis  equa- 
tion (1)  gives  the  sl()|K'  at  any  |M)iiit.  and  equation  (2)  defines  the  neutral  axis. 


The  sioiie  at  the  free  end  (x     1)  -■  ,., ,  I  ,    +  -.:  I . 

I'  rl        p\ 


The  dcllecfion    " 


13.  Springs,  -ffl)  F'lnl  Sprint/s.—  U  two  force.",  oarh  equal  to 
P  Init  acting  ill  opposite  dinrtioiis  in  flic  same  straight  line,  are 
applied  to  tlie  ends  of  a  strainlit  iiiiifonii  strip  of  flat  .steel  spring, 
the  spritiR  will  assume  on(>  of  the  forms  shown  Im'Iow,  known  as  the 


SPRINGS. 


465 


clastic  ctirre.    This  curve  is  also  the  form  of  the  linear  arch  best 
suited  to  withstand  a  fluid  pressure,  Chap.  XII. 


cB 


Fio.  467. 


Fio.  468. 


Fio.  469. 


Fio.  470. 


Fin.  472. 


Fio.  473. 


Consider  a  point  B  of  the  spring  distant  y  from  the  line  of  action 
of  /'.    Then 

FI 
Py-b<'nding  moment  at  fi  =  -^, 

I!  lifing  the  radius  of  curvature  at  B,  and  /  the  moment  of  inertia 
(I  tlu>  section. 

If  E  and  /  are  both  constant, 

/?j/  =  a  constant 

i-  the  equation  to  the  elastic  curve. 


466 


THEORY  OF  STRUCTURES. 


(b)  Spiral  Springs  (as,  e.g.,  in  a  watch).— Let  Fig.  474  repre- 
sent a  spiral  spring  fixed  at  C  antl  to  an  arbor  at  A,  and  subjected 

at  every  point  of  its  length  to  a  bending 
action  only. 

Consider  the  equilibrium  of  any  portion 
AB  of  the  spring. 

The   forces   at   A    are   equivalent    to  a 
couple   of   moment  M,  and   to   a   force  P 
acting  in  some  direction  AD. 
This  couple  and  force  nmst  balance  the  elastic  moment  at  B. 

Therefore  M  +  Py  =  EIx  change  of  curvature  at  B, 

y  being  the  distance  of  B  from  the  line  of  action  of  P,  or 


Af-fP,=£/(l-i^), 


Ro  being  the  radius  of  curvature  at  B  before  winding,  and  R  that 
after  winding. 

Let  ds  be  an  elementary  length  of  the  spring  at  B. 

Then,  for  the  whole  spring, 

I{M  +  Py)ds  -  EIl(^j  -y)°  EIIide-dBo), 

or  Mids  +  Plyds  ^'Elx  total  change  of  curvature  between  A  and  C, 
ami  Ms+P>>y'EI{0-Oo), 

.s-  l)oing  the  length  of  the  spring,  y  the  distance  of  its  C.  of  G.  from 
A  I),  0  the  angle  through  which  the  spring  is  wound  up,  and  ^o  the 
"unwinding"  due  to  the  fixture  at  C.  With  a  large  number  of 
coils  the  distance  between  the  C.  of  G.  and  ^4  may  be  assumed  to  be 
tiil  and  theny  -=0. 

Also,  if  the  spring  is  so  secured  that  there  is  no  change  of  direction 
relatively  to  the  barrel. 

<?o  =  0,    and    Mx-EIO. 

\jo\.  the  wmdmg-up  Ik'  effectetl  by  a  couple  of  moment  Qq^M, 
Q  btiiig  a  tangential  force  at  the  circumference  of  a  circle  of  radius q. 
The  distance  through  which  (i  mm-v^  (or  deflectiori  of  Q) 


SPRINGS.  467 

=9^=^s,    since    M  =  -I, 

}  being  the  skin  stress,  and  c  the  distance  of  the  neutral  axis  of  the 
si)riiig  from  the  skin. 

Thus,  if  b  is  the  width  of  a  spring  of  circular  or  rectangular  section, 

c  =  :^,  and  hence 

2qf 
the  deflection  =rgS. 

1  1  M       Md 

The  work  done  =  «  ^  ><  deflection  =  ^  —3^  =  "o" 

"2  E(^~  2   E<?~2E  c^  ' 

k-  bring  the  square  of  the  radius  of  gyration,  A  the  sectional  area  of 
the  spring  and  V  its  volume. 

Jfc2      1 

In  case  of  spring  of  rectangular  section  73-3  • 


II     ((    ((      (< 


"  circular 


P    1 


Again,  the  spiral  spring  in  Fig.  475  is  wholly  subjected  to  a  bending 
action  by  means  of  a  twisting  couple  of  moment  M  =  Qq  in  a  plane 
[Hrpendicular  to  the  axis  of  the  spring.  Any  torsion  in  the  spring 
itself  is  now  due  to  the  coils  not  being  perfectly  flat. 


Fro.  475. 


Let  /?o  =  radius  of  a  coil  before  the  couple  is  applied, 
"ft  -     "      "  "     "   after     "        "      "       " 

««-«-«(i-s;)-^'p 


108 


THEORY  OF  STRUCTURES. 


0  being  the  angle  of  twist;  or 

Qq.<     M.I       It       s 


=  (.V-A'o)2r, 


.V  l)oiiig  the  numlx^r  of  coils  iM'forc  the  couple  is  applied,  and 
A„   •'       '•         "        "     ■'      after     ''        "      "        " 
The  distance  through  which  Q  acts,  i.e., 

the  'Reflection,"  =(/'>-7f, 
'       Ac 

,    ,  MO     fV  k^ 

anil  the  work  done  =  -77  =  "TTtf  "ia"' 

1  py 

=  -  -7—  for  spring  of  rectangular  section, 

O     ij 

A  111 

~S   E 
c.   Simple  rectangular  spring. 
By  Ex.  11, 

^3  EI  ~3  Ed 
Wl    M    2}    Vim 
I   ^  I 


(I       I  (      I  i 


circular 


iw 


since 


IxP 


Fio.  470. 
Hence 


A,       ,r,_''!?^2//^_l_/2M/_fr 
Also,  n  J      ^^    j^  ^,^-y    ^    -g^,, 


V=9 


(2) 


and 


irj    pv 

the  work  done  =  -7,- =T^T, (3) 


d.  Sprinf]  of  conxtnnl  depth  hid  trintifiulnr  in  plan. 
Let  6,  be  the  breadth  at  a  distance  x  from  the  fi.\ed  end.    Then 
bj-    l-x  ?L- 


b        I   ' 
and  /  at  the  same  [M)int 


Fio.  477. 


Therefore 


ix^'  Er    ^'     Ebd-i' 


SPRINGS. 


469 


Intpgrating  twice,      ^  =  j^^^w^, 


ami 


Therefore 
,\Jso, 


-^1    2 

y    Ebdf  ' 

^~  Ebd'    Ed 

^^~   6/    Ed      6E      3E' 


.    (4) 


or 


V  = 


3WJE 


r2     ' 


(5) 


•         •        • 


(6> 


and         the  work  done  =  —^  =  -^tt 

y.B.— The  results  (1)  to  (6)  are  the  same  if  the  springs  are  com- 
IKiuml;  i.e.,  if  the  rectangular  spring  is  composed  of  n  simple  rect- 
aiisiilar  springs  laid  one  above  the  other,  and  if  the  triangular  spring 
is  coiiiposed  of  n  triangular  springs  laid  one  above  the  other. 

{(■)  Spring  of  comtanl  width  but  parabolic  in  elevation. 

L(it  c/x  be  the  depth  at  a  distance  x  from  the  fixed  end.    Then 

(dr\'    l-x 
\df   "    I   ' 

and  /  at  the  same  point 

M/    d}r^A-x\\ 
~  12  ~  12  W   / 

d^y     W 


I 


i« 


Fio.  478. 


Tlurcfore 


i2Tr  /' 


^,-^a-x)-^^^.(/-x)-» 


'//y 


iitrirniting  twice  an<l  remembering  that  y-  and  y  are  each  nt/  when 


_12n'J[ 
y~    E    bit 


r,lS(/-J-)'-2/*(/-r)  +  i/*|, 


Sir  /••<    4//' 

•^~  E  biP^-^iEd-  • 


all' I  Iience 

Also,  H  J  = -^y-  T^  ^  =  y  A''^'  =  3  A?^' 


.    .    .    (7) 


470 


and  therefore 


THEORY  OF  STRUCTURES. 


The  work  done  =  - 


WJ 


V  = 
6  E  ■ 


3WJE 


In  the  examples  a  to  e  on  springs  it  will  be  observed  that  in 
each  case   the  energy  expended  per  unit  of  volum  •  in  bending  a 

spring  is  proportional  to  ■^,  and  if  the  energy  is  just  sufficient  not 

to  produce  a  permanent  set  it  is  called  the  resilience  of  the  spring. 
In  this  cas3  /  is  the  greatest  stress,  tensile  or  compressive,  which  the 
material  can  stand  without  taking  a  set. 

Again,  in  Chap.  V  it  was  shown  that  the  resilience  of  a  cylindri- 

cal  spring  subjected  to  pure  torsion  is  proportional  to  ^^q  being 

th-  greafest  shear  stress  which  the  material  will  take  without  per- 
manent set,  and  G  its  coefficient  of  rigidity. 

A  table  at  the  end  of  the  chanter  gives  the  values  of  these  re- 
siliences for  different  materials,  but  it  is  important  to  bear  in  mind 
that  the  value  of  a  material  for  a  spring  depends  not  only  upon  its 
resilience  but  also  ujwn  other  characteristics,  such  as  its  magnetic 
pro[x^rties,  its  hardening  and  temj)ering  projierties,  its  deterioration 
by  rust,  the  effect  of  time,  etc. 

(/)  Carriafje-apring. — Assume  a  uniform  resilience  and  suppose 
the  spring  to  be  made  of  n  strips  each  of  thickness  t. 

Let  I  be  the  length  of  the  top  strip; 

X     "         "        "     overlap  =  ^. 
^'°-  ^^®-  First,  considering  the  strips  free, 

1 


let  -jT  l)e  the  curvature  of  the  top  strip  when  unloaded; 

"      "       "      loaded; 


Hi 


sth 


unloaded. 


The  cur\'ature  of  the  sth  strip  when  loaded'- n   r-.' 


SPRINGS. 
Hence,  if  /  is  the  maximum  stress, 
112/1 


471 


R     Ri    Et     R,    Ri+sf 
Next,  let  the  strips  be  connected  together  so  as  to  form  a  spring, 
•111(1  let  -  be  the  curvature  of  the  top  strip  when  the  spring  is  un- 
loaded. 

The  curvature  of  the  sth  strip  = 


1 


ri+sf 

Hence   if  F,  is  the  force  at  the  end  of  the  sth  overlap  which  will 

1 
develop  a  B.M.  F.x  sufficient  to  change  the  curvature  from  ^  to 

1 


ri  +  d 


,  then 


"      Vft    RiRi+st    ri+sti 

2/f      Et(^ 1_U 

°  t  [''^  2\Ri+st    n+st/j' 


Rut  JF,=0  and  therefore 


/'^2\Ri+st     ri+s</J       ' 


1'  denoting  the  algebraic  sum  of  the  several  values  of  the  expression 
inside  the  brackets  obtained  by  putting  s  =  l,  2,  3,  .  .  .  n-1. 

If  st  is  small  as  compared  with  Ri  and  ri,  the  portion  between 
linickets  mav  be  written 


f  1   /,      .sK-»      1  /,  ,s<\"M 


472 
Hence 


or 


THEORY  OF  STRUCTURES. 

nEt/J_     l\  _n{n-\)  Et^  /  1 \\  _ 

""'"*"   2   Vt-i     r,/  2         2   W     ri2/~ 

/2/      1      71-1    /   \  «-l,    n 


a  quadratic  equation  giving  r\. 

Suppose  that  the  plates  of  which  the  spring  is  made  are  to  have 
the  same  strength  throughout  as  the  overlap. 

Let  W  be  the  weight  at  the  centre  of  the  spring,  and  let  6  be 
the  width  of  each  plate. 


The  moment  of  resistance  of  each  plate    =■ 


6    ' 


Then 
Also, 

and  the  deflection 


"  then  plates  =  ^^. 

WI    nfbfi         ,     ,    3  117 
T^    «       and    f=^ 


6 
E_2[ 


P 
i'  =  — 
8ft 


"2  nbfi' 


f   P    3  TT'    73 


4  Et    S  E  nbfi- 


h^ 


inch-pounds 


The  resilience  of  a  wfU-nuulo  carnage-sprmg  is  ^ 

/o  Ijeing  the  proof-stress. 

14.  Allowance  for  the  Weight  of  a  Beam.— A  Ix'ain  is  sometimes 
of  sufli  Icngtii  that  its  weight  becomes  of  importance  as  compared 
with  the  load  it  has  to  carry,  and  must  be  taken  into  account  in 
dctcriiiiiiing  the  diniciisioiis  .if  tlic  Itcam. 

The  necessary  provision  may  be  made  l)y  increasing  the  width 
of  the  beam  dcsijiiieil  to  carry  tlic  external  load  alone,  the  width 
1^'iiig  !i  dimension  of  the  first  order  in  the  expression  for  the  elastic 
moment. 

Assume  that  the  weight  of  the  In-am  and  the  external  load  arc 
reduced  to  ('(juivalent  uniforndy  tlistributed  loads. 


iM 


ii! 


ALLOWANCE  FOR  WEIGHT  OF  BEAM. 


473 


Let  ir,  lie  the  oxtcrnal  load; 

b,  "     "    breadth  of  a  beam  designed  to  support  this  load 

only ; 
B,    "    "  weight  of  the  beam; 

ir    "     "   total  loail,  the  weight  of  the  beam  being  taken 
into  accomit; 
b    "     "    corresponding  breadth  of  the  beam; 
B    "     "  "  weight    "     "      " 


Then 


aiul 


W-B^We, 

b     B     W      W-B ^ 

brB^W.^W.-B,     We-B; 


F. 


Ex.  28.  Apply  the  preceding  results  to  a  cast-iron  girder  of  rectanguiar  sec- 
tion renting  upon  two  supports  30  ft.  apart.  The  girder  is  12  ins.  deep  and  carries 
(1  II  iiij< irmly  distributed  load  of  30,000  lbs. 

Take  4  as  a  factor  of  safety;  be  is  given  by 

120000       M' 
■~2^'^    I  ' 

wliw  ('=30,000  lbs.,    d -12  ins.,    and    Z=  360  ins., 


Ill  therefore 


Ictice 


6t-5ins. 

B,=^,^X 30X450  =.5625  Ibfl., 
144 

TF.  -  B.  =  30000  -  5625  -  24375  lbs., 

,     30000X5     p,  .  „ 
^— 24375- =*^^^'"^-' 


B^!!5^X5625=6923A  lbs., 
24375 

Tl'  =  ir.  +  B=36,923T'ilb8. 


15.  Beam  Acted  upon  by  Forces  Oblique  to  its  Direction,  but 
Lving  in  a  Plane  of  Symmetry.— In  discussing  the  equilibrium  of 
Ml  !i  a  l)eam  the  forces  may  lie  resolved ' into  components  parallel 


;iiii!  |)C 


MTi)endicular  to  the  beam,  and  their  respective  effects  super- 


-174 


THEORY  OF  STRUCTURES. 


Let  AB  be  the  beam,  Pi,  Pz,  Ps,  ■  ■  ■  the  forces,  and  ai,  02,  as, 
their  respective  inclinations  to  the  neutral  axis. 


Ii1 


J  !il 


fih 


/.    p* 

P. 

Fig.  480. 

Divide  the  beam  into  any  two  segments  by  an  imaginary  plane 
il/.V  perpendicular  to  the  beam,  and  consider  the  segment  AMN. 

It  is  kept  in  equilibrium  by  the  external  forces  on  the  left  of  MN, 
and  by  the  plastic  reaction  of  the  segment  B^I^'  upon  the  segment 
^.U.V  at  the  plane  MX. 

The  resultant  force  along  the  beam  is  the  algebraic  sum  of  the 
components  in  that  direction,  of  Pi,  P2,  Pi,  ...  , 

=  Pi  cosai+P2Cosa2  +  .  .  .  =  J(Pcosa)=H,   suppose. 

If  this  force  acts  at  a  distance  h  from  the  neutral  axis,  it  is  equiva- 
lent to  a  couple  of  moment  Hh  and  a  ."single  force  H  with  it?  line  of 
action  coincident  with  the  neutral  axis.  Thus  the  corresponding 
maximum  and  mininmm  stresses  developed  in  the  beam 


„,w     H    H(     hy       \ 


A  being  the  sectional  area  of  the  l)eani.  /  its  moment  of  inertia,  ^• 
the  radius  of  gyration,  and?/  the  distance  of  the  extreme  fibres  (or 
skin)  from  the  neutral  a.xis.  The  upper  or  lower  sign  of  tiie  Pr.^ 
term  which  i-i  the  stress  due  to  heiiding,  is  to  be  taken  acconling  as 
it  i-!  the  ."^ame  or  opposite  kin<l  of  stre.'^s  as  that  represented  by  the 
second  term,  which  is  the  stre.'^s  due  to  the  direct  load. 

Again.  P,  sin  ir . ,  Pj  sin  o  j,  are  the  components  of  the  forces  at 
right  angles  to  the  beam,  and  li  Pi  P2,  .are  resi)ectively  the 
distances  of  these  forces  from  .1/.V,  they  are  etjuivalent  to  a  couple 
fif  moment  1  (Pp  sin  a)=M  and  a  single  force  2(P  sin  a)  =5,  which 
is  evidently  the  shearing  force.  This  force  develops  a  mean  tangential 
stress  in  the  section  at  .l/.V,  which  necessarily  distorts  the  beam, 


FORCES  ACTING  ON  A  BEAM. 


475 


but,  generally  speaking,  the  distortion  is  sufficiently  small  to  be 
<lisregardcd  without  appreciable  error.  ' 

,  w         My 
The  skin  stress  due  to  M  =  ± Mj-  =±'^-j^- 

Hence  the  total  maximum  stresses  developed  in  the  beam  are  the 

algebraic  sum  of 


!(4^>) 


and    jj-,. 


It  will  be  observed  that  this  result  involves  two  intensities,  the 
one  due  to  a  direct  pull  or  thrust,  the  other  due  to  a  bending  action. 
The  latter  is  proportional  to  the  distance  of  the  unit  area  under 
consideration  from  the  neutral  axis.  It  is  sometimes  assumed  that 
tho  same  law  of  variation  of  stress  holds  true  over  the  real  or  imaginary 
joints  of  masonry  and  brickwork  structures,  e.g.,  in  piers,  chimney- 
stacks,  walls,  arches,  etc.  In  such  cases  the  loci  of  the  centres  of 
pressure  correspond  to  the  neutral  axis  of  a  beam,  and  the  maximum 
ami  minimum  values  of  the  intensity  occur  at  the  edges  of  the  joint. 

Ex.  29.  A  horizontal  beam  of  length  I  and  sectional  area  A  rests  upon  sup- 
p,irt.-<  (It  the  ends  and  carries  a  weight  W  r  'niddle  point.  It  is  also  acted 
ujiDii  hi/  a  force  H  in  the  direction  of  its  lenyi. 

If  the  line  of  action  of  H  coincides  with  ae  axis  of  the  beam,  the  maxi- 
mum and  minimum  stresses  developed  at  the  middle  pomt 


'a      Ak'     AV    H  4k'l 


Thus  the  minimum  stress  will  be  nil  and  the  maximum  stress  doubled  if 


H  4/t' 


.,     W     4A- 
i.e.,.f     -^  =  j^. 


^     k-     d       .W     d 
For  a  .-ircular  section  of  diameter  a,     ~~  =g  a"«  ~H  "'H 


"    "  rectangular  section  of  depth  d, 


k 


d       ^W 
.-and- 


2  d 
"3    l 


F.x  30.  Let  the  beam  be  acted  upon,  in  the  direction  of  its  length  only 
liy  .1  force  H  with  its  line  of  action  at  a  distance  h  from  the  neutral  axis. 
The  maximum  and  minimum  stresses  developed 


-f(*^') 


ji»^^^ji|iMij. 


476 


THEORY  OF  STRUCTURES. 


Thus  the  minimum  stress  is  nil  and  the  maximum  stress  is  doubled  when 

1^  =1  or  /(  =  — ,  i.e.,  when  It  is  one  ciijhth  of  the  diameter  for  a  ci'rcu/ar  section, 

and  one  nuth  of  t  le  dep  h  for  a  rectangular  s.'rtion. 

Ex.31.  .1  Klraiijht  uronijhl-iron  bar  i.s  rnpahlc  oj  sustaining  as  a  strut  a 
weight  u\,  and  as  a  hram  a  uright  w.^  at  the  mitldle  point,  the  deflection  being 
umall  as  compared  with  the  transvernc  dimensions.  Ij  the  bar  has  simultaneously 
to  sustain  a  weight  was  a  ftrut  and  a  weight  w'  as  a  beam,  the  weight  being  placed 
at  the  middle  «/  the  pan,  show  that  the  beam  will  not  break  if 

w  +  ~w'<u\. 
w. 

Let  ^4  =  sectional  area  of  bar.    Then 


7  =  max.  allowable  compressive  stress 
•4 


w_ly 
4   /• 


developed  by  bending 


Therefore 


U'2 


4    /      4k'' 


Again,  the  total  maximum  stress  due  to  w  and  v/ 


w      w'  ly 

'^a'^1  1' 


Therefore 


w  w'  III  w, 
— I —  —  <  — 
^4    /      .4 ' 


or 


U'  +  i/''— <U',. 

w. 


Ex.  .32.  The  inclined  beam  0.4,  30  //.  in  length  and  carrying  a  uniforndy 
distributed  load  oj  100  lbs.  per  jont  oj  length,  is  supported  at  A  and  rests  against 
a  smooth  irrtical  surjace  at  0. 

Till'  resultant  weight  is  vertical  and  acts  throuph 
the  (vntrc  ('  of  0.4 ;  the  reaction  H,  at  O  is  hori- 
zontal. 

Let  the  directions  of  these  two  forces  meet  in 
li.  For  equilibrium  the  reaction  R,  at  .4  must 
also  pass  throuph  li.  ' 

Let  the  vertical  throuph  ('  meet  the  horizontal 
through  .1  in  I). 

The  triangle  .1/?/)  is  a  triangle  of  forces  for  the 
three  forces  which  meet  at  R.  and 


/?,       \n      AD      1 


V3^ 

o 


i! 


OBLIQUE  FORCES  ACTING  OS  A  BEAM.  477 

111,-  anjilp  OAD  being  30°.     Therefore 

i?,=1000V3'=  1732  lbs. 

Consider  a  seetion  .1/A'',  jx>rix>n(licular  to  the  beam,  at  a  distance  x  from  0. 

Tlic  only  forces  «ii  the  left  of  MN  are  //,  and  the  weight  ujxjn  OK.    This 

lit  is  l(K)x  lbs.,  and  its  resultant  acts  at  the  centre  of  OM,  i.e.,  at  a  distance 

'^  f.oin  .MX. 

The  component  of  li,  along  the  beam 

„  o     2000  cos' 30     ,,^,. 
=  «,  cos  30°  =— —  -;-„„  =1500  lbs. 
'  2     sin  30 

The  component  of  fl,  perpendicular  to  the  beam 

2000 


/?,  sin  30°  ==^cos  30°  =500v^  =866  lbs. 


The  component  of  lOOx  lbs.  along  the  beam  =  100j;  sin  30°=.50x  lbs.; 

"  "     "       •'     perpendicular  to  the  beam  =100x  cos  30° 

=86.6j;  lbs. 

I  Icnce  the  total  compression  in  |x)unds  at  NM  =  1500  +50x  =C'i, 

"       "    shear  '     "     =  860-80.6x=Sx. 

Ihc  K.M.  at  A'  =8C6x  -43.3x'=.1/x. 

C       y 
;inii  tlic  maximum  stress  developed  in  the  beam    ="7"=*=  J-^^^> 

.1  liiiiiij  liie  sectional  area  of  the  beam  and  /  its  moment  of  inertia. 

riu'se  expressions  may  be  interpreted  graphically  as  already  described, 
(  ,.  Sx  l)oing  represented  by  the  ordinates  of  straight  lines,  and  M^  jy  by 
til'-  (ifiliiititcs  of  parabolas. 

; ;,  f<if  example,  consists  of  two  parts  which  may  be  treated  independently. 
jir.nv  OE  and   AF  perpendicular  to  OA,  and 
n-^pcctivcly  equal  or  profxirtional  to 


1.500 
A 


and 


2.500 
.1   ■ 


loin  EF.  The  unit  stress  at  any  point  of 
till-  l)c;iin  due  to  direct  compression  is  represented 
\'\  i!ii  ordinate  (drawn  parallel  to  OE  or  AF) 
fioii!  that  point  to  EF. 

''l">ii  the  line  GG'  drawn  through  the  middle  point  B  perpendicular  to 


n.t.t:il 


kc  HG-^RG',  equal  or  projHirtional  to  j  ^^^20  cos  30°=4.3.30y-.     .\ccord- 

iiti  .1-^  l!ie  stress  due  to  the  beniliiig  action  at  any  jwint  of  the  beam  is  com- 
pr --ivc  or  tensile,  it  is  represented  by  the  ordinate  (drawn  parallel  to  OE 

i  .1  /'i  from  that  point  to  the  parabola  OGA  or  OG'A;  G  and  G'  respectively 

i»  iiiv:  the  vertices,  ai     GG'  a  common  axis. 


478 


THEORY  OF  STRUCTURES. 


By  superposing  these  results,  the  parabolas  EHF,  EH'F  are  obtained, 
the  ordinates  of  these  curves  being  respectively  proportional  to  the  values 
of  /»  for  the  compressed  and  stretched  parts  of  the  beam,  i.e.,  for  the  parts 
above  and  below  ihe  iieutral  surface. 

i6.  Beam  Acted  upon  by  a  Bending  Moment  in  a  Plane  which  is 
not  a  Principal  Plane.— Let  XOX,  FOF  be  the  principal  axes  of  the 
plane  section  of  the  bea-n. 


Let  the  axis  MOM  of  the  bending  moment  M  make  an  angle  a 
with  OX. 

M  may  be  resolved  into  two  components,  viz., 

M  cos  a  ■■=  X    and    M  sin  «  =  F. 

These  components  may  be  dealt  with  separately  and  the  restilts 
superpoeed. 

Thus  the  total  stiess,  /,  at  any  point  (x,  y) 

-stress  due  to  A' +  stress  due  to  7  =  '7^ +  -7—-/, 

1  z         '  y 

If,  ly  bring  the  moments  of  inertia  with  respect  to  the  axes  XOX, 
YOY ,  rospectivoly. 

If  the  point  {ly)  is  on  the  neutral  axis,  then 


ct 


tan/9"^=  -  y,  =  -  7-  tana. 


0  being  the  angle  between  the  neutral  axis  and  XOX. 


FLASGED  GIRDERS.  ETC. 


479 


17.  Flanged  Girders,  etc. — Beams  subjected  to  forces,  of  wh'ch 
thi'  lines  of  action  are  at  riglit  angles  to  the  direction  of  their  length, 
arc  usually  termed  Girders;  a  Semi-girder,  or  Cantilever,  is  a  girder 
with  one  end  fixed  and  the  other  free. 

It  has  been  shown  that  the  stress  in  the  different  layers  of  a  beam 
increases  with  the  distance  from  the  neutral  surface,  so  that  the 
111  )st  effectiA-e  distribution  of  the  material  is  made  by  withdrawing 
it  fro.n  the  neighborhood  of  the  neutral  surface  and  concentrating 
it  ill  those  parts  which  are  liable  to  be  more  severely  strained.  This 
consideration  has  led  to  the  introduction  of  Flanged  Girders,  or 
Tnixses,  i.e.,  girders  consisting  of  one  or  two  flange''  (or  chords), 
united  to  one  or  two  webs,  and  designated  Single-webbed  or  Double- 
webbed  {Tubular)  accordingly. 


Fia.  484. 


Fio.  485 


The  web  may  be  open  like  lattice-work  (Fig.  484),  or  closed  and 

continuous  (Fig.  485). 

T    T     X     X    '11 


Fio.  486. 


Fio.  487. 


Fio.  488. 


Fio.  489. 


Fio.  490. 


The  principal  sections  adopted  for  flanged  girders  are: 

Ihe  Tee  (Figs.  486  and  487),  the  /  or  Double-tee  (Figs.  488  and 
4n(i),  the  Tubular  or  Box  (Fig.  490). 

Chumfication  of  Flanged  Girdfrs.— Generally  speaking,  flanged 
Siinlrrs  may  be  (Hvided  into  two  classes,  viz.: 

I.  Girders  with  Horizontal  Flanges.  In  these  the  flanges  can  only 
convey  horizontal  stresses,  and  the  shearing  force,  which  is  vertical, 
inii-t  ite  wholly  transmitted  to  the  flanges  through  the  medium  of  the 

If  the  web  is  o|)en,  or  lattice-work,  the  flange  stresses  are  trans- 
niiiicd  through  the  lattices,  or  diagonals. 

If  tiip  web  is  continuous,  the  distribution  of  stress,  arising  from 
tb  transmission  of  the  shearing  force,  is  indeterminate,  and  may  lie 


4S0 


THEORY  OF  STRUCTURES. 


IM      T 
Fig.  491. 


in  certain  curves;  but  tlio  stress  at  every  point  is  resolvable  into 
vertical  and  horizontal  components.  Thus  the  portion  of  the  web 
adjoining  the  llanges  iM'ai  a  part  of  the  horizontal  stresses,  and  aids 
the  flanges  to  an  extent  dependent  ui;on  its  thickness. 

With  a  thin  web  this  aid  is  so  trifling  in  amount  that  it  may  be 
disregarded  without  serious  error. 

II.  Ciirdeis  with  one  or  both  Flanges  Cur\ed.     In  these  the  shear- 
ing stress  is  borne  in  part  by  the  flanges,  so  that  the  web  has  less 
duty  to  iHTform  an<l  recjuires  a  proportionately  less  sectional  a  ea. 
Equilibrium  of  Flanged  Girders.— AB  is  a  girder  in  equilibrium 
^       ^,  under    the    action    of    external    forces, 

and  has  its  upjx^r  flange  compressed 
and  its  lower  flange  extendetl.  Suj)- 
pose  the  girder  to  Im^  divide<l  int'i  two 
segiiients  bv  an  imaginary  .tical 
plane  MX.  Consider  the  segment  AMX.  It  is  kept  in  equiiit)riuin 
by  the  external  forces  on  the  left  of  MX,  by  the  flange  compression 
at  -V  (=r).  by  the  flange  teiL^iion  at  M  (  -T),  and  by  the  vertical 
and  horizontal  web  forces  along  MX.  Tiie  horizontal  web  forces 
may  lie  disregurd"(l  //  the  urh  is  thin,  while  the  vertical  web  forces 
jtass  through  M  and  .V  and  con.secju  Mtly  have  no  moments  about 
these  points. 

Let  d  be  the  efffctire  depth  of  the  girder,  i.e.,  the  distance  between 
the  points  of  application  of  the  resultant  flange  stresses  in  the  plane 
MX. 

Take  monients  about  M  and  .V  successively.     Then 

r(/  =  the   algebraic   sum   of   the   moments   about    M  of 
the  external  forces  uptjii  .l.U.V 

=  the  R.M.  at  MX^M. 


Similarly, 
Therefore 


Td    M. 
Cd-M-Td,    and    C    T. 


Hence  the  flange  forces  at  any  vertical  section  of  a  girder  with 
hnrizd!!!:'.!  fl'injres  ar(>«<jual  in  mii!rnitud<'  !>ut  opposite  in  kind.  The 
flange  force,  whether  compiessive  or  ten.sile,  will  Im-  denoted  by  F. 


FLASGED  GIRDERS,  ETC. 


4S1 


l.ft  /i,  fz  be  the  unit  stresses  at  MN  in  the  lower  and  upper 
flanges  respectively; 
oi,  02  "  the  sectional  areas  at  MN  of  the  lower  and  upper 
flanges  resix!ctively. 
Then 

-'^I  =  |l  =  M  =  Fd  =  a,f  id  =  aaf  2d, 
V       R 


mill  the  sectional  areas  are  inversely  proportional  to  the  unit  stresses. 

This  assumes  that  F  is  uniformly  distributed  over  the  areas 
fl,.  oj.  so  that  the  effective  depth  is  the  \-ertical  distance  between 
(Tilt res  of  gravity  of  these  areas.  Thus  the  flange  forces  at  the 
(•.iiins  of  gravity  are  taken  to  be  equal  to  the  maximum  forces, 
ami  the  resistance  offered  by  the  web  to  bending  is  disregarded. 
Till'  t'lTor  due  to  the  former  may  lx>come  of  importance,  and  it  may 
lie  tniiiid  advisable  to  make  the  effective  depth  a  geometric  mean 
iiciwccii  the  depths  from  outsiile  to  out.side  and  from  inside  to  inside 
(if  the  flanges.  

Tims,  if  these  latter  depths  are  hi,  hi,  the  effective  depth  =  V''/ii/i2- 

Ix.  3.3.  .1  flanged  girder,  of  which  the  effective  deplh  is  10  ft.,  rents  upon  tuo 
.<(/./),.;/.<  SO  //.  apart,  and  earrien  a  uniformly  distributed  load  of  2500  lbs.  per 
liniiil  jiMit  Delerrnine  the  flange  force  at  10  //.  from  the  end,  and  find  the  area 
•■j  tl„  flitnge  at  this  ptrint,  so  that  the  load  on  the  metal  per  square  inch  may  not 
!/(..,/  1(1,000  lbs.  in  tension  and  80(K)  lbs.  in  compression. 

\'Uf  vertical  reaction  at  each  HU|)i)ort 


SOX '.'.WO 


=  100,000  lbs. 


111.  I.  firo       FXIO-.U  -=  lOOOOOX  10-J500X  10X5  -875,000  ft.-lbs. 
;ii!l  /•'=S7")<X)11)S. 

riuw  thi"  Mjctioiiiil  area  of  tension  ''""K^' "■  ,,w^jq~  8.75  sq.  ins. 
ami  ' conip         "      -  ,^^-10.94  sq.  ins. 

1  \  .{».  .1  continmuH  lattice-girder  is  supported  at  four  jxiintit,  each  of  the 
,«"'.  P,i,is  iHi.mi  140  ft.  It  IH.  IM  len(;th,  2J  ft.  .1  i>i.  in  dej>lh,iind  utiyhtny  .08 
(<".  /•.  r  liitiiU  fool.     On  one  occasion  an  cxcemive  load  lifted  the  end  of  one  of 


■■wnf 


4g2  THEORY  OF  STRUCTURES. 

the  side  xpans  off  the  abutment.     Find  the  eonsequent  intensUy  of  stress  in  the 
bottom  flange  at  the  pier,  where  ,7.s  seetional  area  is  127  sq.  ins. 


Therefore 


/Xl27X-'2i  =  l40iJX.Gaxi(l40H)- 
/=2.39  tons/sq.  in. 


i8.  Examples  of  Moments  of  Resistance  of  Flanged  Girders,    (a) 

Douhle-tee  section. 

Firi^t,  suppose  the  web  to  be  .^^o  thin  that  it  may  be  (hsregarded 

without  sensible  error. 

Let  the  aeutral  axis  pass  through  C,  the  centre  of  gravity  of  the 

section. 

Let  m,  as  be  ihe  sectional  area.s  of  the  lower  and  upper  flanges 
respectively,  and  assume  that  each  flange  is  concentrated  at  its  centre 

line. 

Let  /ii,  A 2  be  the  distances  of  these  centre  lines  from  G. 

Let  hi+h2  =  c[. 

Approximately,  I  -  iihi^  +  a2li2^. 

Also,         {ai+a2)hi=a2d    and     (01+02)^3= aid. 

•        Therefore      I  =  a.(^^;^)    ^^^i^T^J   =7^^ 
Fio.  492. 

Again,  if  /i,  fs  are  respectively  the  unit  stresses  in  the  metal  of 
the  lower  and  upper  flanges, 

fi  U 

the  moment  of  resistance  =  — I  =  fiaid  =  ^I=-  f2*2d. 

-=the  B.M.  at  the  section. 

If  ai-a2  =  n,  then  jx^h^j,  suppose,  and 

the  moment  0/  resistance  ^iiA. 

Second.    Let  the  web  Iw  too  thick  to  Ih'  neglected. 

As  Iwfore,  let  the  neutral  axis  pass  through  (1,  the  centre  of  gravity 

of  the  section. 

Let  r;,,  a.  1h>  the  sectional  areas  of  the  lower  and  upjK'r  flanges 
respectively,  mul  assume  that  each  flange  is  concentrated  at  its 
centre  line. 


MOMEXTS  OF  RESISTANCE  OF  FLANGED  GIRDERS.         485 

Let  03,  04  be  the  sectional  areas  of  the  portions  of  the  web  below 
luul  above  G  respectively; 

Let  hi,  h2he  the  distances  from  G  of  the  lower  and  upper  flange 

centre  lines. 

Let  hi  +h2'=d. 
Approximately, 

/hi^    hi\   .      ,  o  .      /hi'  .  h^\ 


I  =  ai/i,2+03(^+^)  +02/122  +  04(^2+^) 
=  (a.+^)h?+(a.+^)h|. 


c t> 

^3 


Fig.  493. 


am 


/l  /2 

the  moment  of  resistance  = -r-I  =-r-I 
'  hi      n-i 

=  the  B.M.  at  the  section. 

.\gain,  let  A'  be  the  sectional  area  of  the  web  and  take  momenta 

about  G.    Then 

A' 
aihi  ^ — (hi  — hz)  =a2h2, 


or  aif  1  + — (f  1  —  f  2)  =  azf  2. 

.4'  d 

Ifoi-aa-.A,    and    a3=04  =  -:7,    then    /ii=A2  =  2> 


and 


Ilitice 


aiul 


/i=/2=/,  suppose. 
,     /.     A\d2     /^     A'vd2 


/       A\d2 

=  (^+6)7' 


the  moment  of  resistance  =  O  v'^  ■*"  "o"/  2" 


=  f(A  +  ^)d. 

/ 1"  iiig  the  stress  in  either  flange. 

It  is  important  to  rcmcMilKT  that  /i  and  /a  are  the  stresses  de- 
vrinjK'd  at  the  centre  lines  of  the  two  flanges  and  that  the  greatest 


484 


THEORY'  OF  STRUCTURES. 


stresses  are  devolojwd  ut  the  points  most  distant  from  the  neutral 
axis.      Thus  the  maximum  stresses  in  the  section  are  /i(l  +  2W 

and  /i-  (l  +  ry  r)  •  U  and  (2  being  the  thickness  of  the  lower  and  upper 

tiangcs  respectively. 

Thus  the  web  aids  the  girder  to  an  extent  equivalent  to  the 
increase  which  would  be  derived  by  adding  one  f<ixth  of  the  web  area 
to  each  flatige.  In  practice  the  web  is  usually  considered  as  aiding 
the  tlange  to  the  extent  of  one  eighth  instead  of  one  nixth  of  its  area. 
Approximately  this  makes  allowance  for  the  rivet-holes  cut  out  of 
the  web  in  making  connections  with  the  flanges  and  stiffeners.  Some 
speciHcations  altogether  disregard  the  effect  of  the  web  in  resisting 
IxMiding  and  a  somewhat  higher  flange  stress  is  then  allowable.  The 
former  is  probably  the  preferable  plan,  as  it  encourages  the  use  of 
thicker  webs,  which  may  add  considerably,  especially  in  expo-sed 
.situations,  to  the  life  of  the  structure. 

Again,  if  the  weight  of  the  material  in  such  a  Innim  remains  con- 
stant, M  increases  with  d.  At  the  same  time  the  thickness  of  the 
wel)  diminishes,  its  minimum  value  Ix'ing  limited  by  certain  practical 
considerations  (Art.  19).  Hence  it  foUows  that  the  distribution  of 
material  is  most  effective  when  it  is  concentrated  as  far  as  possibh 
from  the  neutral  axis. 

Tlie  principles  of  (>conomic  construction  require  a  l)eam  or  girder 
to  Ix'  designed  in  such  a  manner  as  to  \^o  of  uniform  strea;,'th,  i.e., 
cMpially  strained  at  every  point.  -Vn  exception,  however,  is  usually 
made  in  tlie  case  of  (inilkr  iK'ams  or  girders.  The  fibres  of  this 
materi.'d  are  real  hlnes  and  offer  the  most  "ffective  resistance  in  the 
direction  of  their  h-ngth,  .^o  that  if  they  are  cut,  their  remaining 
strength  is  due  only  to  cohesion  with  tiie  surrouiuling  material. 
Resides,  there  is  no  economy  to  l)e  gained  by  removing  a  lateral 
portion,  as  the  wiiste  is  of  little,  if  any,  practical  value. 

The  correct  numient  of  inertia  (lo)  with  resjK'ct  to  the  neutral 
axis  of  the  section  shown  by  Fig.  W:\  may  be  conveniently  found  in 
the  following  manner: 

Let  ^,  uii'l  1.2  1m>  the  thickness  of  the  bottom  and  top  flanges 
resjK'Ctively; 
h  1k'  the  depth  of  the  web; 


MOXfEXTS  OF  RESISTANCE  OF  FLANGED  GIRDERS. 


4S5 


/be  the  moment  of  inertia  of  the  section  with  respect  to 
any  main  bounding  line,  say  AB.    Then 
/=  moment  of  inertia  of  bottom  flange  with  respect  to.iB 


+ 


a,/i2 


"      "       "  top  flange  "         "       " 


ami  if  y  is  the  distance  between  AB  and  the  parallel  axis  through  G, 

I(i=I-(a,+A'+a2)y2. 

If  the  web,  instead  of  being  rectangular,  gradually  increases  in 
width  from  top  to  bottom  (Fig.  494),  it  may  be  sub- 
divided into  a  rectangle  of  area  i.'  and  two  triangles 
cMcli  of  area  a". 

Tlicn  the  moment  of  inertia  of  the  web  with  respect 
to  .'IB  1  IG.  494. 

=  moment  of  inertia  of  rectangle  with  respect  to  AB 
-t- 2  X  moment  of  inertia  of  a  triangle  with  respect  to  AB 

Built  Beams. — The  moment  of  inertia  (/)  of  a  built  beam  sym- 
mftrical  with  respect  to  the  neutral  axis  may  be  determined  as 
follows : 

Let  Fig.  495  represent  the  section  of  such  a  beam,  composed  of 
ciiual  flanges  connected  with  the  web  by  four  equal  angle-irons. 

L / J  Let   the  dimensions   be  as  shown  on  the 

-f~  figure. 

Then  /  =  the  moment  of  inertia  of  the  rect- 
angle fhi,  diminished  by  tmce 
the  sum  of  the  moments  of 
inertia  of  the  rectangles  oAj, 
bhs,  and  cht 


XT""'*' . 

II     II 
II     II 

— II?! 


,1 — 


.-Ji 


^a 


fh 


ahi' 


Fio.  495. 


12  ~^W'2 


12 


rht\ 


"4  \ 

+  ^+T2-) 


■  -^  I  fhi3  -  2(aha3  +  bhs^*  +  cht^)  I . 


486 


THEORY  OF  STRUCTURES. 


In  this  value  of  /  the  weakening  effect  due  to  the  rivet-holes  in 
the  tension  flange  has  been  disregarded.  If  it  is  to  be  taken  into 
account,  let  p  be  the  diameter  of  the  rivets. 

The  centre  of  gravity  of  the  section  is  now  moved  towards  the 
compression  flange  from  its  original  position  through  a  distance 


IJL. 
4.1' 


x-ii-Ah\-hl), 


and  the  moment  of  inertia  of  the  net  section  with  respect  to  the  axis 
through  the  new  C.  of  G.  is 


I-A'x^-^,(h\-hl), 


A'  being  the  net  area  of  the  section. 

If  the  web  is  of  the  open  ty[)e  and  if  e  is  the  thickness  and  h^ 

the  depth  of  the  open  part,  then  — ^-  must  be  subtracted  from  the 

value  alreaily  obtained  for  /. 

If  the  beam  is  unsymmetrical  with  respect  to  the  neutral  axis, 
its  Iq  may  be  determined  by  first  of  all  finding  the  moment  of  inertia 
/'  with  resiK'ct  to  an  axis  coincident  with  one  of  the  main  bound- 
ing lines,  as  already  described.    Ther. 

lG=r-Ay2, 

A  being  the  area  of  the  section  and  y  the  distance  between  the  two 
axes. 

Effective  Lcnijth  nnd  Depth. — The  effective  length  of  a  girder  may 
be  taken  to  be  tiu-  distance  from  centre  to  centre  of  l>earings. 

The  effective  depth  dejK-nds  in  part  upon  the  character  of  the 
web,  but  in  the  liuculation  of  flange  stresses  the  following  approxi- 
mate rules  are  .>^ufficiently  accurate  for  practical  purposes: 

If  the  wci)  is  continuous  and  very  thin,  *he  effective  depth  is 
the  full  depth  of  the  girder. 

For  plate  girders  the  effective  depth  is  usually  taken  as  the 
di.stanof  bctwrcii  th(>  erntn's  of  gravity  of  the  flanges.  In  tlie  event 
of  cover-plates  lM>ing  used  thi^  depth  should  not  exceed  the  distance 
back  to  back  of  the  flange  angles.     The  rivet-holes  in  the  tension 


MOMENTS  OF  HESISTAXCE  OF  FLANGED  GIRDERS.         487 


flaii<ro  are  sometimes  deducted  in  finding  the  C.  of  G.  of  that 
flaiiRO. 

For  open-webbed,  pin-connected  and  riveted  girders  the  effective 
depth  is  the  distance  between  the  centres  of  gravity  of  the  upper 
ami  lower  chords.  In  the  former  this  should  always  be  the  same 
as  tlu>  distance  between  the  centres  of  the  pins. 

If  the  flanges  are  cellular,  the  effective  depth  is  the  distance 
hot  ween"  the  centres  of  the  upper  and  lower  cells. 

Mx.  .34.  The  flangex  of  a  girder  are  of  equal  sectional  area,  and  their  joint 
an .(  (s  cijual  to  that  of  the  web.  What  must  be  the  sectional  area  to  resist  a  bend- 
jm/  Miimriit  of  300  in.-tons,  the  effective  depth  being  10  ins.  and  the  limiting  inch- 
stris!<  4  totisf  C 

Let  .l'=area  of  web; 
• '    d  -depth  of  web  - 10  ins. 


Thrri 

anil 
Hiiice 

ami 
Tlirrcfore 


area  of  each  flange  — — 

/  -A''^+  2'^  (Pj    =^A'd»,  approximately. 
300  in.-tons  « moment  of  resistance 

A' -Hi  sq.  ins. 
area  of  section  -2/1'  -22  J  sq.  ins. 


Fio.  496. 


i;\.  3.').  The  thickness  of  the  web  of  an  equal-flanged  I  beam  is  a  certain  frac- 
tfiii  ,./  the  drjUh.  Show  that  the  greatest  economy  of  material  is  realized  when 
ll:i  nnit  of  the  web  is  equal  to  the  joint  area  of  the  flanges,  and  that  the  moment 
oj  risisinitcc  to  bciuHng  ix  ijAh,  f  being  the  coejficient  of  strength,  A  the  tola' 
ficlKindl  orcn,  and  h  the  depth. 

let  a  =area  of  cvh  P.ungc; 
■  .1'=   "     "  wcb=m;i», 
vih  liciiig  tho  thickness  and  m  a  coefficient  less  than  unity.    Then 

2o  +  .1'  —a  minimum  -2o  +  mA'. 

Tiltreforc,  differentiating, 

2da+2mhdh-0 (1) 

Again,  the  moment  of  re8istance-/^a+— -j*-/(o+-r-j  A. 


m 


11 


488  THEORY  OF  STRUCTURES. 

Differentiating,  hda  +  (a  + —^J  dh  "0.  .     .    . 
Hence,  from  eqs.  (1)  and  (2), 

2        2mh  ' 

mh'    A' 
a  •=-—=—    and    A' =2a. 


(2) 


Fig.  497. 
Also, 

and 


or 


,4=.4'  +  2a=4a, 
the  moment  of  resistance  =/{"  +"2^  )h-=—fAh. 


Ex.  3t>.  r/if  lower  and  upper  flanges  of  the  section  oj  a  girder  are  1  in.  and 
1 J  ins.  thick  re.'>pectively,  and  are  each  24  ins.  u-ide;  the  effective  depth  of  the  girder 
is  48  in^  and  the  web  is  i  in.  thick.  Determine  the  position  of  the  neutral  axis, 
and  also  find  the  flange  unit  .Presses  when  the  bending  moment  at  the  given 
section  is  580  ft.-tons.     Using  the  preceding  notation, 

o,  =  24  sq.  ins. ,    a ,  =  36  sq.  ins. ,    and    aa  +  a4  =  24  sq.  ins. 

The  centre  of  gravity  of  the  web  is  at  its  middle  point.    Thus      c 


24/1,  +24{A,  -24)  =36(48 -/i,). 


I go"  144" 

Therefore       Ai=— i^   and   *,  =  -=—,  defining  the  position  of  G. 


Fig.  498. 


Again 


192    1      06  144   1      72 

03  =  —  T,  =y  sq.  ms.,    a,  =-—   -^  =y  sq.  ms., 


267264 


.     /„.     32\/192v'      /        24wl44\'    2672 
and  /=(24  +  y)(-)    +(36  +  y)(-) 

Hence  580xr2=moment  of  resistance  in  in.-tons 

J^  '2fi7'2M_  I,    267264 
^192       7      "144       3     ' 
7  7 

and  therefore        /,  =5  tons/sq.  in.    and    /2=3i  tons/sq.  in. 

Ex.  37.  In  a  douhlc-ftanged  ca.ft-iron  beam  the  thickness  of  the  iveb  is  a  cer- 
tain  fraction  of  the  depth  h,  and  the  maximum  tensile  and  compressive  intensities 
of  stress  are  in  the  ratio  of  2  to  5.  Show  that  the  greatest  economy  of  materid 
is  renlized  when  the  area,s  of  the  boHom  flange,  web,  and  lop  flange  are  in  the  ratio 
of  2.5  to  20  to  4,  and  that  thf  moment  of  resistance  to  bending  is  if  Ah,  where 
I —^j- xmaximum  tensile  intensity  of  stress  and  A  is  the  sectional  area. 


MOMENTS  OF  RESISTANCE  OF  FLANGED  GIRDERS.  489 

Let  the  neutral  axis  divide  ^le  depth  h  into  the  segmente  A,  and  A,     Then 
Jb-jA^l    and    h,+h,'h. 


Therefore 


and 


-AW4„.+25a,+fmA'). 


Fio.  499. 


Hence 


the  moment  of  resistance  ""^'  ^F^'  "7  V**'  +25ai  +  — mfc'j . 

Differentiating, 

4A-da,+25;i-da,  +  (4a,+25o,  +  19mA')dA-0.     .    . 
Taking  moments  about  the  centre  of  gravity, 

(..f)|»-(.+f,»»')|». 


or 

or  2o,-5a,-{m/i'=0. 

Differentiating,  2  •  do,  -  5  •  c?a,  -  3mA  •  dA  -  0. 

Also,  at+a,  +  mh*  •=  a  min. 

Therefore  da,+da,  +  2mhdh-0.      . 

Hence,  byeqs.  (1),  (2),  and  (3), 

doi  "da,  —  —  mhdh 
and  4o,  +  25a,- 10mA' -lOA', 

r  lx;ing  the  area  of  the  web. 
r.ut  2a,-5a,-JA'. 

Therefore  a,  =  »;4'  =  V««r 

A;.'Hin,  A-o,+i4'+o,-H-A'. 


(1) 


(2) 
(3) 


490 
Hence 


THEORY  OF  STRUCTURES. 


the  moment  of  resistance  =—  (5/1'  +5A'  +  i^^'f 


=  3.1'/i=.4/.|; 

where/ -4^X2. 

Ex.  38  A  beam  36  ft.  between  bearings  is  a  hollow  tube  of  rectangular  sec- 
ti'in  and  coti.tisis  of  a  24" Xi"  top  plate,  a  24" X  J"  bottom  plate,  and  two  side 
plntrs  each  35"  X  i".  The  platcx  arc  riveted  together  at  the  angles  of  tlie  interior 
r,ctangle  by  means  of  four  W  X  4"  X  J"  angle-irons,  the  d-in.  side  being  horizontal. 

First.    Disregard  effect  of  rivet-holes.    Then 

\2     12  V2  12  12     ^  12  ' 


and  the  moment  of  resistance 


=^X  16341  =40S5i  in.-tons. 


.\iso,  if  W  is  the  safe  uniformly  distributed  load, 
R'X36. 


8 


and 


X12=B.M.  in  in.-tons -4085i 


n'  =  7.5.653  tons. 


S----^iR 


Fio.  500. 


Second.  Take  the  effect  of  the  rivet-holes 
into  account  and  assume  them  to  be  of  1  in. 
diameter, 


The  gross  area  A    of  the  sectim  =""  sq.  in. 


7lCt 


I /    ( t    it 


■■  i\t  sq.  in. 


The  distance  x  through  which  the  centre  of  gravity  moves  towards  the 
compression  flange  is  given  by 

X  =  }  t',  (36'  -  34')  =  ?  J  in.  =  .46  in. 
The  /  with  respect  to  the  new  centre  of  gravity 

=  16341  -76(2a)'i'3(36'-34')  =15712.2. 
The  new  minimum  moment  of  resistance 

--„^^„X  15712.2  =3830  in.-tons, 
18.46 

and  if  W  is  the  new  safe  uniformly  distributed  load, 
IPX  36 


8 


i:  1 


Therefore 


X 1 2  -B.M.  in  in.-tons  -3830. 
IF' =70.93  tons. 


MOMENTS  OF  RESISTANCE  OF  FLANGED  GIRDERS.         491 

The  pitch  of  the  rivets  connecting  the  ang'es  with  the  upper  plate  is  4  ins., 
1(1  It  is  assumed  that  there  is  an  effective  width  of  5i  ins.  in  shear  for  each 


nvct. 


The  distance  between  the  neutral  axis  and  the  upper  face 

=36-18.46  =  17.54  ins. 
At  the  surface  RR  the  intensity  of  shear  q  is  given  by 
8X11=^(2-11X17.29) 


or 


g- 18.862  X-ptons/sq.  in. 


At  the  neutral  surface  00,  where  the  intensity  of  shear  is  greatest  and 
equal  to  3i  tons  per  square  inch, 

.3ixl=j(24x.jXl7.29  +  2x6x^X16.79  +  2X3iX2Xl4.79 

1      17  04\ 

+2Xl7.04XoX— TT-) 


=  yX505.1658. 


Therefore 


3i/  3^X15712.2 


and 


108.86      ,„  ,      ,      . 
q  =  18.862 X  r-712^  '^       ton/sq.  m. 


Iki.cc  if  the  plate  and  aig'.r  'aces  are  close  together, 

V  id'3i=.lCX4x5i 

and  d  =  1.02  ins.; 

if  tho  plate  and  angle  faces  are  not  close  together  so  that  the  rivets  are  subject 
to  a  1mm  ling  action  by  Ex.  23,  Art.  9, 


aiiil 


J(Vid'3i)=.13x4x5i 
d- 1.18  ins. 


l.x.  30.  Find  the  moment  of  resistance  of  a  section  composed  of  two  equal 
/'.»■(.;».'.•,  each  consisting  of  twn  600-mm.  X  7-mm.  plates  riveted  to  a  1200-mm.  X8-mm. 
in/.  i,l„t,'  h]t  means  of  two  100-mm.X  100-mm.X  12-mm.  angle-irons;  two  70-mm. 
:ii.,„,».XO-mm.  angles  arc  also  riveted  to  the  inner  faces  of  the  flanges,  the  ends 
(>/  Ihr  horizontal  arms  being  24  mm.  from  the  outside  flanges;  the  total  depth  of 
thi  xu-tinn  =3.228  m.,  and  the  interval  between  the  two  f  ':  plates,  which  is  open, 
U  2  m.;  coefficient  of  strength  =6  k.  per  millimetre^ 


E!l'! 


492 


THEORY  OF  STRUCTURES. 


^m 


souo 


^ 


Tl    I  i  !  i 

i  I      III! 


Disregarding  the  effect  of  rivet-holes, 

/  =  ,'8()OO..i22S'  -  ,»j  (L>4.X>0b'  +  61 .3182'  +  9.3060* 


+  10'J..J1.'00  +S8.:J176  +  12.3000  )  -A8.2000 
113W).')0789:j2.S 
12  ' 


Vt-M    I       ' 


and  the  moment  of  resistance 

6    ll.U)i).'>07S».32S 
~1614    12X(1000)' 


km.  =350.3565  km. 


■!0»-<1 
I     !«— -412— »(     I 

y ui H 

Fig.  501. 


Ex.  40.  .1  girder  iij  21  //.  span  has  anection  composed 
of  two  equal  flanges  each  coiinixtinij  of  two  3i"X5"Xi" 
angles  riveted  to  a  3!t"xi"  web;  the  cover-plates  on  the 
flanges  are  each  12"X  i",  and  the  rivets  in  the  covers  alter- 
natf  with  those  connecting  the  angles  and  web;  the  pitch 
of  the  rivets-  is  ^  ins.  Find  the  diameter  and  al.io  find  the  maximum  flange 
stresses,  (a)  disregarding  the  wtakening  effect  oj  the  rivet-holts  in  the  ten.iion 
flange;   {h)  taking  this  effect  into  account. 

The  hxid  upon  the  girder  is  a  uniformlg  distributed  load  of  20,S(k)  lbs.  {in- 
cluding weight  of  girder)  and  a  hmd  of  .")0,(KK)  lbs.  concentrated  ut  each  of  the 
points  di.stant  4J  //.  from  the  oiiddle  imint  of  the  girder. 

{a)  Disregarding  the  effect  of  riveting,  the  neutral  axis  is  at  O,  the  middle 
point  of  the  depth.    Then 

If  q  is  the  intensity  of  shear  at  the  surface  .S'.S', 
9lOi 


l()S!»0,i,,. 


IT'" 


'^xrjvgxiatt-*;^^''*-". 

/  '  /     32 


r-M  - 

Fio.  .502. 


3      S 


At  0,  3i  -  -  7-(12i  X  HHl  +2  x4i  x  i  X  I'J}  4  2  x 3j  x  J  x  171)  -- 
q  10|     2S35 


S  Ut7.')3 
84 


Therefore 


3i  i 


32 


64 
l'.»753' 


or    (y^.tXJOS  tong/sq.  in. 


There  is  one  rivet  to  each  (§  x3JxlO|)  sq.  ins. 

First.     Vssunie  that  there  is  close  contact  along  A'5,  so  that  the  rivets  are 
in  shear  only.     Then 


irrf' 


\. 


and 


3 J  =  working  strength  of  rivet  of  diam.  d 

-i  -3^  ■  10|X.(W03 
d-.48in.  in. 


MOMEWTS  OF  RESISTANCE  OF  FLANGED  GIRDERS. 


493 


Stcond.    If  the  contact  along  .S.S  is  not  dose,  the  rivets  are  subject  to  a 
Ix'iidiiig  action,  and  then 


("|-3i)  =  J  X3i  X  lOi X  .0363  (Kx.  23,  Art.  9) 


3  /nd' . 
4 
and  d^Mi  in.,  say  A  i"- 

•     TiikiniK  into  account  the   effect   of   riveting-,  and  using  a  A-in.  rivet, 
I  •  ri.utra!  axis  is  D;oved  to  G,  where 

9 
n<;  -    ,~^(39i'  -38')  =  „7,v„^.  =  .494  in.,  "  —  being  the  net  area  of  the  section. 


(>4 


Then 


Alsn 


19J  -OG  =  19.3S1  ins.     and     19J  +  ^W/  =  20.369  ins. 

/«  =/o  -^V  (.491)'  --^,  -^(39}'  -3.S')  -  10.-)08.6. 
64  12  10 


Thi-  maximum  B.M.  is  at  the  middle  point,  and  therefore 

B.M.a«.  =  ("'^*^21  +.50000X6)  12  =4,2.-)5,200  in.-lbs. 
—moment  of  resistance  in  in.-lbs. 


Hence  if  fi  and  /c  are  the  flange  tensile  and  compressive  stresses  respeo- 
tivclv, 


iiir.is..(o),  4255200 - 


19.875 


10890i  I ,     and    /.  =-  A  -  7766  Ibs./sq.  in. 


11.  cus,.  (M,  42.-)5200  -.^(j^.^X  10.-.08.6  -  J^-  X  I0.W.S.6, 

sntli.it     /( -^7847  Ibs./sq.  in.  and  ,r«S24S  Ibs./s*!.  in.     -  j--k— i,--J»-*J 

8|  '"*  [   I  r~    )  '\\  t~f~ 
1  A.  11.  Dftrrmine  the  ponition  of  the  neutral  axis,      V.XK  "''t  ,.h  y '  r-  \ 

th,  I'inmnit  ()/  rt-.sistnrirc,  thr  ratio  nj  the  maximum  to 
l>:  .inrnijr  iutrrinily  o]  shrar  n}  the  xrction  nhoum  l>y 
Fill,  "iii:!,  thr  oujficiniln  of  Mrnvjth  prr  square  inch 
'""•'  \\  tnnx  jiir  litisioit  and  omprrisinn  and  3 J  Ions 
/■'  '■"'■.  AIS4I  find  Ml'  diamitrr  d  nf  thf  rivets  li, 
<lir,,i,ir,lin<i  the  wnikeninf/  tffrrt  "/  the  rivet-holes 
'1'/  .';'  M/  the  pitch  to  he   I  inches.      Seutral  axis  is      ^ 

'If  <!.  ll.   centre  c/  gravity  '7  the  section;  a6— 5J|  ins.     '             ".          i 

-,-.        .                                               '  k  -  —  — **         —  •* 
llirntori' 

Fio.  603. 

+  12Xl)-12xlX17i-.5H  X  ix  I7i-ixHx*Xl7J+23xiX10 
f  10lxJx6t+12XJ+2x5JxJx|4t. 


494  THEORY  OF  STRUCTURES. 

and         G// -8.8171  ins.;  also OG-0//-C//=2.CS29 and GA' =9.1829. 

Apain, 

/„=t'2X12x1'  +  12  Kl  x6'-,Vx.5JJ(i)'-5Hx  i(5?)'-2  X  JiX^,(i)'-H 
Xix5i'  +  ix2xix(5i)'  +  j  x2xix3»  +  §x  i(10i)'  +  i>iX  12x  1'  + 

12X1(11)'-2047.S27. 

3347 


Hence 


/g=2047.S27 

334; 
90 


90 


(2.GS29)'  =  1797, 


=  17.4323  sij.  ins.  being  the  area  of  the  section. 

41 
The  least  moment  of  resistance  =  —:—  X  1797-880.64  in.-tons. 

Let  q  be  the  shear  stress  at  the  surface  SS.    Then 
g6-^12xj(8.9329). 

IG 

The  shear  stress  is  greatest  at  Q.    Therefore 

S  1761    S 

3ixi-^(ix7ix3i  +  12xlx8)--j-g    -^. 

S" /o' -28.574  tons,  and 

4    1701 

-«L  I  -A''  12x4x8.9329-.  142  ton/sq.  in. 
6  4    1701 


Hence 


Also, 


28^.574 
j4K'Ta(7P  shear  ft  res*       1 7.4.323 


-.234. 


.l/oJiMium  shear  stress        3.5 
/Vssaming  the  contact  along  SS  to  Ix;  close, 

^'3i-3^4x.l42    and    </ -.787  in.,  say  H  in. 
4 

If  the  contact  is  not  close, 

2.f'^''3ij-3X4x.l42    and    (/ - .9,  say  U  in. 

Ex.  42.  An  aqunlurt  jor  a  "pan  of  20  led  consuls  of  a  rast'iron  ehanntl 
(xnm  30  inn.  wuir  aiid  20  ins.  deep.  Find  the  thifkness 
of  the  metal  so  that  the  water  may  safely  rise  to  the  top  of 


t  a       the  channel,  the  safe  eoeffirient  of  strength  being  1  ton  per 

}■        \      ■  -4  •-       aquare  inch.     Find  the  mle  limHing  span  of  the  chanml 
I  under  its  own  weight. 


Fio  604. 


Let  t"  -tliicknesa  of  metul.    Then 


MOMENTS  OF  RESISTANCE  OF  FLANGED  GIRDERS         495 

(30  +  20  (20  +  0  -  600„„  450   35<  +  t*- 
he  weight  Tl',  of  the  aqueduct  in  tons  = , ,        2000  °"   16~" 


IT,"    "   .ater        "    <«    "'" jr-«^^-*2(i)0 


2000 
=2f|. 


If  G  is  the  centre  of  gravity  of  the  section, 


OG(2x20X<  +  <  30  +  20  =2x2OX<Xl0--X<3O  +  2« 


••ind 


Also. 


OG- 


400 -lot-/' 
70 +2/ 


;   0G  +  (=- 


400  +  5o<  +  < 
70+2( 


1    y 


,                                                1                                     <'     16000     3u+2(. 
/=lx2x20't+2X20Xtxl0'  +  j2(30  +  20<'  + (30+20(4 3— '  +  —3-'^ 


and 


Hence 


16000     30  +  2<       <(400-15/-/n' 
^°"     3      "'"      3"  70+2( 


/  W 

Moment  of  resistance  in  in.-tons  =^,  ^  ^=max.  B.M.  in  in.-tons-       g 


W, 


'20X12. 


The  proper  value  of  t  can  now  be  determined 
following  table : 


trial  as  shown  in  the 


i 
k 

1.7 


in  tons. 

.2744 
.6508 
.3663 
.3693 


B.M. 

in  in.-tnni. 

86  36 
94  56 
89.215 
89.203 


Momer  t  nf  Rmistance 
in  in. -tons. 

66  414 
132  35 

88.4 
89.14 


'a 
384  674 

776.88 
514  566 
518.736 


Hence  .17  in.  is  the  thickness  required. 

The  total  area  of  th'  section  « 1 1 .9.578  sq.  ins. 

If  L  feet  is  the  limiting  length  of  the  aqueduct  under  its  own  weight, 


^•■mo  !•'-»•»  "^  '■ 


-56.4  ft. 


I  A.  43.  .4  fast-iron  girder  130  ins.  Mwcen  supports  and 
IJl  Ins.  deep  had  a  top  flange  2i"Xi",  a  bottom  flange 
111"  ■  11",  and  a  web  i"  Ihiek.  The  girdir  jailed  under 
/w,/.s  (//  17i  tons  plaerd  at  the  two  /mnts  distant  3f  jt. 
jr-.m  aieh  support.  What  were  the  erntral  flange  .itreasc.-<  at 
I'-,  miinirnt  0/  rupture?  What  was  the  eentral  defleetion  when 
!'•  hiad  at  eneh  point  was  7^  tonsT  (£-18,000,000  lbs.; 
If  ,i]ht  0/  girder  -33(W  lb». ;   Urn  -2240  lb».) 

Area  of  top  flange        -2iX   J-ffsq.  in. 
"     "  web  -lOX   i-V  •'    " 

"     "  bottom  flange -10  Xlj-V  "    " 


T 


A 

Flu.  605. 


490  THEORY  OF  STRUCTURES. 

Let  X  =  distance  of  centre  of  gravity  from  A.    Then 

x(ii  +  V+V)=?|XllU+ j'x6J  5     and     j-=H8l  =3.617  ins. 

Also,  moment  of  inertia,  I  a,  with  respect  to  lowest  face  of  section  is  given  by 

/a  =ilO(|)'  +  i^i{10)'  +  y-(f.V)'+T'sS(i)'+t|{ntt)' =660.92573. 
Therefore  /«=660.92.TO- V»''(tlSn'=370.642.  ' 

Again, 

max.  B.M.-(17jX2240X3}X12+i?,^Xl39)  in.-lbs. 

=  1822.519  in.-lbs.  =  n-7/rz  X  .370.642  = -(^  X  370.642. 
3.t)17  )S..)US 


Hence 


/(  =  177%  It.s./sq.  in.     iiiid     /r  =41S:}»)  Ibs./sq.  in. 


These  results  arc  of  courst^  based  on  the  hy|M)thetical  assumption  that  the 
elastic  theory  of  the  transverse  strength  of  l)eams  holds  good  up  to  the  point 
of  failur.'. 

Again,  by  Ex.  12, 
the  deflection  due  to  the  7i-ton  concentrations 

_7iX2240x3iXl2;T39'_45\      23554  in., 
ISOOOOOUX  370.642  V  S  6  / 

tho  deflection  due  to  the  weight  of  the  girder 


:?3tiHxi;i<r 


-=.0176.5, 


:1S4  ISIKKKMWX  370.641! 
and  the  total  central  deflection  =.2.').32  in. 

19.  Design  of  a  Girder  of  an  I  Section  with  Equal  Flange  Areas, 
to  Carry  a  Given  Load. 

.\t  any  point  (list  iiit  .r  from  tlio  niiddlo  of  tlio  girder,  let  y  be  the 
depth  of  tlie  nirder,  .1  the  sectional  area  of  each  fiange,  A'  the  sec- 
tional area  of  the  web,  M,  the  IVM.  and  S,  the  S.F.     Then 

f(A^^y^M„ 

f  beiiiK  tho  safe  unit  stress  in  tension  or  roinprr.ssion. 

It  is  in  accordance  with  cood  jiractice  to  assiuiie  that  the  flanges 

.1'  .  A 

are  assisted  by  the  web  to  the  extent  of  —  instead  of  ^. 


(llliDER  OF  AN  I  SECTION   WITH  EQUAL  FLANGE  AREAS.  497 

U'eft.— Assume  that  the  web  transmits  the  whole  of  the  shearing 
force.  This  is  not  strictly  correct  if  the  flange  is  curved,  as  the 
tliiiige  then  bears  a  portion  of  the  shearing  force.  The  error,  however, 
is  on  the  safe  side. 

Theoretically,  the  web  should  contain  no  more  material  than  is 
absolutely  necessary. 

Let  /,  be  the  safe  unit  stress  in  shear.    Then 

and  the  sectional  area  is,  therefore,  independent  of  the  depth. 

A'    S 
The  thickness  of  the  web  =  —  ^j^, 

y    i-y 


but  this  is  often  too  small  to  be  of  any  practical  use. 

I'.xiK'rience  indicates  that  the  minimum  thickness  of  a  plate 
;vlii(li  has  to  stand  ordinary  wear  and  tear  is  about  |  or  -fg  in., 
while  if  subjected  to  saline  influence  its  thickness  should  be  f  or 
i  ill,  Thus  the  weight  of  the  web  rapidly  increases  with  the  depth, 
ami  the  greatest  economy  will  be  realized  for  a  certain  definite 
ratio  of  the  depth  to  the  span. 

riie  thickness  of  the  web  in  a  cast-iron  girder  usually  varies 
from  1  to  2  ins. 

In  the  case  of  riveted  girders  with  plate  webs  of  medium  size 
all  practical  requirements  are  effectively  met  by  sjxicifying  that 
tlic  siicaring  stress  is  not  to  exceed  one  half  of  the  flange  tensile 
stress,  and  that  stiffeners  are  to  ]tp  introduced  at  intervals  not  ex- 
oee'liiiji  twice  the  depth  of  the  girder  when  the  thickness  of  the 
well  is  less  than  one  eightieth  of  the  depth.  Again,  it  is  a  common 
I'nii  tieal  rule  to  stiffen  the  web  of  a  plate  girder  at  intervals  aj)- 
prnxiiniitoly  equal  to  the  depth  of  the  girder,  whenever  the  shear- 

iiit:  ^tn>ss  in  pounds  per  square  inch  exceeds  12000h- (1+— ^^, 

//  l"'!!!g  the  ratio  of  the  depth  of  the  web  to  its  thickne-ss. 

I'lamjes. — First.  Assume  that  the  flanges  have  the  same  sec- 
tional area  from  end  to  end  of  girder. 


498 


THEORY  OF  STRUCTURES. 


If  the  effect  of  the  web  is  neglected,  und  taking  /  as  the  coeffi- 
cient of  strength, 

Mr 

and  the  depth  of  the  beam  at  any  point  is  proportional  to  the  ordi- 
nate of  the  hcnding-niornent   curve  at  the  same  point. 

For  example,  let  th(>  load  be  uniformly  dist'-ibuted  and  of  in- 
tensity ir,  and  let  /  be  the  span.     Then 


m.-'-Kj-'-). 


and  the  beam  in  elevation  is  the  parabola  ACB,  having  its  vertex 

at  C  and  a  central  depth  C0  =  ~.    The  depths  thus  determined 

are  a  iittle  greater  than  the  depths  more  correctly  given  by  the 
equation 

Second.    Assume  that  the  depth  y  of  the  girder  is  constant.    Then 

.         .1'        Mr 

b     ly 

and  neglecting  the  effect  of  the  w(>b,  the  area  of  the  flange  at  any 
point  is  proportional  to  the  ordinate  of  the  curve  of  bending  mo- 
ments at  the  same  point. 

Let  the  load  be  uniformly  distributed  and  of  intensity  w;  also, 
let  the  flange  be  of  the  same  uniform  width  b  throughout. 

The    flange,  in   elevation,  is  then  the   parabola  ACB,  Fig.  .WS, 

having  its  vertex  at  C  and  its  central  thickness  ^0  =  ^^.    Huch 

iM-ams  are  usually  of  wr(»(inht  iron  or  steel,  and  arc  built  up  by 
means  of  plates.  It  is  impracticable  to  cut  the.se  plates  in  sucli  a 
manner  as  to   make  the   curved  boundary   of  the   flange  a  true 


GIRDER  OF  AX  I  SKCTIOX   MITII  F.QIAL   FLASdE  AREAS.  490 

li;ir;ili()l:i  (or  any  otluT  curve).     Ilonce  the  flange  is  generally  con- 
.-iriK'tcd  as  follows: 

Draw  the  curve  of  bending  moments  to  any  given  scale.  By 
iillcring  the  scale,  the  onlinates  of  the  same  curve  will  represent 
the  (lange  thicknesses.  Divide  the  span  into  segment.s  of  suitable 
Iciifillis. 

I'roni  .1  to  1  and  B  to  7  the  thickness 
(i!  ilic  flange  is  1«  =  7/;  from  1  to  2  and 
7  to  i;  the  thickness  is  26  =  6e;  from  2 
td  ;!  and  G  to  5  the  thickness  is  3c  =  5<i; 
aiiil  from  3  to  5  the  thickness  is  CO. 

AT      1 ' 

Tiie  more  correct  value  of  a[  =  -j y)    is  somewhat  less  than 

tlutt  now  determined,  but  the  error  is  on  the  safe  side. 
•     .\gain,  at  any  section 


EJ1[ 
R 


y 


and  hence    Rocy  the  depth. 


Tims  the  curvature  iliminishes  as  the  depth  increases,  so  that 
a  L'irder  with  horizontal  flanges  is  sufxrior  in  point  of  stiffness 
to  (iiie  of  the  parabolic  form.  The  amount  of  metal  in  the  web  of 
the  latter  is  much  less  than  in  that  of  the  former.  If  great  flexi- 
hility  is  recjuired,  as  in  certain  dynamometers,  the  parabolic  form 
i:<  of  course  the  best. 


I.x.  44.  Design  of  50-ft.  plate-girder  .ipan. 

rill'  (lat.a  to  l)p  vised  are  jw  follows: 

Live  load  40()0  lbs.  per  lineal  foot  of  span,  or  an  axle  concentration  of 

44,000  11)3. 

400 
rcrcoiitap'  to  be  added  to  live  load  for  impact  -  f-.v^jA, 

will  ri'  /, -=letigth  of  span  over  which  the  load  giving  the  maximum  stress  is 
distributed. 
I  )i  -taiice  centre  to  centre  of  Ijearings  =50  ft. 

'      "       "      ' '  pirdcrs  ■=■   7  " 

AH'iwmMc  i!aii2("  stress  =  14,000  lbs.  per  sq.  in. 

shrariiig  stress  in  welw  and  rivet  -  10,000  ' " 

"         Uarinu      "      in  rivets  -20,000" 

"  fibre  stress  in  timber  -  'i.OOO  " 

Wi-  limy  affflume  that  the  rail  distributes  the  axle  eoneentration  equally 
ovtr  ihnc  ties;  hence  each  tie  will  be  Uiaded  as  shown  in  Fig.  519.     .As  the 


( (    It    tt 

tt    ti    ti 


600 


THEORY  OF  STRUCTURES 


ties  will  not  greatly  exrecd  1  ft.  centre  to  rentre,  the  span  for  three  ties  may 
be  taken  as  2  ft.  and  the  impact  allowance  80  per  cent. 
Therefore 

B.M.  =7333X12X1.8  =  158,400  in.-lbs. 

Making  the  tie  7  ins.  wide  to  s  fure  a  good  bearing, 


g  P 

_tiz!z::l_ 

\^ — -V' -  _4 
Fig.  509. 


where  <l  is  the  depth,  so  that 

d=8.2". 

We  shall  use  7"X!)"  ties  10  ft.  long,  set  on  edge,  spaced  at  13-in.  centres 
and  dapiK'd  J  in.  on  the  girders;  also  two  7"XS"  longitudinal  guard  timbers. 
The  dead  load  on  the  s[)an  may  now  be  estimated  as  follows: 

Ties  and  guard-rails 21 S  lbs.  per  lineal  foot  of  span 

Track  and  fastenings 62 '       "     "     " 

Bracing 40 "    "     " 

Maingirderp 600 

Total  dead  load 920  "      "       "       "    "     " 

Live  load 4900, "    "     " 

Impact,  73  per  cent 3571"     "       "       "    "     " 


Total  load 9391   " 

or  say  4700  lbs.  per  lineal  foot  of  each  girder. 


Fio.  510.  Fig.  511. 

In  railway  structures  the  thickness  of  the  web,  for  durability,  should  not 
be  less  than  §  inch.  The  depth  generally  depends  on  economical  considera- 
tions, which  will  ui-ually  Ik-  satisfied  by  making  it  from  one  eighth  to  one 
tenth  of  the  span,  or  in  thi'  case  say  6<)  ins.    Then 

.l'={K(v  I  -24.75.sq.in.s.</r«.s.'<,  or  about  ISsq.  ins.  actual 

This  will  resist  in  shear  18x10000  =  180,000  lbs. 

(;rcate.st  shear  in  girder  =^^4700X25  =  1 17, ">00  lbs. 

(The  web  sc<tion  is  therefore  more  than  sufficient  to  carry  the  shear,  but 
the  thii'kiir—  <  :u't!<>t  1><'  i!l!)ii!!i><hfd,  and  a  decrease  in  the  depth  would  prob- 
ably cause  a  more  than  corresponding  increase  in  the  flange  area.) 

The  effective  lepth  '/,  or  distance  l)elween  the  centres  of  gravity  of  tL- 
top  and  bottom  lianges,  cannot  be  known  exactly  until  the  flange  sectluii  is 


111 ! 

•I  h 


DESIGS  OF  A  50-FT.  PLATE-GIRDER.  501 

(ltt(  rminpd.     It  may  be  assumed,  however,  to  be  5.35  ft.     Hence  if  A  is  the 
an  a  of  the  flange, 

Now.y=        g         =- 1,468,750  ft.-lbs.    Therefore 

1468750=  {A  +-^)  X  14000X5.35, 

and  A  —16.5  sq.  ins.,  net. 

(hie  eighth  instead  of  one  sixth  of  the  web  area  is  taken  as  assisting  the 
tiaiijic,  to  allow  approximately  for  the  rivot-holes. 

Use  two  angles  6"X6"XA" =12.86  sq.  ins. 

One  cover-plate  14Xi -  700  "     " 

Total  gross  area =19.80  "     " 

S('vpn-(>ighth-inch  rivets  will  be  used,  but  a  hole  1  nich  in  diameter  will  be 
(l((lu( ted  to  allow  for  possible  injury  to  the  metal  in  punching.    Therefore 

area  to  be  deducted  from  each  flange  =2Xie  +2X  A"=3.25'8q.  ins., 

and  the  net  area  of  flange  - 19.86-3.25  =  16.61  sq.  ins., 

wliiili  is  sufficient. 

Id  allow  for  irregularities  in  the  edge  of  the  web  the  flange  angles  will  be 
[il.'iicd  fUiJ  ins.  back  to  back.  The  effective  depth  may  now  be  checked,  and 
will  Ih-  found  to  agree  with  the  assumption  made  above. 

To  find  the  length  of  the  cover-plate: 

Flange  angles =  12 .  86  sq.  ins. 

One  eighth  web -  3.10  "     " 

Total  area  beyond  end  of  cover-plate —15.96"     " 

Uctluctiiig  four  rivet-holes  (4X  "e) =  2.25  "     " 

Net  section =13.71  "     " 

l.lToctive  depth  beyond  end  of  cover-plate— 5.25  feet.    Therefore 

H.M.  which  flange  can  resist  =13.71  X  14000X5.25 -1,007,685  ft.-Ibs. 
l.(  t  /  =  distance  from  centre  of  girder  at  which  this  B.M.  is  developed. 

Thou  ,v.-^(--x»),    or    x-i^--- ^ 

IsofTxTooTe^ 

"\  4  4700  "  "•' 


i:{>-ia-<:;i^;i>j 


502  THEORY  OF  STRUCTURES. 

and  the  theoretical  Icnjrth  of  cover-plate  =2X  14  =28  ft. 

We  shall,  however,  extend  it  18  ins.  further  at  earh  end,  so  as  to  get  room 

for  sufHcient  rivets  to  develope  its  strength, 
makinf;  the  total  length  31  ft. 

In  designing  the  comjiression  flange  it 
wotild  not  be  neressary  to  deduct  the 
rivet-holes.  It  acts  as  a  column,  however, 
and  so  it  is  customary  to  make  both 
flanges  alike. 

To  determine  the  pitch  of  the  rivets 
connecting  the  flange  angles  to  the  web,  we  may  assume  that  ihe  entire  ver- 
tical shear  is  carried  bv  the  web  and  is  distributed  uniformly  over  the  web. 

117.")00     ,_„„  ,.  .    , 

At   the  ends  the  vertical  shear  is  117,.500  lbs.,  or      ^^      =1<80  lbs.  per  inch 

of  depth. 

Horizontal  shear  =  1780  lbs.  per  inch  of  length. 

Th(>  load  from  the  ties  nmst  also  be  communicated  to  the  web  through 
the  rivets. 

Th.?  amount  of  this  is  7333  X  A  = -"jfi^  lbs.  per  lineal  inch. 

Thus  the  total  stress  per  lineal  inch  =v/l7S0» +  .564' =  1867  lbs. 

Now  the  value  of  a  J-in.  rivet  in  bearing  on  a  f-in.  plate  is 


Detail  of  Flange 
Fig    512.  Fio.  51.3 


20000XJX|  =6562  lbs., 


which  is  less  than  the  value  in  double  shear. 

65()'' 
Therefore  the  maximum  pitch  =--;;^  =3.5  ins.    We  shall,  however,  space 

them  with  a  3-in.  [ntch,  as  shown  in  Fig.  .")l3.  Towards  the  centre  of  the 
.span  the  shear  decreasi>s,  but  in  this  case  we  may  retain  the  same  spacing 
throughout. 

The  tlaiigcs  need  not  be  spliced,  as  suflficicntly  long  angles  are  obtain.able. 
The  web,  however,  must  have  a  splice  at  mid-span,  which  .should  contain 
enough  rivets  to  develope  the  strength  of  its  entire  net  section.  Two  splice- 
plate-i  (.1 )  13i"X  i"X.J4",  one  on  each  .side  of  the  web,  with  two  vertical  row 
of  rivets  on  each  side  of  the  splice,  Fig.  olO,  will  be  used.  ,\lso,  since  these 
plates  do  not  extend  to  th-  edge  of  the  web  and  are  therefore  ill-suited  to 
resist  landing,  we  shall  use  tvo  plates  (B)  .V'xr'XSfi"  on  each  flange, 
•whose  net  section  in  accordance  with  good  practice  is  50  iM>r  cent  greater 
than  the  web  section  counted  as  assisting  the  flange. 

Web  .stiffeners  cannot  be  designed  rationally.  It  will  l)e  in  keeping  with 
Kood  prartirc.  howcv.r,  t»  use  l"x;5i"Xi"  angles  rivctc.l  in  pairs  to  each 
side  of  the  web  at  inter\als  not   exceeding  the  depth  of   the  girder,  or  say 


5  ft 


Wind  and  lateral  bracing  for  spans  so  small  neetl  not  be  calculated,  aa 


F. 


^ 


riuii  of  Eiid  Stiffenert 
and  Bearing 

Fi(i.  514. 


RELATIONS  BETWEEN  DEFLECTION,  SLOPE,  ETC.  503 

tl„.  -smallest  angles  permitted  by  most  standard  specifications  will  be  suffi- 

ci,.,>t.     We  may  use  31"X3i"X  i"  angles  throughout. 

n.arina  Platrs.-The  girder  will  Ix'  about  52  ft.  long  over  all,  so  that  the 

w,.i.'lit  supported  at  each  end  will  be  4700 X2(i  =  122,220  lbs. 

iakin.'  300  lbs.  per  square  inch  as  the  safe  load  on  good  granitoid  or  lime- 
stone,  the  area  required  ^vill  be  122,200-^.300=407  sq. 
ins.  We  shall  use  for  the  fixed  end  two  24"X'H"xr' 
plates  riveted  to  the  girder  and  for  the  sliding  . nd  one 
24" X  18" XI"  plate  riveted  to  the  girder  and  a  similar 
i-in.  plate  on  the  masonry,  the  contiguous  surfaces  being 
planed  so  as  to  slide  readily  with  changes  of  tempcra- 

To  distribute  the  load  evenly,  end  stiffeners  of  suffi- 
,.„.nt  section  to  carry  the  above  load  as  a  column  will  he  placed  as  shown  in 
Vi'Z.  .')14. 

20   Relations  between  the  Deflection,  Slope,  and  B.M.  Curves.— 

(•on>ia..r  an  elcmont  KLL'K'  of  thickness  dx  and  bounded  by  the 
v.Mtical  planes  KL  and  K'U.  It  is  kept  in  equiUbnum  by  S  and  M, 
th..  SF  and  B.M.  at  the  section  KL,  by  S+dS  and  M  +  dM,  the 
S  F  and  B.M.  at  the  section  K'U,  and  by  the  load  wdx  (which  in- 
<.lur„.s  the  weight  of  the  element),  tr  l)eing  the  intensity  of  the  load 
a  A-  Since  dx  is  indefinitely  small,  it  may  be  assumed  that  the 
loa.l  wdx  is  uniformly  distributed  and  that  its  line  of  action  is  the 
iiiiilillc  line  w. 


Taking  moments  about  v, 


Jx 


M-S-^  +  M+dM-{S+dS) 


,„    (i^        r.  "7"*"" 


.( 


wdx 

I  K 


■*jM+dM 


S-t-dS 


3 


ur 


dM     „ 


Fig.  515. 


.dx 


.riM.>f:arding  the  term  dS%  which  is  indefinitely  small  compared 
with  the  remaining  terms. 


Also, 


-S-uxix+S+dS~0, 


or 


dS 
dx 


=  w. 


Therefore 


d'M    dS 
dx2~dx    ^" 


....    (A) 


504 


THEORY  OF  STRUCTURES. 


If  the  beam  is  loaded  with  a  number  of  weights  concentrated  at 
different  points,  then  between  any  two  consecutive  weiyhts,  w  is  merely 
the  weight  of  the  beam  per  lineal  unit  of  length,  and,  as  far  as  the 

concentrated  loads  are  concerned,  -7-  is  nil,  so  that  S  is  constant 
between  such  weights. 

Again,  the  deflection  (y),  the  slope  ((?  =  tan  ^=  j~),  and  the  bend- 
ing moment  (A/)  are  connected  by  the  equations 


d2y    d^     M 
dx2~dx~EI 


(B) 


Comparing  equations  {A)  and  (B)  it  is  at  once  observed  that 

M 
the  values  of  y,  0,  and  -pr.  are  obtaineil  from  one  another  by  a 

process  of  graphical  integration  precisely  similar  to  that  by  which  the 
relative  values  of  M,  S,  and  w  are  found.  Thus  any  property  con- 
necting the  last  three  quantities  must  also  be  true  for  the  first  three. 
In  other  words,  the  mutual  relations  between  curves  drawn  to 
represent  the  deflection,  slope,  and  bending  moment  must  be  the 
same,  mutati.s  mutandis,  as  those  between  the  curves  of  bending 
moment,  shearing  force,  and  load. 

Thus,  divide  the  effective  bending-moment  area  into  a  number 
of  elementary  areas,  Fig.  516.  by  drawing  vertical  lines  at  convenient 


T,    T,. 


Fiu.  .516. 


»--<--. 


Fio.  517. 


IMI'- 


distances  apart,  and  take  tho  vorticn!  linp.t  12,  23,  .  . 
Fig.  i")!?,  to  represent   the.se  anvis. 

Let  the  pole  0  be  at  the  distance  El  from  the  line  In. 


n  —  l,-n. 


!' 


GliAPUICAL  DETERMINATION  OF  SLOPE  AND  DEFLECTION.   505 

If  the  widths  of  the  areas  are  made  sufficiently  small  a  funicular 
ilcllcction  curve  CP1P2P  can  !:.t  drawn  in  exactly  the  same  manner 
a:<  the  B.M.  curve  was  dra^vn  in  Art.  6,  Chap.  II. 

It  was  shown  that  any  two  tangents  to  the  B.M.  curve  intersect 
in  a  point  which  is  vertically  below  the  C.  of  G.  of  the  corresponding 
load  curve.  Hence  it  must  follow  that  any  two  tangents  P\T\, 
F'T',  intersect  in  a  point  Ti  which  is  vertically  below  the  C.  of  G., 
(/I,  oj  the  effective  B.M.  area  DEFG. 

It  was  also  shown  that  the  B.M.  at  any  point  is  the  intercept 
of  the  vertical  through  the  point  between  the  closing  line  CC  and  the 
B.M.  curve.  Hence  it  follows  that  the  deflection  at  any  point  is 
the  intercept  between  the  closing-line  and  the  deflection  curve,  of 
the  vertical  through  that  point. 

.\gain,  through  the  pole  0  draw  OH  parallel  to  the  closing-line 

intersecting  In  in  H.     Then,  since  EIO  =  jMdx,  the  angle  P1T1T2 

between  any  two  tangents  is  equal  to  the  corresponding  B.M.  area 
divided  by  EI.     It  is  also  evident  that  the 


slope  ii  at  1  = 


\H 
EI' 


and  that  the 


nH 
slope  tnSitn=pj. 


Tlierefore 


.      .      In 


rtiiicli  {lives  the  total  change  of  slope. 

In  the  case  of  a  cantilever,  the  last  side  of  the  funicular  polygon 
i?  obviously  the  closing-line. 

21.  Graphical  Determination  of  the  Slope  and  Deflection. — If  p 
is  the  radius  of  curvature  at  any  point  of  the  deflected  neutral  axis, 

l_^_d^_M 
p'dx^'dx'EP 


Therefore 


Mdx 


fdx_dy_         rMdx 
J   p  -dx'"-J    EI 


^06 
■and 


THEORY  OF  STRUCTUHES. 


Thus  it  may  be  supposed  that  the  cantilever  or  beam  carries  a 

uniformly  distributed  load  of  intensity  -,  and  by  the  process  of 

integration  the  curves  of  slojie  and  deflection  can  be  obtained. 
Let  .1  be  the  effective  B.M.  area  between  any  two  points  P  and 

Q  of  the  deflected  neutral  axis. 

Let  the  tangents  at  P  and  Q  meet  in 
the  i)oint  T  and  produce  QT  to  meet  the 
vertical  through  P  in  R. 

The  i)oint  T  is  vertically  below  the  C. 
of  Cf.  of  the  area  A.  Let  z  l)e  its  hori- 
zontal distance  from  PR,  and  take  PR^d.  If  0  is  the  change  of 
curvature  betwwn  P  and  Q,  i.e.,  the  angle  PTR,  then 

/•MdxA 

^^J  EI  ~Er 

assuming  that  E  and  /  are  constant. 
Also, 


5  =  P/f  =  »Z  =  gJ. 


Ex.  45.  .1  cantilever  OA    <>/  length  I  trilh  a  c 

weight  \y  nl  A. 

The  H.M.ana  i^-  the  triiUigleO/iC  (Fig.  519), 

o 


and  OC  =  Wi 
Therefore 


and  the  slope  at  A 

Abo 
and  the  deflection  of  A 


A~hwr, 
1  wn 


-e- 


At''>i\VVll'\Wl* 


Fio.  S19. 


-Y- 


3    El 


F.x.  40.  -4  cantiltver  OA  of  Ungth  I,  earnjittg  a  uniformly  distributed  load 

''   Tlio  U.M.area  is  OBC,  the  vurxe  UC  U^.ng  a  parabola  with  !'»  v.rtex 
at  H. 


GRAPHICAL  DETERMINATION  OF  SLOPE  AND  DEFLECTION.  507 

^  1 

1      I  ...n      n'  —  — n 

3    2 


Also, 

Therefore 
and  the  slope  at  A 

Again, 
ami  the  defleotion  of  A 


oc-f. 


aJ-^'iJwP    o 


6  EI 
A       1     »3,     1     „ 

■^'"e"  4     8     ' 


Fig.  520. 


-Y 


i^wl* 
°8  EI" 


Ex.  47.  .1  hnriznntal  beam  OA  of  length  I, 
resting  ujmn  supimrlH  at  0  and  A  and  carrying 
a  uriijht  \V  at  Uk  middle  point  B. 

The  H.M.  area  is  the  triangle  OCA,  the 
vortical  distance  of  C  above  D,  the  middle 
point  of  the  horizontal  line  0.4 

beinR-y. 


Considering  one  half  of  the  beam, 

iwii    m* 

^'2  4  2  "  16' 

1  m* 


ami  the  slope  at  .4 

Also, 
aTHJ  I  lie  maximum  deflection 


^e~ 


I«  El 
_        2   I     _ 
16"*  3    2~48' 


1         2/1 
>l*-AlV/'5   Ty-T^WP, 


48  El 


].K.  4S.  A  hnriiont^d  Itrnm  0.1  0/  length  I,  renting  upon  supports  at  0  and  A, 
mill  currying  n  weight  W  at  a  point  distant  n  from  each  end. 
I.cl  ')  1h'  the  length  of  the  intermediate  Hegment. 
Ihc  H.M.  area  in  the  trajK-zoid  OCDA,  the  vertical  distance  CE  (-DF) 

I.r,hi.'    ll'„. 

(  oiiiiidering  one  half  of  the  beam, 

A-iWa'  +  >,\yal,~\''(a+b). 


ami 


Wa 


the  •lope  at  0  -  »  -  .^^.y  (o + 6) 


E08 


THEORY  OF  STRUCTURES 


Also         Az=Wa'ia  +  iWab{a  +  ^) 

=  !^(:8a'  +  12a6+36'  , 
and  the  max.  deflection 


If  the  tvo  weights  are  concentrated  at  the  points  of  trisection, 

23  Wa' 
o— 6    and  the  max.  deflection  =.^  -pj- 

"-pT,  approximately. 


Ex.  49.  A  horuontal  beam  OA  of  length  I,  resting  upon  supports  at  0  and  A, 
and  carrying  a  weight  W  concentrated  at  a  point  B  distant  a  from  0. 

The  B.M.   area  is  the  triangle  OCA,  the 
^  „   ,    .       Wa(l-a)     -, 
vertical    distance  CB  being  - — ^ "-"n 

suppose.    The  most  deflected  jwint  F  is   at  < 
some  point  between  B  and  .4. 

Let  the  horizontal  tangent  at  F  meet  the 
tangents  at  O  and  -4   in  T,  and  Tj  respec- 
tively, and  let  «,,  z,  be   the  horizontal  dis-  f'o  •'•23. 
tances  of  T,  and  T,  h.m  0  and  .4  respectively. 

Let  .1/,  Ix"  the  length  of  the  B.M.  ordinate  ED,  vertically  above  F.    The 
length  of  AD  is  evidently  J?,. 

Consider  first  the  portion  of  the  beam  between  O  tind  F,  and  let  A,  be  the 
B.M.  area  OC'ED.    Then,  sinc«>  the  horizontal  distance  of  the  centre  of  gravity 

of  the  triangle  OC'A  from  0  is  -g-, 


A,*.  -  ~M,l  -^  -  f  .V.J*,(/  -  Ut  +  4'.) 


-4A/,/(/  +  o)  -f.U^.(J-«,)  ~EIY. 


Consider  in  the  soronil  place  the  (wrtion  of  the  beam  between  A  and  D. 
Then 

\M,lt.t,-iM^,'-EIY. 


Therefore 


|.W,/(/+«)-|.»/^,(/-«,  -fA/,f,« 


GRAPHICAL  DETERMINATION  OF  SLOPE  AND  DEFLECTION    509 


and 


or 


9    z,      M     l-a 

ft'-,V(i'-a'). 


Hence 


Also,         JM,    2^W-a    2  /  l-a       ~T\~3~- 

ice 

the  max.  deflection  -F-|^i^ !_E2(/._a.)l. 

ARain  let  *  be  the  vertical  distance   between  G  and  F,  and   let  t'  be 
th(>  horizontal  distance  of  the  centre  of  gravity  of  the  area  BCED{  -A*) 

from  G. 


Then 


Therefore 


M 

--\l-ay-iMMl-a-t;) 


-^\l-a)'-^\l'-a')+EIY. 


£/(r-*)-^'o«-a)-|-  ja'(/-a)' 


which  gives  the  deflection  of  the  point  at  which  IT  is  concentrated 
Again,  A,- Jj»/a  "  Jv,^}^  _l.»r|(/'-a'), 

A,Jw^JLl^-A.J^W^(l-am-a). 


and 
Hence 


A'-.jW,/-a-A,-^(/-a)(/-2o). 
the  slope  at  O-g-  -^i  -T(/-a)(2/-o) 


:|iil 

",l-' 


IB'/} 

(I  «l        <l     J   _^     "     "/It       _tv 


510 


THEORY  OF  STRUCTURES. 


Ex.  50.    A  beam  OA  of  length  I,  supporUd  at  0  and  at  A ,  and  carrying  a 

unijormly  dixlribuial  load  oj  intensity  u: 

The   n.M.   art'u  is    DBA,   the    curve    DBA 
being  a  parabola  with  its  vertex  at  a  distance 

—  vertically  above  (',  the  middle  point  of  OA. 

Considering  one  half  of  the  beam, 


Fio.  524. 


A  = 


2  m7] 

3  S 


and 


5; 


1^ 


Therefore 

and 


the  slope  at  0  (or  at  -4)  =2;j^ 


the  max.  deflection  Y 


EI        "384El' 


Ex.  51.  A  horizontal  beam  OA  of  length  I  fired  at  the   supports  0  <ira  A 
and  carrying  a  uright  U'  concentrated  at  the  middle  point  B. 

The  fixture  of  the  ends  introduces  negative  bending  moments  at  0  and  at 
A,  and  the  H.M.  diagram  consists  of  the  two  lines  BC,  DC,  the  point  C  being 
vorticidly  above  the  middle  point  (1  of  0.1.  .\t 
the  points  E  and  F  the  B.M.  is  nil,  and  the 
beam  may  Ix^  supposed  to  lie  made  up  of  two 
cantilevers  EO,  FA,  and  of  an  intermediate  span 
EF  resting  ujxm  the  ends  E  and  F. 

Thus  if  OE'X, 


OB-AD~iWx. 


Fia.  525. 


Again,  since  the  neutral  axis  is  horizontal  at  0  and  at  A,  the  total  change 
of  curvature  between  O  and  .1  is  algebrakally  nil,  and  therefore  the  total 
effective  B.M.  area  must  also  Iw  algebraically  nil. 

Hence  the  B.M.  area  above  0.4  -B.M.  area  lielow  OA, 

nr  the  triangle  CEG  -the  triangle  OBE, 

and  £  io  the  middle  point  of  OG;  so  that 

x-j     and    OB(~AD-CG)-iWl. 
The  B.M.  area  OBE-  ^Wl'  and  the  corresponding  t-^.    Therefore 


^i!!li 


:  :  i 


GRAPHICAL  DETERMINATION  OF  SLOPE  AND  DEFLECTION.  511 

1  TFi* 
the  slope  at  E  —^  -^r, 


and 


1     WP 
the  deflection  of  ^^^oei'  £/• 


The  B.M.  area  ECG  - aiWP  and  the  corresponding  «  — g.    Therefore 

1    WP 
the  deflection  GH  "55^  -gj 

I   Wl' 

anil        the  total  maximum  deflection  -deflection  of  E  +GH  -  —  gj  • 

The  points  E  and  F  at  which  the  B.M.  is  nil  and  at  which  the  curvature 
nptcssarily  changes  from  positive  to  negative  are  called  points  of  inflection. 

Ex.  52.  A  horizontal  beam  OA  of  length  I,  fixed  at  the  supports  0  and  A 
and  carrying  a   uniformly   distributed   load  of 
iiit(ii''ity  w. 

The  B.M.  diagram  is  now  the  parabola  I 
/}A"(  T'V),  with  its  vertex  at  C  vertically  above  ' 
the  middle  point  G. 

The  points  E  and  F  are  ( vidently  points  of  a 
iiijhrtidn. 

Lot  EF~2x. 


Fio.  526. 


Then 


OE-^-x, 


frtjf* 

theB.M.  CG'-j, 


and         theB.M.OB{-AD)-wi(^-x)  +|(^-*)'-f  (?-*')  • 


Also,  as  in  the  preceding  example, 

the  area  ECG  -the  area  QBE 
the  area  BECK  -the  area  OGKB, 


and  therefore 


or 


i{^q(;i-.')]{-'S-)l 


■o  that  »-^-i-tX. 289. 


512 


THEORY  OF  STRUCTURES. 

n 


The  total  max.  deflection  =  deflection  of  E  due  Uiwi^-x\  unifonnly  distributed 
+       "  "       "    uxatendofO£ 


ju(i-x)*    J    u.x(^-x) 


8'     El 


ud 


El 


5  w(2xy 
"384    El 


-3^(3i' -  8f 'X  -  24te' +  Oex*) 


I   wl* 
'384  EI" 


22.  Beams  Supported  at  More  Points  Than  Two.— Consider  the 
rth  span  OX  of  a  girder  resting  upon  a  number  of  supports  in  the 
same  horizontal  plane. 


K:r^ — ^N 

Fia.  627. 


0-^ 


Mr 


1 


Rr 


Let  Rr-\,  Rr  be  the  reactions  at  0  and  at  X; 

Mr-\,  Mr  he  the  Ix^ntUng  moments  at  0  and  at  X; 

M,  be  the  B.M.  at  any  jwint  .V  (i.e.,  x,  y)  of  the  neutral  axis 
due  to  the  load  on  OX ; 

M  "    "    B.M.  at  A'  dm  to  the  load  on  OX. 
Taking  moments  about  A', 


and  therefore 


R^llr='M  +  Mr-Mr-l, 
MMr       Mr- 1 


Hence  the  shear  at  the  (r-l)th  support  for  the  rth  span 

=  the  reaction  at  the  same  sujiport,  supjwsing  the  span  an  inde- 

jx'H<lo?it  girder,  i.e.,  cut  at  its  supports. 
+  the  difference  of  the  forces,  or  reactions,  eciuivalent  to  the 

moments  at  the  supjKirts. 


A    ii 


i  .1  i 

■  i 


BEAAfS  SVPPOr.TKD  AT  MORE  POINTS  THAN  TWO. 


513 


Af;aiii,  the  B.M.  at  N 


=  tlio  moment  at  the  same  point  supposing  the  span  an  indepen- 
dent girder, 
t-the  reactions   equivalent  to  the  moments  M,_i,  Mr,  multi- 
plied respectively  by  the  segments  Z^  — x  and  x. 
Lot  OX,  Fig.  528,  be  the  rth  span,  and  let  OBX  be  the  curve 
of  hciiding  moments,  suppof^ing  OX  an  independent  girder,  i.e.,  cut 
at  O  and  A'.     On  the  same  scale  as  this  curve  is  drawn,  take  the 
verticals  OE  and  XF  to  represent  M,_i  and  M,  respectively  in 
iiiajiiiitude,   and  join   EF.    The   curve   OBX   corresponds   to   the 

/M  \ 

port  ion  l-j-x  —  Mx)  of   the  above 

equation,  and  the  line  EF  to  the 

ri'inaiiidcr,  i.e.,      ,     (Ir—x)  +-j-x. 

The  actual  bending  moment  at 
any  point  of  OX  is  represented  by  the  algebraic  sum  of  the  ordi- 
iiatcs  of  the  curve  and  line  at  the  same  point,  which  will  be  the 
intercept  between  them,  since  they  represent  bending  moments  of 
njijiosltc  kinds. 

Let  .1  be  the  effective  moment  area,  or  the  algebraic  sum  of 
ihe  areas  for  the  load  and  for  the  moments  at  0  and  A',  shown  shaded 
ill  tlie  figure,  and  let  x  be  the  horizontal  distance  of  its  centre  of 
;:ra\iiy  from  0. 

let  .1^  Ix'  the  area  for  the  load  alone,  i.e.,  the  area  of  the  curve 
'>li\.  and  let  z^  be  the  horizontal  distance  of  its  centre  of  gravity 
from  (>.    Then 


Fio.  528. 


Ax  -  ArZr  +  Mr..  , ^  + 1  i^r  "  M,.  i)^.^ 
=  Ar^r  +  iMr-^Ir^  +  hM^r^. 


'•"^'   '-'^  f.^t  the  toad  up<m  tkf  rtk  span  OX  be  a  weight  P  concentrated  at  a 
pniiit  I  ■  ilisiiuit  p  from  0. 

Ill'  It.M.  diagram  ronmi«ts  of  the  two  straight  lines  OB,  XB,  if  it  ii  Msumcd 


THEORY  OF  STRUCTURES. 


514 

Pp(lr-p) 

that  OX  is  an  independent  girder.    Also  the  vertical  distance  BC j-^ 

Therefore  .U=area  OBX ^^  ^^^=^Pp(/.-p). 

Again,  join  B  to  the  middle  point  K 
of  OX,  and  let  GL  be  the  vertical  through 
the  centre  of  gravity  of  the  triangle.    Then 
zr=OL~OC  +  CL=OC+iCK 


=OC  +  |(OA--OC)- 


Ir  +  P 


Hence         .4,2r=iPp(^'-p')• 
Similarly,  for  the  (r  +  l)th  span, 

Ar+,Zr+,=iQq(lr+,'-q'), 

«r^,  being  the  horizontal  distance  of  the  centre  of  gravity  of  .4,+,,  the  moment 
area  of  a  load  Q  concentrated  on  the  (r  +  l)th  span  at  a  distance  q  from  the 
(r  +  l)th  support,  the  span  being  considered  an  independent  girder. 

Ex.  54.  Let  the  load  upon  the  rth  span  OX  be  a  unilormly  distributed  load  of 

intensity  uv.  ,   .  «        i-    n 

The  B.M.  diagram  is  now  the  parabola  OBX,  with  its  vertex  B,. vertically 

above  the  middle  point  C,  Fig.  530.     Then 

„  f'r/r' 

Therefore 

^r  =area  OBX  =g  -^— /,  =^u)r/r". 


Also, 


Ir 
ir=.,-. 


Fio.  630. 


Hence  .4rJr=2>4Uv/r*. 

Similarly,  for  the  (r  +  l)th  span, 

Ex.  .i.").  Lit  a  unilormly  (lislrihuted  load  «/  intensity  w,  cover  a  length  2a(  </r) 
of    th<   rth  siHin,  and  lU  'z   In-   the  distance  nj   its   centre  of   gravity  from  the 

The  loud  may  l)o  fiupiM)scd  to  consist  of  a  num»x>r  of  indefinitely  small 
elements,  and  its  effect  may  Ix"  obtained  by  suporixising  the  several  effects  of 

these  eleiiieuts.  ,      ,        ^  i.  ^ 

Let  .//'  iH-  the  length  of  such  an  element  at  p  from  the  (r-l)th  support- 

Then,  by  Ex.  r>3. 


.4r-'r 


■L 


'Jjlhl-E  (;,>-p')  -2aurUlr'-i'-a'). 

It  It 


CONTINUOUS  GIRDERS. 


515 


23.  Continuous  Girders. — When  a  girder  overhangs  its  bearings, 
or  is  supported  at  more  than  two  points,  it  assumes  a  wavy  form 
and  is  said  to  be  continuous.  The  convex  portions  are  in  the 
isainc  condition  as  a  loaded  girder  resting  upon  a  single  support, 
tlic  upper  layers  of  the  girder  being  extended  and  the  lower  com- 
prcsscil.  The  concave  portions  are  in  the  same  condition  as  a 
Idadod  girder  supported  at  two  points,  the  upper  layers  being  com- 
pressed and  the  lower  extended.  At  certain  points,  called  poin/s 
oj  lOtUrary  flexure,  or  points  of  inflexion,  the  curvature  changes 
sifiii  and  the  flange  stresses  are  necessarily  zero.  Hence,  apart 
from  other  practical  considerations,  the  flanges  might  be  wholly 
severed  at  these  points  without  endangering  the  stability  of  the 
pirder. 

24.  Theorem  of  Three  Moments.— Let  0,  X,  V,  the  (r-l)th, 
rtli,  and  (r  +  l)th  supports  of  a  continuous  girder  of  several  spans, 
be  depressed  below  their  true  horizontal  position  O1O2O3,  through 
the  vertical  distances  </,._  1  ( =  OiO) ,  dr(=OiX),  and  dp+i(  =  03F), 
(/,-!,  dr,  and  d,+i  being  necessarily  very  small  quantities. 


■j- ir-r- 

» 


D: 

I 


--%*-    ^r+l 


rr!:^:::;j 


Fio.  531. 


Let  OCXDV  be  the  deflection  curve,  and  let  the  tangent  at  .Y 
iiMit  the  verticals  0\0  and  O3V  in  E  and  F,  and  the  tangents  at  O 
aii.l  V  in  Ti  and  T2. 

Take  0,£  =  j/,_,,  OsF-ijr^i  and  let  n,(=OTxE),  O.^.i^FTiV) 
I"  tlie  changes  of  curvature  from  0  to  A'  and  from  V  to  A"  respec- 
tively. 

Let  .1,,  .4,+ 1  be  the  effective  moment  areas  for  the  spans  OX, 
\  1   resjM'ctively. 

Let  /,,  x,j.|  be  the  horizontal  distances  from  O  and  F  respec- 
ii\  I  ly  of  the  points  T'l  and  T2,  which,  as  already  proved,  are  vertically 
Lh  (i\v  the  centres  of  gravity  of  the  corresponding  effective  moment 


i  1  i 

i 

i  ■> 

i     : 

i  :!- 
1 

516 

areas.    Then 

Ir  'r+1 

Again,  as  already  shown, 


THEORY  OF  STRUCTURES. 


Vr-l      Vr+l      dr        dr 

or    -J— +1 — =T'^1 — • 

It  h  +  1        'r      h+l 


Ar 


and 
Therefore 


J/p_  1  -  dr-  1  =  OE  =  XfOr  =  ^Xr 

A    1 

rfr  +  l  -J/r+1  =^^'  =Jr  +  l^r+l  = 'eT^'"*'^' 


l_MrXr      i4r  +  lJr  +  l  \       yr-1        j/rj^       ^Tul  _^tl 
£/l     /,      +         /,  +  i        /  /r  /r  +  1  Ir  k^l 


dr  —  dr-\       dr  —  dr^^X 


Ir 


/r  +  l 


Hence  bv  Art.  22 


El(^'-f;^+^-^^)-A^j-+\Mr-.lrHM^r 


2r+l 


'r+1 


and  therefore 

M,-ilr+2M,a,  +  lr  +  UT  Mr+llr+1 


■'*.f-'*-;;^-»<H--^)' 


which  is  the  analytical  expression  of  the  theorem  of  three  moments 
in  its  most  general  form. 

Exs.  .').3  and  .')4  give  the  values  of  ArZr  and  Ar+xZr+i  for  con- 
centrat"d  and  for  miiformly  distributed  loads. 

If  either  of  tlie  supports,  e.g.  A',  should  lie  above  O1O3,  then 
the  sign  of  d^  is  finintirc. 

A  similar  relation  holds  tru(>  for  every  three  consecutive  supports, 
so  that  for  a  onntinuous  girdf>r  of  n  spans  there  are  n-\  equations 
connecting  the  n-\  \  bending  moments  A/ 1,  M2,  M3  .  .  .  M„,  Mn+i 
at  the  several  supports. 


THEOREM  OF  THREE  MOMENTS. 


517 


There  must  be  two  further  conditions  before  these  equations  can 
be  solved,  and  they  are  usually  provided  by  the  method  adopted 
for  carrying  the  ends  of  the  girder. 

//  the  ends  rest  freely  on  the  supports, 

Mi=o    and    Mn+i=o; 

//  the  girder  is  fixed  at-  1  and  rests  freely  atn  +  1, 

2Mi+M2=-^'    or    =-i-pi(U-p)(2l,-p); 
4  n 

according  as  the  load  upon  thei  first  span  (h)  is  uniformly  dis- 
tril)uted  and  of  intensity  wi,  or  consists  of  a  number  of  weights 
^1,  P2,  P3  .  •  •  concentrated  at  points  distant  pi,  p2,  ps  •  •  •  , 
nspectively,  from  the  fixed  end,  the  symbol  I  denoting  algebraic 

sum. 


Also, 

//  both  ends  are  fixed, 


Mo+i  =0. 


aM. 


w,l 


and 


,+M2= 7^    or    =-i-Pi^(li-p)(2li-p), 

4  *i 


2M„+,+M„=-^^    or     =-i-Qj^2(l„-q)(aln-q), 


n\  Ix'ing  the  intensity  of  the  uniformly  distributed  load  on  the 
n\h  span,  and  q  the  distance  of  the  concentrated  load  Q  from  the 
(11  ^\)th  support.  These  ^.xture  conditions  can  easily  be  proved 
as  follows: 

It  is  assumed  that  the  supports  at  1  and  2  are  in  the  same  hori- 
zi'iital  plane. 

First.    I^t  tc,  be  the  intensity  of  the  load  uniformly  distributed  over  the 
first  span  (/,),  and  let  /?,  be  the  vertical  reaction  at  1. 
At  any  point  (x,  y), 


518 


THEORY  OF  STRUCTURES. 


Integrating  twice,  and  remembering  that  ^  and  j/  are  each  lero  when  »  -0, 

dy        x'     w,x' 

-EIy-R,^—j^+M,-^. 
Also,  J/  -0  when  »  -  J,.    Therefore 


or  fl,I,+3A/,-    1^. 

Takinf;  moments  about  2, 

i2,i,+A/,-A/.— ^. 


Hence 


aM, +M,=  - 


'.1.* 


Second.    Let  a  weight  P  be  concentrated  at  B  distant  p  from  1. 
From  ItoB.    At  any  point  (x,  t/)  in  IB, 


Integrating  twice  and  remembering  that  ^  and  y  are  each  zero  when  x  -0, 


-£/^  =  R.^'  +  M,x 


If  Ofi  and  ya  are  the  values  of  j-  and  y  at  D,  i.e.,  when  x  -p, 


-£/OB-R,|-  +  Af,p 


THEOREM  OF  THREE  MOMENTS. 
From  Bto2.    At  any  point  (x,  y)  in  B2, 

-J?/§-«,x-P(x-p)+af,. 

Integrating,      -El£-R,^  —^(x-py+M,x+e,. 

b' 
At  B,  -  EI9»  -  ft,^  +  M,p + e„ 

and  therefore  c,  -0,  so  that  the  slope  equation  becomes 

x'    P  x' 

Integrating,        -EIy~R,j-^(.x-py  +  Mi-j+e,. 

At  B,  -EIya'-'Rij+  Miy  +  «.. 

and  therefore      c,  -0,  so  that  the  deflection  equation  becomes 

-£/v-ft.|-^(x-p)'+3f,|'. 

But  y  is  also  at  zero  at  2,  i.e.,  when  z  =  ?,.    Therefore 
n    P  I,' 


or 


Taking  moments  about  2, 


Hence 


aM,  +  M,  =  -  P,-,(l,  -  p)  (si,  -  p) . 


I.x.  56.  (o)  The  bridge  over  the  Garonne  at  Langon  carries  a  doubk  track,  is 
ahoni  tm  ft.  in  length,  and  consists  of  three  spans,  AH,  BC,  CD.  The  two  main 
qii-ili  rx  are  continuovs  and  rest  upon  the  abutments  at  A  and  D  and  upon  piers 
(It  II  mid  C.  The  effective  length  of  each  of  the  spans  AB,  CD  is  208  /,'.  6  int., 
and  „!  the  centre  span  BC  243  ft.  The  permanent  load  upon  a  inaia  girder  is 
1-77  Iha.  per  lineal  foot,  and  the  proof  load  is  2688  lbs.  per  lineal  foot.  Find 
the  rmdions  at  the  supports  (1)  when  the  proof  load  covers  the  span  AB;  (2)  uAefl 


520 


TH^m^Y  OF  STRUCTURES. 


the  proof  load  cowrx  «,..  .  /A<n  BC;  (3)  when  the  proof  load  covers  the  apana  AB 
and  BC;  (4)  when  the  proof  lou.l  co.ers  the  whole  girder. 


Ml  -0 
A 


'»  Ri 


U 


c 

—K— 


M«  — 0 
D 


I.  * 


BT73 

T 


Fio.  532. 


Take  BC^a-  243' ,  and  /t  B  =  208i'  =  ft  -  TD.  Assume  the  B.Ms,  at  B  and 
at  C  to  be  right-handed.     The  B.Ms,  at  .4  and  at  D  are  ru7. 

Firu/.  Let  .\I,',  M,'  be  the  B.Ms,  at  B  and  C  due  to  the  dead  load  of  1277 
lbg./hn.  ft.    Then 

23/,'(6 +0)  +  M.'a  -  -  ~-(a'  +  6") 


•nd 


1277 
A/,'o  +  2M,'(a  +  6)  -  -  ~  (o«  +  {.•). 


Therefore  3f ,'  -  -  ^^  ~^  -  -  6522303  f t.-lbe.  -  M»' 

4     3a  +  2o 


1277 
-/?,'208i-i^'(208i). 


Hence 
and 


«,'- 101845  lbs. -A' 
/?,'-3l95«5  "    -ft'. 


Points  of  inflexion      For  A  B  (or  CD)  the  point  of  inflection  is  at  x  feet  from 
the  end  support,  x  being  given  by 

1277 
B.M.-0-/?,'x~~x», 


or 


x--^--159'.5. 


The  mnximum  pointitf  B.M.  In  AB  (or  CD)  is  at  79f  ft.  from  the  end 
support  and  its  vahic  is  -^  v.,'--  ^4,(M11,240  ft.-lbs. 
For  BC,  measuring  z  from  B,  the 

B.M.  -0-ff,'(208i  +x)+R,'i--J-(2(»i+x)\ 


x*-242.0H4x-t- 10215.15-0    and    x-54'.15    or    188'.83. 


BRIDGE  OVER  THE  GARONNE. 


521 


The  maximum  positive  B.M.  in  BC  is  at  121 J  ft.  from  B  (or  C)  and  its  value 
is  ■2,'MiM7  ft.-lbs. 

If  the  proof  load  of  2688  Iba.  covers  the  whole  girder  the  B.Ms,  and  reactions 
3'Jfi5/_1277+2688\ 


ar( 


1277  \ 


Then, 


ft. 


.,__      )  times  the  corresponding  reactions  first  obtained. 
3965 


1277 

3965 
'l277 

3965. 


X 101845 -316223  lbs. -ft., 
X  31 9565  -  992227  lbs.  -  ft., 


JH^^^X  -6522303 -  -20251320  ft.-lbs.  -Mu 

3065 
The  maximum  positive  B.M.  mAB  (or  CD)  -  j^  X  4,061 ,240  f t.-lbfc 

-12,60&,900  ft.-lbs. 

306.5 
Thp  maximum  potitive  B.M.  on  BC  -  y^  X  2903347  -  9,01 4,700  ft.-lbs. 

Fig.  533  shows  the  S.F.  and  B.M.  diagrams  for  the  two  cases,  the  dotted 
lim-s  being  the  diagrams  when  the  proof  load  covers  the  whole  girder. 


Fro.  A33. 

.<rr<ind.    Let  M",  Mt"  be  the  bending  moments  due  to  the  proof  load  of 
;^^>vs  IbM./lin.  ft.  on  the  span  AB.    Then 

2.W."(o  +  6)  -••  i"."'o  -  -  ~V 


522 
and 


THEORY  OF  STRUCTURES. 
A/",a+2iV,"(o  +  6)-0. 


.l^erefore   Mr—l^^li'Ub'—''''''^''''-''^ 


and 


Mm"'-AI," 


-ff,"X  208.5 -2688 
-+ 1,956,885  ft.-lba. 


,(208.5)' 


2(a  +  6) 

-  ft/' X  208.5. 

Hence      H,"- 296340  lb*    and    ft." -9386  lbs. 

Again,  ft."(208.r  +243)  +ft,"243  -M," -  -7,271,879. 

Therefore  ft."- -47364  lbs. 

But  Ri"  +  R"  +  A"  +  ^♦" - 2688 X 208.5 - 560,448  lbs. 

Hence  ft," -302,086  lbs. 

Superposing  the  results  (or  the  dead  load  and  for  the  proof  load  on  AB, 

ft  -ft,'  +  ft."-398,ia5  lbs.;  ft.  -  ft.' +  ft." -272,201  lbs.; 

ft  -  ft  '  +  ft  "  -  621 ,651  lbs. ;  R*  -  R*'  +  «•"  - 1 1 1 ,231  lbs. ; 

\l  -  V.'  +  M."  -  -  3,794,182  ft  -lbs. ;   3/.  -3//  +  A/."  -  -4,565.418  (t.-lbs. 


Points  o/  in^xim.     At  such  points  the  B.M.  -0. 
FarAB. 


396.')  , 
B.M.-0-ft,x 2~x' 


•ad 


op 


The  maximum  positive  D.M.  is  at  39^5-100.42  ft.  from  A  and  its  value  is 


^a^-'^'^^^'"'**'^-''* 


Far  CD.    Measuring  x  from  D,  the 


Md 


1277 
B.M.-0-fta-^*» 

•  -0    or     -j'!,^J-174'.2l. 


-i.  .:.i 


BRIDGE  OVER  THE  GARONSE.  523 

The  maximum  posUive  B.M.  in  CD  is  at  87M05  from  D  and  its  value  Ja 
1    «,' 


2  1277 


•4,844,300  ft.-lbs. 


Fio.  534. 


For  EC,    Measuring  x  from  C,  the 


1277, 


BJ«.-0-ft.(208i+x)+/?u: — ^ix+20^)\ 

or  x'-183.52x  +  7150.15-e, 

tn.i  X-66M3     or     127'.39. 

The  maximum  potitive  B.M.  is  9r.76  from  C,  and  its  VHlue  is 

1277 
/?.X300.26  +  «,X91.76 2-(300.26)« -810.735  ft.-lbs. 

T'iT.  ri34  shows  the  8.F.  and  B.M.  diagrams  for  this  case. 

7  mr.l.    Let  M,'",  Mt'"  be  the  B.Ms.  due  to  the  proof  load  of  2888  lbs./Un. 

(i.  oil  th(>  span  BC.    Then 


Rr-R*'".    Rr-R*"\    M,"''M,' 


TlHTcfore 


A/,'" --672 


3a.-«-'2b 


and 


-^f,"'-  -8414019  ft.-lbi.-/?,"'xa08i 
R'"  -  -4Q35A  lbs.  -«.'". 


Alwj, 


/t,"'-3M0«7lb6.-At" 


624  THEORY  OF  STRUCTURES. 

Superposing  the  results  for  the  dead  and  proof  loads, 
R,  -ft,'  +  «,'" -61490  lbs.  -Rt, 
R,  -  R,'  +  fl,'"  -686512  lbs.  -R^ 


Fig.  535. 

PoinU  of  inflexion.     For  AB  (or  CD)  the  point  of  inflection  ia  «t  x  feet 
from  the  end  support,  x  being  given  by 

IJ.M.-0-/?,x-'-^x',    or    ^-^2^^ -96.3  ft. 


M'' 


Tlie  maximum  positive  B.M.  in  AB  (or  CD)  is  48.15  ft.  from  the  end  sup- 
I    f{  > 
port  nnd  its  value  is  r,  j  2^7  ~ '  .•♦**0,430  ft.-lbs. 


For  BC.    Measuring  x  from  B,  the 


3065 


B.M.-0-«,{208J+*)+ftri;-1277-208i(104l  +  x)-^««, 

or  x'-243.a'5x  +  7.'>34.1-0, 

and  x-38'.3    or    204'.  7. 

The  mnximum  poxitiw  B.M.  is  121)  ft.  from  B  (or  C),  jid  ita  value  is 
ff ,  ■  .W)  +  ff , ■  1 2 1  i  -  1 277  •  2) W J  •  22r>i  -    :,-U 21 4)  •  - 14,330.098  f t.-lbfc 
Fig.  535  shows  the  S.F.  and  U..\f.  diagrams  for  thi'^  rase. 


BRIDGE  OVER  THE  GARONNE. 


525 


Fourth.  Let  M,"",  .V,""  be  the  B.Ms,  due  to  the  proof  load  of  2688  lbs./  lin. 
ft.  on  the  two  spans  AB,  BC. 

Jl/,"" -A/,' +jV," +3/,'"-  -22,208,201  ft.-lbs. 

AU""-Mm'+M,"+Mm"'-  -12,979,437     " 
H,  =«,'  +  «," +  «,'"- 357,830  lbs,;         ff,-ft,'  +  i.,"+A"'-639,148  lbs. 
/^  =  «,'+«,"  +  «,'"- 988,598  "   ;         Rt-R/  +  n,"  +  Rt'"  -  70,876  " 


Fio.  536 

l''inl.<t  nf  inflexion.     For  AB  the  point  of  inflection  is  at  z  feet  from  A, 
I  UmK  given  by 


fl.V/-0-//,x-    .,   x",    or    X- 


2R, 
3'H55 


-180.5  ft. 


Tlir  maximum  (Kwitive  H.M.  is  00.25  ft.  from  .4  and  its  value  is 
}j  gjjj^- 16,146,600  ft.-lbs. 

I'lir  I  l)  the  poiii      f  inflection  is  x  feet  from  I),  x  being  jtiven  by 

r'77  2/? 

H.M.-O-V-    .j  x',    or    x-,.^^*^ -Ill  ft. 

TIk-  nmximum  positive  H.M.  is  .55.5  ft.  from  D  and  its  val'.e  is 

J  j^'y- 1,966,880  ft.-lbfc 


■i*I, 


526  THEORY  OF  STRUCTURES. 

For  BC  the  point  of  inflexion  is  x  feet  from  B,  x  being  given  by 

B.M. -0 -/?,(208.5  +  z)  +  ff^  -  ~(208.5 +i)', 

or  i*-262.156ar- -5839.15, 

and  X  -24.7  ft.    or     -=237.65  ft. 

The  maximum  positive  B.M.  is  at  131.15  ft.  from  B,  and  its  value  is 

22,184,000  ft.-lbs. 

Fig.  536  shows  the  S.F.  and  B.M.  diagrams  in  this  case. 

Ex.  57.  The  weights  on  five  wheels  passing  over  a  contintums  girder  of  toe 
spans,  each  of  50  ft.,  taken  in  order,  are  as  follows:  15,000  lbs.,  24,000  Ibi., 
24,000  lbs.,  24,000  lbs.,  24,000  lbs.  The  distances  of  the  wheels,  centre  to  centre, 
taken  in  the  same  order,  are  90  ins.,  56  ins.,  56  ins.,  856  ins.  Let  it  be  required 
to  place  the  wheels  in  such  a  position  as  to  give  the  maximum  bending  moment  at 
the  centre  pier. 

The  pier  must  evidently  lie  between  the  third  and  fourth  wheels. 

Let  X  be  the  distance  in  inches  of  the  weight  of  15,000  lbs.  from  the  nearest 
abutment.  The  remaining  two  weights  on  the  xpan  are  respectively  z  +90  iiu. 
and  z  +  146  ins.  from  the  same  abutment. 

The  two  weights  on  the  other  span  are  142  — x  ins.  and  198— z  ins.  respec- 
tively from  the  nearest  abutment. 

Hence,  if  M  is  the  bending  moment  at  the  centre  pier, 

15000  ''4000 

-iMxGOO--^  z(600'-z»)  +'^^  (z  +  90)|600'-(z+93)M 

''4000 
"•"7)00  f-'  +  l '♦6)1000' -(z  +  146) 'I 

24000 
+  ,TjJj-X(142-z)|600'-(l42-z)«| 


''4000 
+  —,  X(198-z)|fi00'-(198-z)«|, 


which  reduces  to 


xH'AHV  —  x')  1 

-•W-         ,„.       -+34.')fiOOO---Kz+90)»  +  (z  +  l40)'  +  (142-zj'  +  198-x)«!, 

But  M  is  to  Ix;  a  niaximuni,  Hnd  therefore 

rf.W     „    VMr-.ix'      1  ,         ^.  .  , 


which  reduces  to 
and 

Therefore 


SWING-BRIDGE. 

«'  +  1843.2j:- 167923.2 

x-87ins.-7J  ft. 

B.M.max.-  -3446581  in.-lbs. 


827 


Kx.  58.  A  svdng-bridge  cmsists  of  the  tail  end  AB,  and  of  n  span  BC,  of 
kriijth  I  ft.,  the  pivot  being  at  B.  The  ballast-box  of  weight  W  extends  over  a 
hiKjIh  AD  (  =2c  ft.),  and  the  weight  of  the  bridge  from  D  to  B  is  w  tons  per  lineal 
loul.    If  DB  =x,  if  pis  the  cost  per  ton  of  the  bridge,  and  if  q  is  the  cost  per  ton 

vj  the  ballast,  show  that  the  total  cost  is  a  minimum  when  x  +  c-(-  ~''\  and 

ihiit  the  corresponding  weight  of  the  ballast  is  u^(x+c).      Draw  the  S.F.  and 

B.M.  diagrams  for  the  bridge  when  open  and  when  closed,  taking  W-lOwe 
ami  l  =  ix  =  8c. 

For  weight  of  ballast-box,  take  moments  about  B  when  bridge  is  open. 

Then 

_,,    ,    .     u»x'    wl' 
W(c+x)+-^ 2". 

2   c+x  ' 
.\lso,  differentUting,       dW---  —^^^-^dx. 

-        (X+C)' 

Again,  total  cost ~W^  +  w(l+x)p-SL  min. 

Therefore  q-dW +wp-dx-0. 


''"     w  dx"^'!      (x+c)'    » 


from  which 


and 


(x  +  c) 


-'(•^) 


McTlIC,  too, 


w  r  -x'      p 


"--ir.v-fc")- 


N./'.  mill  li.M.  diagrams.     Bridge  open. 
•  I  I'l  />at  z  from  .1, 

1"  I'll!  0  lit  A  and  -  lOuv  at  D. 


528 


THEORY  OF  STRUCTURES. 


1 

■ 

i 

1' 

It, 

D  to  B  at  z  from  A, 

S,--6wc-2u'z, 

being  -lOuc  at  D  and  -\4wc  at  B. 

B  to  C  at  z  from  ^l,  since  R,  at  B'=20M)r, 

S. I2u>c  +  20wc-wiz-4c)~l2wc-wz, 


l»einR  8ur  at  B  and  0  at  C. 
Again,  -1  to  D  at  z  from  il, 


a  parabola  with  its  vertex  at  A, 
At  D  the  n.M.  =  -10uc». 
D  io  B  at  z  from  ^4, 

3/,  -  -  lOwciz  -c)  -|-(«-2c)», 

a  parabola  whose  vertex  is  on  the  left  of  X  at  a  point  measured  8c  horizon* 
tally  from  and  40ur'  vert  i  rally  above  -4. 

At  D  the  B.M.  -  -lOuC,  and  at  B  the  B.M.  -  -32u'c». 


1*-  *  O-  -»+• — X — »»- 


4B^ 


Fio.  537. — S.F.  and  B.M.  diagrams  with  bridge  open. 
C  to  /i  at  z  from  C, 

a  paraKihi  with  its  vcrte.x  at  C. 
At  /i!he  M.M.--:{'.'mt». 
Hriilijc  rloKvd.    The  bridge  is  now  rontinuowt  over  three  supporta, 

-2.U.(8c  +  4f)-+y(8c)'+    /      ■■'~^*(16c»-/')+      /      "'^.(Ifir'-t"), 

./o  Ju 


SWING-BRIDQE. 
or  -24J/k:-172u«'. 

I'litTofore  M,  —  -*fuc*=R,ir  —  i  Owe  ■  3c  —  2wc  •  c 

-/^.iSr-   Hue- 4c. 
Thus  ff,-fi,\Hr,     /i-.^s/iuc,    and     A.-lOHwc 

.1  tn  D  at  i  from  A, 

S,^R,-  iiwi  ■=  6,*,  we  —  5w2, 

Imimc H/j  ur  at  .1  and  -3ilur  at  D. 
1)  to  li  at  «  from  .4, 

Si—Ii,  -Hwf  —  wz—  —Hlwe—m, 

Uwa  -3J|h<"  at  D  and  -Sifu*  at  B. 
H  to  C  at  «  from  A, 

S,-Rt+R,-l2wc-u.iz-4c)-SHu>e-wt, 
bt'iiiK  Ulur  at  B  and  -S^uv  at  C. 


.-  '1 1 


Fin.  53K. — S.F.  and  It.M.  diagrama  with  bridge  cloa«d. 
Again,  .1  to  i>  at  «  from  A, 

M,  "  R,x  — ,z fi^w*  -Ju't', 

apar.'ilKila  with  its  vertex  liVfC  measured  horiiontaliy  and  *fff)ftBe*,  measured 

viTtii.illv  from  A. 


II:  'I 

if:  ^^ 


!l'\ 


530 


THEORY  OF  STRUCTURES. 

At  D  the  B.M.  -2^\u•c\    Thus  there  is  no  point  of  inflexion  in  AD. 
£)  to  B  at  «  from  .4, 


43 


w 


M.~R,z-10ux{z-c)—^{i-2cy--i2^wM  +  8wc*-jt\ 

which  may  be  written 

„      11065     ,        w  /^S  ^  \ ' 
^^•- TT52"*         TW+V  ' 


43 


a  parabola  with  its  vertex  on  the  left  of  A,  at  ^c  measured  horiiontally  and 

.  'ac'  measured  vertically  upwards  from  A. 
lio2 

43 
At  BtheB.M. ---gM*'- 

The  B.M.  is  nil  when 


43  u) 

Mt-0-- ^wcz  +  Svx* -  2«'i 


43 


or 


,i  +  -^ic-16c",    i.e.,  when    « -2.59c. 


C  to  fi  at  <  from  C, 


wt*    „  5  w«* 

Mt-Ra — 2"  "is"** — T' 


a  parabola  with  its  vertex  at  3Ac  measured  horizontally  and  VW««*  n»e«»- 
ured  vertically  from  C. 

Jo 

The  B.M.  at  B-3A««-8c-^(8c)'-  — gU«'. 
The  B.M.  is  nil  at  point  given  by 


3f,-0-3A««« — 2*'    °'"    '""A* 


Ex.  59.  1  con<inuou«  girder  of  two  spans  AB,  BC,  carrying  a  load  of  uni- 
form  intensity  w,  las  one  end  A  fixed,  and  the  other  end  rests  upon  the  support 
at  C.  If  the  bending  moments  at  A  and  B  are  equal,  show  that  Me  spans  are 
in  he  raiio  of  \^  to  VJ,  and  find  the  reactions  at  the  supporU,  W^  being  the 
load  upon  AB  and  W,  that  upon  BC. 


Therefore 


•"•"'"•"~4        3?, +2/, 


EXAMPLES. 


531 


ALo,       2M,+M,---'^,    or    Jtf,-Af. ffih*-—[ik^ 


Hence 


IWA'^WJ,' ^W^^    or    ZWJt,^2W.l,. 


4      3A+2/, 

""'^  7;■2»^■2^."2i,•  °'  /.■\2"»F,- 

Taking  moments  about  B, 

Therefore  «!  ~~2' 

and        «t-2*^'-4  i,'V«.+2/./ 

tr./6i,V+«.'-i.'\     ir./6t,+3f.-tl.\     3™. 
"  4  ^  (3i.+2yi,»    ;"4\    3i.+2J,    J     8     ' 

TF     5 
Hence,  too,  R,~W,+W,-Rt-Rt~Y +s^ 

Ex.  60.  A  continiumB  girder  of  four  equal  spans,  and  wUh  one  end  fixed 
at  the  first  support,  carries  a  uniformly  distributed  load.  How  mv£h  must  the 
third  support  be  raised  to  make  the  reactions  equal  (a;  at  the  third  and  fourth  sup- 
p/rtv;  (b)  at  the  fourth  and  fifth  supports? 


Ml 


Mi 


X 


M. 


u* 


■i 

Fio.  539. 


1 


Take  each  span -l(- 100  ft.)  and  let  w  be  the  intensity  of  the  load,  so 
th:it  ul  =  194  tons. 

From  condition  of  fixture 

trP 
2.V.+Af.--^. 

I4>t  d  he  the  amount  bv  which  the  third  support  must  be  raised.    Then,  by 

...      »r     Bid 
the  ceneral  equation,  and  taking  IS  — »-. 


532 


TIIEUHY  OF  STRUCTURES. 


wP 


.V,+4.U,  +  .U3--— +GA', 


il/,  +  4.U3  +  .l/,=  -^-12Ar, 


M:,  +  4M, 


vP 


+  6iV. 


Hpnoe 


e         144  :?:?         '>SH 


M, 


30 


4-.'(i.. 


41 


^'cP-~\,    .U,=  -:^»/'  +  ^Ar, 


38S 


9 


388 


07 


Taking  moments 


74fi         426 
R^l  +  «.'  =  2«/'  -^ .1/3  =  —  »•/'  - 7^ ^'^ 


R^l.R;2URjJ'f.M.JI^.P.^N. 


(1) 

(2) 

(3) 


(a)  If  Ri-'Rt,  those  throe  equations  give 
774..     774  A' W 


3T8"''  = 


U7 


A'=^ 


97     /' 


(6)  If/?,  =  /^.,eqs.  {l)an(^(2)give 
287    „     1182,,     1182  Eld         ,      .      287  «7' 
3"88"''='  97'^=  OTT"    "^"^    ''-4728W"- 

The  corresponding  reaetions  and  R.Ms.  ean  be  easily  deduced. 

FiX.  61.  .1  su-iiig-hridtfr  MiCi),  440  /^  in  length,  ha,s  e<,ual  arms  AB  and 
CD,  and  rests  m/x(N  ro/irr.s  at  B  and  C  which  run  in  a  circular  path  of  22  //. 
diameter.     Each  arm  i.\  a  tru.in  oj  9  panels. 

Take   .1 «  -  209' =  a  -  CD  and 
«('  =  '22'-6. 

Consider  the  effeet  of  a  load  P 

oncentrated  at  x  ft.  from  .4. 

Fio    540  Then 


4- 


A 

H O— 


'   B    C 


.4,1 _<.____-J         tv.1 

R|  «j  "♦      tons  ei 


2.V,(a  +6)  +3/.6-  -P-(o'-x') 


and 


SWlSC-liKIUGE 

(.V^>  +  2.1/3(6 +  «)=0. 


633 


Therefore 


9^     -  •-     -     2023120'^^  ' 


uiid 


^^*~     -  h(4«'+.S«6+3/>») 
.V, 


'2(a  +  b)" 


3/, 

21' 


Putting  A  -a,^***^,,,,  the  vahies  of  M,  and  -U,  in  foot-tons,  corresponding 


to  thf  conrentratioii  of  P  at  the  lat,  2d,  3d 
Ix'  tabulated  as  follows: 


.  and  8th  panel-points,  may 


Value  of  - 


.U2. 

Ma. 

1080.4  = 

-10  399P 

+  80-4  = 

+ 

4952P 

3234.1  = 

-20  019P 

+  154^  = 

+ 

.9533P 

453f).t  = 

-28.078P 

+  216.1  = 

+  1 

.3371P 

5460.1  = 

-33  798P 

+  260.4  = 

+  1 

6094P 

5S80.4  = 

-36  398P 

+  280.4  = 

+  1 

7332P 

5fi70.4  = 

-35  095P 

+  270.4  - 

+  1 

.6711P 

4704.4  = 

-29  118P 

+  224.4  = 

+  1 

.3866P 

285G.1  = 

-17.679P 

+  136^4  = 

+ 

.8419P 

Again,  taking  monuents  and  simplifying, 

10 ..        „    M, 


^'=^)-'^" 


/?•= 


43S9- 


The  values  of  the  rcartions,  corresponding  to  the  concentrations  of  P  at  the 
^1,  .'il,  3d  .  .  .  and  Sth  panol-points,  are  given  in  the  following  table: 


■"'{ 

R,  in  Tons. 

Kj  in  Tons. 

.S39IP 

.6561P 

6S20P 

1.2713P 

.5323P 

l.S()47P 

.393SP 

2  2156P 

2703P 

2  4620/' 

.  10.54P 

2  -.057P 

0S29P 

2  3(W7P 

O260P 

1  S153P 

a  in  Tontf. 

«4  in  Tons 

4976P 

0024  P 

9.-)78P 

0046P 

1  3434P 

00<i4P 

I  6171P 

0077P 

1  74I5P 

0()S3P 

1  6-'.)lP 

oasop 

1   -MWir 

.0066P 

845  )P 

.0040P 

-1 


Adding  all  these  reactions  together,  the  result  is  8P  tons,  which  verines 

the  calculations. 


534 


THEORY  OF  STRUCTURES. 


Similar  values  for  the  B.Ms.  and  rrarfitms  arc  obtained  for  the  arm  CD, 
and  by  means  of  the  preceding  tables  the  values  may  Ix'  found  for  every  dis- 
tribution of  the  panel  load.  Let  1,  .2,  3, .  .  .  10  denote  the  1st,  2d,  .  .  .  16th 
panel-points  from  the  |X)int  of  support  A.  The  following  table  of  B.Ms,  is 
ro^'  eaeiiy  prepared: 


Load  P  at 

Ml  in  Ft.-tom. 

.Ml  in  Ft.lonn. 

-    10  SJHJP 

+   49.-)2P 

1  and  2 

-   30  41.SP 

1  44WP 

1  to    3 

-  5.S  miP 

2  7S.-i«)P 

1  "    4 

-  92  294P 

4  3»r)0P 

1  "    5 

-12S  t)!)2P 

6  12S2P 

1  "    6 

-lfi.1  7S7P 

7  79<»3P 

1  "     7 

-192  iJaiP 

9  IfCiOP 

1  "    8 

-•2\o  rmp 

10  027.SP 

1  "    9 

-209  7421 P 

-    7  «.-,r2P 

1  "  10 

-JOH  3.V).-)P 

-  .30  7092P 

1  "  11 

-•200  0H44P 

-   71  Wi42P 

1  "  12 

-204  9.-)12P 

-108  •2fi22P 

1  "  13 

-2(W  .14 ISP 

-142  ()002P 

1  "  14 

-•202  (xmp 

-170  13H2P 

1  '•  15 

-•201  0.-.I4P 

-100  l.')72P 

1  "  10 

-'200  .V.f.2P 

-200  5.Vi2P 

The  H.Ms,  in  the  lant  line  are  equul,  and  this  mast  necci-surily  Ijc  the  case, 
as  ait  the  panel-points  are  loaded. 

A  sintilar  table  of  reactions  in  tons  may  also  U'  prepared  as  follows: 


l.<>a<l  /'at 

K, 

«, 

«>. 

«. 

Km  p 

t>.'>«>l  P 

-       4970P 

()0-24P 

1  and  2 

1   .-.211  P 

1   9274P 

-   1  4.'i.-.4P 

0070P 

1  to    A 

2  o.-v;{4P 

3  7:i2IP 

-  2  79SSP 

0134P 

1    *     4 

2  4 172/' 

5  9I77P 

-  4  4109P 

021  IP 

I  "     'i 

2  7I7.-.P 

S  41  (Kip 

-  0  1.174P 

0^2»4P 

1  "     fi 

2  HS29P 

10  »|Ck3P 

-  7  K3ti.-,P 

»j;i74P 

1    *     7 

2  'M\:,sp 

13  'JJtHW 

-   9  •2'2!»7P 

(M40P 

1  •'     M 

2  •Ht23/' 

l,-.  tl.r.3/' 

- 10  07.->«»P 

04«0P 

1  "     9 

2  9'Hi:iP 

14  IS94P 

-  s  jvmp 

074.5P 

1  •'   10 

3  (MlJItP 

12  7'.Mi2P 

-  .n  9.vi«p 

l.'»74P 

1    '   11 

3  (llt»<l/' 

II   1I7IP 

-   3  4.'i«!»P 

3'.>28P 

1  "   12 

3  <tl'»2/' 

9  37^^' 

-   0  9MS(|p 

.'i931  P 

1    '   13 

3  (t2i.ftP 

7  7.V*.-,/' 

+  1  •227r,p 

iWOOP 

I  "   14 

3  iVXiM' 

(i  n.MP 

3  0323 /• 

1   .'>I92P 

1  "   1". 

3  0379/' 

r>  I.-73P 

4  mmp 

2  •.?0I2P 

1  •'   Hi 

3  ()l(k3P 

4  9.->97/' 

4  9.597P 

3  tMorjp 

Ad(hnK  the  la.<t  line  toffi'ther  it  it  found  that  l)u>  total  sum  uf  the  reac- 
tions is  \i\P.  whii'h  verifies  the  riilculations 


M,\. I  STAGES  AND  DISADVANTAGES  OP  CONTiNUOUS  GIRDERS.  ^^ 

The  B.M.  and  8.F.  diagrama  for  all  distributions  of  the  panel  load  can  now 
-jwily  drawn. 

25.  Advantages    aad    DiiadTUitagM    of   Continuoui    Girden. — 

II  •  advantages  claimed  for  continuous  girders  are  facility  of  erec- 
tion, a  smaller  average  B.M.  and  therefore  a  saving  in  the  flange 
iiialcrial,  the  concentration  of  the  maximum  B.M.  over  the  piers, 
aixl  the  removal  of  a  portion  of  the  weight  from  the  centre  of  a 
?|.;iii  towards  the  piers.  Circumstances,  however,  may  modify  these 
ahuntages,  and  even  render  them  completely  valueless.  The 
tlaiino  stresses  are  governed  by  the  position  of  the  points  of  inflexion, 
which  under  a  moving  load  will  fluctuate  through  a  distance  de- 
1«  ndont  upon  the  number  of  intermediate  supports  and  upon  the 
naiiiro  of  the  loading.  In  bridges  in  which  the  ratio  of  the  dead 
liial  to  the  live  load  is  small  the  fluctuation  is  consiflerable,  so  that 
f.r  :i  .^K'nsible  length  of  the  main  girders  a  passing  train  will  sub- 
j.rt  local  memljers  to  stresses  which  are  altc.  lately  positive  and 
inL'ative.  This  neces-sitates  a  local  increase  of  material,  as  each 
111.  iiiImt  must  \ye  designed  to  bear  a  much  higher  stress  than  if  it 
urrc  strained  in  one  way  only. 

Anuin,  the  web  of  a  continuous  girder,  even  under  a  uniformly 
ii-iriliutcd  (load  load,  is  theoretically  heavier  than  if  each  span  were 
ml.  pendent,  and  its  weight  is  still  further  increased  when  it  has 
In  resist  the  con»plex  streww-s  induced  by  a  moving  load. 

Hence  in  such  bridges  the  blight  saving,  if  there  be  any,  cannot 
1..  aiil  to  countt-rbalancc  the  extra  laljor  of  calculation  and  work- 
1!  in>hip. 

In  Kinlers  subjected  to  a  dead  load  only,  and  in  bridges  in  which 
tip  ratio  of  the  dea«l  load  to  the  live  loml  is  large,  the  saving  be- 
111,  s  more  marked  and  increases  with  the  number  nf  intermediate 
-  !  l.nrt.s,  iM'ing  theoretically  a  maximum  when  the  number  is  in- 
1,1  lie.  This  maximuum  economy  may  be  approximated  to  in 
11  K  tire  by  making  the  end  s|)ans  alwut  four  fifths  the  intemietliate 

-I'.'IIIS. 

Ill  the  calculations  relating  to  the  Tlieorem  of  Three  Moments, 

It  li  IS  Ui'n  asHunie<i  that  the  ({uantity  El  is  consiant,  wl;ile  in  reality 

/    >\vn  for  mild  steel,  may  \ary  10  to  15  per  cent  from  a  mean 

'i.'.  and  /  may  vary  still  more.    It  does  not  apiiear,  however. 

t'  ii  tliis  variation  has  any  appn>ciable  effect  if  the  depth  of  the 


536 


THEORY  OF  STRICTUUKS. 


uinltT  or  truss  diangos  (jradiially,  but  the  ('ffect  may  become  very 
marked  with  a  rapid  change  of  depth,  as,  e.g.,  in  the  cast>  of  swing- 
bridges  of  the  triangular  tyjie. 

The  grapliical  metliod  of  treatment  may  still  lie  employed  by 
substituting  a  rehired  curve  for  the  actual  curve  of  moments, 
formed  by  changing  the  l(>nglhs  of  the  ordinates  in  the  ratio  of  the 
value  of  EI  at  a  datum  section  to  EI. 

It  is  often  founil  economical  to  increase  the  depth  of  the  girder 
over  the  piers,  which  introduces  a  local  stiffness  and  moves  the 
points  of  inflexion  farther  from  the  supports.  A  point  of  inflexion 
may  Im-  made  to  travel  a  short  distance  by  raising  or  depn-ssuig 
one  of  the  supiKirts. 

In  order  to  insure  the  full  advantage  of  continuity  the  utmost 
care  and  skill  are  recpiin'd  lM)th  in  design  ami  workmanship.  Al- 
lowance has  to  Ik«  made  for  the  excessive  expansion  ami  contructioii 
due  to  changes  of  tem|H'rature,  and  the  piers  and  abutments  mast 
U"  of  the  strongest  and  In'st  description,  so  that  there  shall  Ix-  no 
settlement.  Indeed,  ihe  difficulties  and  uncertainties  to  1m'  ilealt 
with  in  tlic  construction  of  continuous  girders  are  of  such  a  s«Ti(>U!< 
if  not  insurmountable  character  that  American  engineers  have 
almost  entin'ly  di-carded  their  us*-  except  for  draw-spans. 

Much,  in  fact,  is  men-  guesswork,  and  it  is  usual  in  practice  to 
Ix"  guided  by  ex|Krience,  which  confines  tin-  |K)ints  of  inflexion 
within  certain  siilc  limits. 

I'nder  these  circumstances  it  niay  prove  ilesirable  tu  jii  the 
jmitits  itj  iujUxioii  iilisnhitthi.  and  lln'  advantages  of  lining  .-o  ;ire 
(o)  that  the  ealeulalioii  of  the  well  stresses  ln'conus  I'asy  and  del'nite 
instead  of  In'ing  complie:ile.|  :itiil  e\eri  indetenninat  ■;  '/»)  tiint 
reversed  stresses  (for  wliicli  pill-trusses  are  le.ss  adapted  than  rivet^il 
lrii»-es)  iire  ;iIiiiosl  eiilire!\  avdhled;  ((■)  that  the  stress<'s  an-  nei 
sensililv  iitleeled  by  slight  ine(|ii.ilities  in  the  levcl.s  of  »he  supjMirls; 
(d  that  the  siraiiiiiig  due  to  a  change  of  t<'ni|ieralure  takes  plate 
under   iimre    li\<irali'e   eoiiditioiis. 

The  //('('    may  he  elTeeied  ill  Ihe  following  manner: 

>i:i   \  li,iii;i-  ii;,iy  I  e  iiilroduceil  at   the  ^eleeted  |Miinl. 

11;, •  iHiMJii  1)1  doing  SO  is  very  obslous  when  circunistiUHfH 
require  a  vide  eeiilr     -pall  and  two  short  side  spans. 

{b)  If  the  wel)  is  o|»eii,  i.e.,  lattice-work,  the  |Miint   of  inflexion 


.U.4A7.VC'Af  BEXDING  MOMEXTS  IN  COSTINCOVS  GIRDERS.  537 

ill  tlic  upjMT  flange  may  bo  fixed  by  cutting  the  flange  at  the  sdocleJ 
[Hiiiit  and  lowering  one  of  the  supjwrts  so  as  to  produce  a  slight 
npcuiiig  Ix'twecn  the  seven-d  parts.  The  |X)sition  of  the  point  of 
i!,lli  xion  in  the  lower  flange  is  then  defined  by  the  condition  that  the 
:i!p  iiraic  sum  of  the  horizontal  coni|)onents  of  the  stresses  in  the 
iliafTonais  intersected  by  a  line  joining  the  two  points  of  inflexion 
is  zero. 

It  must  \tv  remembered,  however,  that  this  fixing  of  the  points 
I ;'  iiirt«'xion,  or  the  cutting  of  the  chords,  destroys  the  profx'rty  of 
loiitimiity,  and,  inaeed,  is  the  essential  distinction  between  a  con- 
muious  girder  and  a  cantilever.  A  combined  ^.antilever  and  girder 
;  i.-scsscs  all  the  ud.antages  and  none  of  the  disadvantages  of  a 
"  ntiiiuous  ginh-r. 

1  our  methods  may  be  followed  in  the  erection  of  a  continuous 
irii-ilr-r,  viz.: 

1.  It  may  be  built  on  the  ground  and  lifted  into  place. 

_'.  it  may  Ik>  built  on  the  ground  and  rolled  endwise  over  the 
!'!!  r<.  As  the  bridg*  is  pushed  forward,  the  forward  end  acts  as  a 
iMiitiicvcr  for  the  whole  length  of  a  span  until  the  nex(  pier  is 
ri:i(liii|.     This  Pietiuxl  of  erection  is  coiiii.ion  in  France. 

.!    It  may  lie  built  in  position  on  a  .scaffold. 

J  Much  span  nuiy  Ik-  erected  .««'parately  and  continuity  pro- 
iiirl  by  .Kfcuri'ly  jointing  consecutive  end.-i,  having  drawn  to- 
iT'MMr  the  upjx-r  flanges.  A  nior*-  effective  distribution  of  tV 
I!  I'lTial  is  often  made  by  leaving  a  little  sfmce  Ix-tween  the  flaiig<>8 
■•■'■'■  fortninp  a  wedge-,>;liii|)ed  joint. 

if}.  Maximum  Bending  Moments  at  the  Points  of  Support  of  Con- 
iiniH'us  Girders  of  n  Equal  Spans. 

i '  I  the  figure  n'pr-M'nt  a  continuous  girder  of  n  spans,  I,  2,  'A, 
It     I   Ix'ing  the  «-l  intermediate  supports. 


n 


1 


A 


A 

r-1     r    r+l 

A 

A 

AAA 
iiii    541 

A 


n     1 

A^ 


r 


11 1 


*  \SK  I      ,'Wuriie  all  the  spans  to  be  of  the  same  length  /.  and 

I.  ";.  ir„-i,  u'n  be  the  inten^ties  of  loads  uniformly  dis- 

itt.l  (i\ir  the  1st,  2d,  .  .  .  (n  -  l)th  atid  nth  sfmns,  resjiectiveSy. 


638  THEORY  OF  STRUCTURES. 

By  the  Theorem  of  Three  Moments, 

4mi+ma--4(w'i+«'2), (O 

mi+4m2  +  m3--j(«'a+«'3). (2) 

(2 

ma-l-4m3-^»n4= -|"(w3  +  tt'4)i (3) 

m3  +  4m4+rns= -j(w4  +  «'») W 

m4  +  4m8  +  m«= -j(ws+«'«) w 


m«-3  +  4m„_2  +  mn_i=  -T(w'n-2  +  "'»-»)» 


(n-2) 


ni«-2  +  4TO„-i 


-  -tC"'»- !+«'«) 


(n-1) 


/»!    UII 


(I  m„  arc 


lK)th  z(>ro,  an  tlio  ginler  is  8upj>08ed  to  be  resting 
upon  the  abutuwiits  at  0  and  ii. 

From   tlicsc   (n-l)  «'(iiiation.<,   thf   U-ndinp  moments  mi,  wij, 


m„.\ 


niav  Ih'  found  in  U 


Tins  of  the  dihtrihuti'd  loads. 


Kliminating  in^  from  2  and  .'{, 


nil 


I'jWa-  Jm4-   -7lf"'2  +  "'3)-4(U'i  +  W4)|. 


(Xi) 


Kiiminalinp  wia  fn>n>  4  and  Xi, 


mi  t  ViwM  ^  lam 8 


O/'a  4- H":,)  -  4(  H-s +  «'«)  +-1>'>("'4 +  «'•)!•      (X») 


i  I "  li 


Kliniinaling  wu  from  .1  and  Za, 


h 


\(w 


, 4. „.  ) _4(„.,  +  ,,.4)  +  \rt(Ui -♦- H's)  -56(ir5  +  «>) |. 


(-t,) 


m 


MAXIMUM  BENDING  MOMENTS  IN  CONTINUOUS  GIRDERS    539 

Finally,  by  successively  eliminating  mn,  m«,    ...  m,_a, 

p 
i±On-im,_i= -j|(u'a+t«3)-4(u'3  +  uu)  +  15(tP4  +  ira)-.  .  . 

tlic  upper  or  lower  sign  \mng  taken  for  the  terms  within  the  brackets 
iiccording  as  n  is  odd  or  oven,  and  the  coefficients  an-i,  o»_a,  a*-!, 
.  being  (pven  by  the  law, 

On-i  —4a,.  2  — a,_a, 
o,_a  =  4ai.-s-o«-4i 


as  =  404-03  "=209, 
04  "=403  —  03=  .56, 
03-403-01-=  If), 
a2-4ai  —  4, 
o,  =  l. 


Tonimencing  with  cijuations  n-3  and  n-2,  and  proceeding  as 

lirl'iirr, 

p 
".    i"ii  fWn-i  -  -7ta,_2(u'i  +»ca)-a,_3(u'a+«'3)+o„_4(M's  +  «'4) 

-.    .    .   ±l.')(«'n-4  +  W„-3)T'*(»'»-3  +  "'.-3)±(«'«-2  +  W— l)t,  d) 

il '  iip|KT  or  lower  sign  l)oing  taken  for  the  terms  within  the  brackets 

;i(iiiicliiiK  lis  n  is  odd  or  oven. 

Si!\  liig  the  two  equations  y  and  z, 

P 
'■'  1  II-', _  I  - 1)  ^  -  7  l«n-  lO.-afi  <-  (On-  io,_2  -a,_ia»_8  -  !«>)» 

-f/j,_inn-3-o»-io,-4-3)«'3  +  .        ±(3a«_i+a,_4-o«-3)u'»-.» 

±((1,  .|+0n.-.l-   "»-2"f.     I  ^nii-2)W'iil, 

p 

'li  i    ±»U_i(a»„-i  -1)=  -^1  -a„-au'i  4  fo,_i+fl,-3-o„-3)«7a 

-(3o,-|  +0,-4-  0,-3)"'3T('0„_ia,_3-rt„-|fl,_4-3)U),_8 

t  !f»,_,rt,_a  -Or -!«■,- Il-  1)"'-    I  ±o,..in„    jtr.t, 


540 


THEORY  OF  STRUCTURES. 


Hence,  since  w,,  tr^,  .  .  .  w'«  are  positive  integers,  the  value  of 
m„  will  be  greatext  when  u'l,  k'z,  «'<,  K'e,  'Cs,  ...  are  greatest  and 
u'3,  ics.  U'l,  .  .  .  are  least;  and  the  value  of  w„_,  will  be  greate.it 
when  ?(•„,  «•„_,,  ir,_3,  »r„-4,  ...  are  greatj  '  and  Wn-2,  «'n-4, 
jr„-6,  .  .  •  ar<'  !«'«•'*<  I"  o*''*"""  «ord.s,  the  Iwnding  moments  at 
the  1st  and  (n-l)th  interniechate  supjwrts  have  their  maximum 
values  when  the  two  spans  adjaeent  to  the  supjwrt  in  (juestion, 
and  then  every  alternate  span,  are  loaded,  and  the  remaining  spans 

unloaded. 

»i2,  m3,  .  .  .  Wn-2  "»ay  now  be  easily  determined. 
Thus,  by  eq.  (1), 

P 

PI  4 

-  -T  \  (wi  +  ""a)  -  '-,  — ^iflB-iam-aWi 

4     ^  fl    n-l         * 

+  (fln-l"n-2-««-l''n-;»-l)"'2-   .    .    .| 
P  X 

=  --,   ,      — i  J(fl^-i  -  1  -4rtn_in„_.2)U'i 

4(yB-i-i) 

+  (cP„-.x     \      \a„-in„-2-^4n„-ia„-a  +  i)wa  +  .  .  ,j 
But  a„-i-  4fl„-2    "..-:«•     Therefore 

w,=  -  ,     ..--—,     \-((i„    if(,.-:i4-l)i(i  +  (3«„-i<i„-3  +  3)ir2  +  .  .  .1. 

4(«-„    1      1) 

,11(1  is  (irol'-^l  when  v..  "•,.  » 6.  ""?.  ...  are  greatest  and  «•,,  w,, 
SiiiiilMrly.  by  e<is   1 1 1  ami  (2), 


//*■, 


/2  P 

4  4 


l(fi„_«a,-i+4)«-, -(3a.-ia,-4->  «2)"-2 


MAXIMUM  BENDISa  MOMENTS  IN  CONTINUOUS  GIRDERS.  541 

ami  is  ffreates  when  wi,  Wa,  W4,  w^,  wg,  ...  are  greatest  and  Wa, 
i/'s.  1C7,  W9,  .  .  .  are  least. 

Thiis  the  general  principle  may  be  enunciated,  that  "  in  a  horir 
:iiiilnl  rontinuous  girder  of  n  equal  spans,  uith  its  ends  resting  upon 
In;,  olmtments,  the  bending  moment  at  an  intermediate  support  is 
iimitrst  when  the  two  spans  adjacent  to  such  support,  and  the  alternate 
/.«;  -v  counting  in  both  directions,  carry  uniformly  distributed  loads, 
tlu-  riDKtinder  of  the  spans  being  tinloaded." 

Ca.sk  II.  The  principle  deduced  in  Case  I  also  holds  true  when 
till   loiuls  are  distributed  in  any  arbitrary  manner. 

("oiusidor  the  effect  of  a  weight  w  in  the  rth  span  concentrated 
at  a  |)oint  distant  p  from  the  (r-l)th  support. 

By  the  Theorem  of  Three  Moments, 

4mi+»i2=0 (1) 

»w,+4m2  +  m3=0, (2) 

»n2  +  4m3+m4=0, (3) 


;w,_3+4m,-i+mr= -ti-yCP-p")"-^,  suppose  .    .     .     (r-1) 

1 

w,.i+4mr+TO,.n-=  -v>i     \l^-{l-p)^ 

'  --«'|(/-p)(2i-p)--B,  suppose,  .     .    (r) 
nj,  +  4m,+i4-wi,+3  -0,       (r+1) 


^'IM 


••-rfl 


nin-s  +  4r»„-2  +  'nfi-i  ""O. (n— 2) 

m,_i+4ni„-i  =0 (n-1) 

By  cMuations  (1),  (2).  (3),  .  .  .  (r-2), 

11  1  ^1  nr-i 

m,- --7-nia-Tl'"3--ki[»»4-.  .  .=  +r — "'"'-a'"  ±1-— ' 


16" 


56 


an-2 


Or- 1 


542 


THEORY  OF  STRUCTURES. 


the  upper  or  lower  sign  being  taken  according  as  r  is  even  or 
odd. 

By  equations  (n  - 1 ),  (n- 2),  (n- 3),  .  .  .  (r  +  1), 

»J,+2  ,   *",  +  ! 


On-r-l 


-T- 


On-r-l' 


The  coefficients  a  are  given  by  the  same  law  as  for  the  coeffi- 
cients a  in  Case  I.    Thus 

Or-a  ,  Om-r 

m,_  2  =  — Mr- 1    and    m,+ 1  — m,. 

Or- 1  flii-r-fl 

Substituting  these  values  of  »n,_2  and  m,+i  in  the  (r— l)th  and 
rth  etjuations, 

and  mr-i+mJi "^^^)=. -fl-m,_i+m,«, 


where 


5-4--^ —    and    c-4  — ■ 

Or-l  On-r-t-l 


Honce,  solving  the  last  two  equations, 

Ac-B 


n»r_i-- 


6c  -1 


and     »»,-  — 


bc-l' 


The  rat'"*  -'^  and  -"-''-  are  each  less  than  unity,  and  hence 

(tr~l  On-r+l 

b  and  c  aiT  each  <4  and  >3. 

It  may  now  easily  lie  shown  that  Ar  —  B  and  Bb  —  A  are  each 
IK)»itivi'.     Ih'nce  wi,_i  and  w,  are  l)oth  of  the  same  sign. 

Tho  iH'tiding  moment  m,  at  any  intermediate  supjwrt  on  the 
left  of  r  -  I  is  given  by 


nta 


+  — *-ntp- 1  if  7  and  r  are  the  one  even  and  the  other  odd, 


rtr-l 


MAXIMUM  BESDtSa  MOMENTS  IS  CONTINUOUS  GIRDERS.  543 


wiu-  — ^-m,-i  if  a  and  r  are  both  even  or  both  odd. 

"  0,-1  ^ 

Thus  the  bending  moment  at  the  9th  support  is  increased  in 
tlic  former  case  and  diminished  in  the  latter. 
If  9  is  on  the  right  of  r, 

m^  =  +  *~'"*"'m,  if  q  and  r  are  both  even  or  both  odd, 

0»-r-fl 


or 


»i,  =  -  *  *'*''m,  if  q  and  r  are  the  one  even  and  the  other  odd» 

and  the  bending  moment  on  the  9th  support  is  increased  in  the 
foriner  case  and  diminished  in  the  latter. 

Thus  the  general  principle  may  be  enunciated,  that,  "in  a 
horijumtal  cmitinuoiM  girder  of  n  equal  spann,  witii  ita  ends  resting 
uixtn  two  abutments,  the  bending  moment  ai  an  intermedin^  support 
is  ip-eatesl  when  the  ttvo  spans  adjacent  to  such  support,  and  the 
(iltirnate  spans  counting  in  both  directions,  are  loaded,  the  remaindtr 
I'j  the  spans  being  unloaded. 

Case  III.  The  same  general  principle  still  holds  true  when  the 
two  end  spans  are  of  different  lengths. 

i:.g.,  let  the  length  of  the  first  span  be  kl,  k  being  a  numerical 
c.»rticient,  and  let  2(1  +A:)-x. 

Kq.  (1)  now  becomes 


mix+mj-0. 

Fl^Hx>eding  as 

before, 

Ml         Wa 
61          fca 

wis        m^ 
ba'      b,'^' 

tllf- 

ifX'fficienUi  bj 

,  b^bs,  .  .  . 

being  given  by  t 

viz 

•» 

^2  =  /, 

/»,»4fe3 

-6,-4x-l, 

64-463 

-6a  =  l.'u--4, 

6,-46, 

•        •         • 

-62-5&r-15, 

544 


THKOKY  OF  STRUCTURES. 


The  two  s«'ts  of  (••M>flici('iits  (a)  and  (b)  are  idontical  when  x  =  4, 
aixl  wIk'Ii  j->4,  all  the  (•(K-lIicifUts  h  cxcopt  the  first  (6i  =  l)  are 
mimcrically  iiicrrascil. 

Ilciu'c  the  saiiic  Kciit'ral  n'sults  rtill  follow. 

A.^.—Thc  «'(iualions  giving  m,  arc  simple  and  easily  applicable 
ill  i)ractiee.     They  may  In-  written 


m 


,    ,  +-^''i-'f     ■[''  if  (/  is  on  the  left  of  r, 
'         «,_  1  w  -  1 


and  ,.,,=  ±""— ^V^^ifvisontherightofr. 

"  n  -  r  +  I     "'         * 


If  there  are  several  weights  on  the  nh  span, 


A==l'^{r--ir)     and     Ji  =  l-ir^j(l-pH2l-p). 

F.X  fi''  Thr  vimlurt  ovrr  Ihr  Okw  consisUt  of  two  rnd  upaiix,  each  of  94  /<., 
,„„l  }>■•■  in  ,rm,;l,al,  sf^ws,  mrh  ,./  T-'ti  //.  The  platjorm  w  carrial  by  Iw,  nuiin 
^inlrrs  irhirh  ur,  o>„lmuoi4K  from  rmi  to  rml.  The  total  deiui  load  u}^m  the 
\,,rdrr<  ouv,  Ix  I'lk  n  at  ,mc  ton  (../  JOtJO  //»«.)  /it  lineal  foot. 

lKiii.tr  the  siipixirts,  tukc-ii  in  order,  by  the  letters  a,  b,  c,  d,  e,  f,  g,  A.  and 
l.t  i;  1h>  r  (iniriii  t<>  liiid  the  inaxiiiium  U-ndinR  niomcnt  at  d  when  the  bridge 
i^  .  il.j.-.t.a  to  an  ad.litional  pr.K.f  load  of  li  tons  imt  lim-al  foot. 

Tl..-  -nan-,  ab,  ol,  dr,  jg  of  ea.h  (fit.l.T  carry  1  \  tons  |Kr  lineal  f(K)t. 

ri,..  s,v,ns  /«•,  r/.  i,h  of  .-ach  jtinlor  larry  J  ton  i«r  lineal  foot. 

Denoting  tlu-  »«Mi.ling  moments  at  a,  b,  c,  d,  r,  /,  y.  h.  respectively,  by 
m  w     the  internMHliate  siwins  liy  /,  the  end  sjwns  by  kl,  and  rerncm- 


'xTinu  that  m,  -0-'",-  ^^''  '"'^'^ 


2m»(il+l)+m,--jaii  +  i), 


ij  +  4mj  +  m.-  -  j{i  +  li). 


TO,  +  4m.  +  m,-  -  rdJ  +M)i 


m. 


I* 
+  4»u  +  m,='  -jdi+i). 


m 


,  +  4m,  +  m,-=  -^  (|  +  1}). 


hi 


»n,  +  2r«,a  +  l)--.  (li+*-i). 


VIADICT  OVER  THE  OSSE. 


But  /,  =  iV«  - 1,  very  nearly. 

Thi-n'fore 


7m,  +  2m,-  -^  ^, 


^'7 
m,  +  4»ii  +  m4- -  J  .7, 


/'  5 


/•   7 
in*  +  •*»!,  +  m,  "■  —  7-  — , 


/'   7 
tWj  +  4»i,  +  m^  ■"  —  J   — , 


2m,  +  TnUf  • 


'-1  I? 
4     4" 


545 

.    .    .  (1) 

...  (2) 

...  (3) 

...  (4) 

...  (5) 

...  (6) 


Fn.m  eqs.  (1),  (2),  (3), 


/'   347 
97m«  +  26m.--'-  ^. 
4      a 


From  eqa.  (4),  (5),  (6), 


;>   279 
2fiTO4  +  97m,- -J   -J-. 


lltMice  m«,  the  maximum  required, 
r      19151 


-  -f)05.5  ft.-tons. 


4    8X8733 
Taklic  or  RRsii.iF.NrKs  or  Si-nivu  Materiaj^w.* 


llM-« 

'  :i-i  -nil.  iiiilmrdcncd 
'  -.M  -'<v\,  h:ir<lcned. 

'"I'l-i" 

<il.',»  

'iuil   Ilirlal 

Mll,|M,...l 

Mill   ■••!.  Iitinipned. . . 

I'lin-nh.r-lironie 

''Wi.M.-iii  Iron 


Maximum 

Tfliinile  iir 

('4impreti)«ivr 

StreM  in 
ThouiwnilK 
of  Pnuncin 
which  will 
not  Protlurc 
Set. 


6  95 

8() 
I9U 
4  3 
4  S 
(i  2 

3S 

70  5 

19  7 

24 


fin 

MilU<»nH 

r 

n(  I'liUnfl^ 

l»r 

jt: 

Hquare 

Inch. 

2  «2 

9  2 

:«> 

107 

3« 

rm 

15 

fi2 

K 

1  2« 

9  9 

2 

:«) 

2(1 

:«) 

S3 

H 

13  H5 

20 

10 

Maximum 

8lirar 

SintM  in 

Thouxanila 

a  in 

Milliona 

of  Piiumlr 

iir  Pounilf 

which  will 

not  Produce 

Set. 

per 

Square 

Inch 

5  2 

3  4 

«vt 

11 

145 

13 

2.9 

5.6 

4  15 

3.7 

26.5 

11 

53 

11 

U,5 

5  25 

20 

10.5 

30 


4 

186 
809 
.75 

2.33 

32 

12M 

20 

19 


•  Perry's  "Applied  MechanioB." 


I*  ;jt 


MICROCOPY    RESOLUTION    TEST   CHART 

ANSI  and  ISO  TEST  CHA..^   Mo   2 


^       /IPPLIED    "    ACE      Inc 


...ti^^t^r  N««   •'o'k  >-.! 


546 


THEORY  OF  STRUCTURES. 


EXAMPLES. 


1.  A  flat  spiral  sprinp  .2  in.  wido  and  .03  in.  thifk  is  subjeetod  to  a  bond- 
ing niDmcnt  nf  10  iii.-lhs.     Find  its  ritdius  of  curvatun',  A'  Ix-inp  3fi,0(X).000  jhs. 

AiK.   l.Crj  ins. 

2.  A  stniidit  strip  of  tonijH-rcd  steel  .7  in.  broad.  .1  in.  thick  (ropre.-ent- 
ing  the  tlriiHi  o'i  a  beanii,  is  subjei-ted  to  a  iM-iidinfj  moment  oi  KK)  in. -lbs. 
Find  its  radius  of  curvature.  ■^'t"-  -1  ins. 

3.  Whit  is  the  fireatest  stress  in  a  bar  which  is  subject  to  a  lK>nding 
moment  of  4000  in. -lbs  (11  if  the  section  is  a  circle  of  ]  in.  radius;  ('_•)  if  of 
I  form,  '2  ins.  doi'p  and  1  in.  wide,  the  web  and  llanpes  each  boiiif;  f  in.  thick? 

Am.  (1)  5.1  tons:  (2)  3.1f)  tons. 

4.  \^"Xl"  steel  bar  i.s  In-nt  into  a  circle  of  .")0  ft.  radius.  Find  the 
maximum  .stress  induced  in  the  materiid,  the  coefficient  of  elasticity  Inking 
'JS,(KK),(KK)  lbs. 's(|.  in.  .l"--  S7.')0  lbs. 

5.  .\  square  bar  J"Xj"  is  subjected  to  a  iM-nding  moment  of  3.")()  in.-lbs. 
What  is  the  pri-atest  stress  in  the  bar,  and  the  radius  of  the  circle  into 
which  it  is  Ix-nt,  E  boiiig  taken  at  2,0(K),(XH»  Ibs./sq.  in.? 

.\iis.  4978  lbs.;  V2.-}  ft. 

6.  A  >."  K\"  bar  is  l)ent  into  the  arc  of  a  circle  of  10<K)  ft.  radius.  Find  the 
moment  of  resistance  and  the  maximum  stress  develojx'd,  E  IxMng  30,(MK),()()0 
Ibs./sq.  in.  •l""'-  '-•">0  Ibs./sq.  in.;  iOS\  in.-lbs. 

7.  .\  round  stool  rod  J  in.  in  diameter,  resting  U|K)n  supports  ,1  and  /?, 
4  ft.  apart,  projects  1  ft.  In-yond  .1  and  9, ins.  lieyond  li.  Tlie  extremity 
beyond  .1  is  load'd  with  a  weiirht  of  r_>  ll)s.,  and  that  iM-yond  D  with  a  weight 
of  ir>  lbs.  Neglecting  the  weight  of  the  rod.  investigate  the  curvature  of  liio 
riKl  between  the  sup(Hirts,  and  calcul.itc  the  greatest  deflection  U'tween  .1 
and  n.  Find  also  the  greatest  iiitm.sity  of  stress  in  the  rod  duo  to  the  tuD 
applied  forces.     (E  -3(t.(KKI,(KM)  !lw.) 

Ans.  Noutral  axis  between  A  and  B  is  are  of  a  oirrlc;  .45  in.;  11,730 
Ibs./scp  in, 

8.  Taking  (i  -<'i.')00  toiis/s(|.  in.,  find  the  <leflection  ilur  /«  shrnr  of  a  '2"  ■ '."' 
steel  bar,  10"  long,  lixod  at  one  end  and  loaded  at  the  other  with  a  weight 
of  J  t.iiis.  •*"••*•  -^[^  •"■ 

g.  At  a  certain  point  a  bar  is  subjected  to  a  UMiding  moment  of  40(K)  in.-lbs., 
the  greatest  stress  in  the  material  U-iiig  10,(KK)  lbs.,S(|.  in.  If  the  bar  i.s  of 
circular  section,  lind  its  diameter;  if  it  is  square  in  section,  fiiul  the  side.  If 
its  c.Mflicieut  of  eia.sticity  is  ^s.OOO.OtK)  lbs.,  find  the  curvature. 

.\iis.   l,.V»ins. ;  1.34  ins.;  rndius  oj  lurmlurc.  ISC.'J  ft.  and  150.2  ft. 

to.  A  horizontal  iM-am  of  depth  d.  breadth  />,  and  lenuth  12  ft  r<-«t«  u|Mm 
supports  at  the  ends.  A  weight  W  at  the  centre  <lellccts  the  lH<am  .1  in. 
whiii  the  side  of  length  U  is  verticil.  An  addition.il  weight  of  12.50  lbs.  is 
rc.|uired  to  produce  the  same  detlc<tion  when  the  side  of  hiigth  il  In  vertinil. 
If  (/  .'/.  .-.lid  li  E-  l,.'uO,tXJt)  lbs.,  fiiid  thr  :,irtiuiial  ares  of  the  l>f-<ini  st»!  '!» 
value  of  IF.  •>""•  72».|.  ins.;  41«i  lbs. 


EX  A  \fPLES. 


547 


ir  A  band-saw  is  J  in.  wide,  .02  in.  thick,  and  passes  over  two  pulley.s 
rach  \2  ins.  in  diameter.  If  the  tight  tension  is  100  lbs.  find  the  maximum 
inlensity  of  stress  in  ;  leb-nd.     (£'=30,000,000  lbs.)     Ans.  60,000  Ibs./sq.  in. 

12.  .\  steel  St.  .  ,"'X.l"  has  an  initial  curvature  of  .f)025.  Find  the  H.M. 
and  the  corresponding  maximum  stress  develojx'd  in  the  material  which  will 
ui)  straighten  the  strip;  {li)  give  it  a  curvature  in  the  opposite  direction  equal 
to  tile  initial  curvature,  E  being  .3f),()00,000  Ibs./sq.  in. 

-U/.s.  («)  l,j  iti.-lbs.;  4.")00  Ibs./sc).  in.;   {[>)  30  in.-lbs.,  9000  lljs./sq.  in. 

13.  A  laminated  spiing  of  3  ft.  span  has  20  plates,  each  37.")  in.  thick  and 
■.'.!».")  ins.  wide.  Calcuhite  the  deflection  when  centrally  loaded  with  .">  tons. 
(/:.'  =  1 1  ,(i00  tonsAsq.  in.)  .1  us.  2.4.J  ins. 

14.  -V  laminated  spring  of  40  ins.  span  has  twelve  plates  each  .37.^  in.  thick 
and  3.40  ins.  wide.  Calculate  the  deflection  when  centrally  loaded  with  4  tons. 
(/•;  =  Il,(iOO  toiis/sq.  in.)  Ans.  3.9  ins. 

15.  A  laminated  spring  of  7."i  ins.  span  has  thirteen  plates,  each  .39  inch 
tliii  k  and  3..'5  ins.  wide.  Calculate  the  deflection  whcti  centrally  loaded  with 
I  ton.     (^"11,(K)0  tons/sq.  in.)  .4n.<f.  .I  ins. 

16.  The  section  through  the  back  of  a  hook  is  a  trapezium  with  the  wide 
.-icic  inwards.  The  narrow  side  is  1  in.  and  the  wide  side  2  ins.;  the  depth 
ol  tlic  s»Mti(Hi  is  2J  ins. ;  the  line  of  pull  is  1  J  ins.  from  the  wide  .side  of  the  sec- 
linii.  Calculate  the  load  on  the  hook  that  will  prtxluce  a  tensile  skin  stres."? 
nf  7  tons/sq.  in.  .In.,.  3.87  tons. 

17.  A  .steel  3"Xi"  eye-bar  has  a  4-in.  pin-hole  and  an  S-in.  head.  The 
sinws  along  the  bar  is  30,000  lbs.,  but  its  line  of  action  is  J  in.  from  the  axis 
ni  llie  bar.  Find  the  maximum  and  minimum  stresses  in  the  main  Inidy  of 
I  lie  bar,  and  also  in  the  metal  at  a  diametral  section  through  the  pin-hole  at 
n,'lit  angles  to  the  line  of  stress.     Ann.  26,06f)|  lbs.;  0;  12,I42J,  7,S.")7i  llis. 

18.  The  direction  of  the  30,(KX)  lbs.  pull  on  a  0"X1"  eve-bar  is  jiarallef 
I'l  ami  I  in.  from  the  axis.  Find  the  maximum  and  mininunn  stn'sses  iti 
111.'  material.  ,l„,s.  10,(KK)  lbs.  an.l  0. 

19.  If  the  pin-holes  for  a  bridge  eye-bar  were  drilled  out  of  truth  sidf  way.H 
end  the  main  body  of  the  bar  were  "» ins.  wide  and  2  ins.  thick,  wii.it  pi(i;><.rti<tii 
AMiild  the  maximum  stress  In-ar  to  the  mean  over  any  cross-section  of  the 
1 .11  iit  which  the  mean  hne  of  forw  was  |  ^n.  from  the  middle  of  the  section? 

.l/(.s-.  23  to  '20. 

JO,  Calculate  the  maximum  and  minimum  stresses  in  a  menilM>r  of  1  itcuiar 
-'lion  I J  ins.  ill  diiuneter  when  subjected  to  a  pull  of  tMKH)  lbs.  and  a  bciid- 
111-  nionient  of  40(K)  in.-lbs.  Ans.   l.'i,4(»2  lbs.  and  N074  Ibs./sq.  in. 

it.  A  bar  of  steel  2i"X|"  in  section  is  subjected  to  a  pull  of  l,"i  toiw. 
Ill  •  iiican  line  of  the  load  pasws  J  in.  from  the  centre  of  gravity  of  the  s«'c- 
'  n  in  the  direction  of  the  width  of  the  bar.  Find  the  greatest  ami  least 
in-is  on  the  bar.  Am    1."i..3(l  and  3.84  tons/sq.  in. 

a.  The  horizontal  section  of  a  crane-hiMik  is  a  rectangle  3..")  in-,  in  width, 

!lic  thickness  of  the  hook  Ijeing  1.3  ins.    The  stress  in  a  hnriKontal  section 

-  not  to  cxcce<l  (i  tons  s«|.  in.     Find  the  maximum  load  which  can  lie  raised, 

■■^■■■-  'vTiHonta!  di-tanee  betw.t-n  the  tcjln-of  the  wrtion  and  thf  liri.of  ...ti.m 

ii  ihe  load  being  2.5  inn.  Anx.  5.1(W»  ton  . 


548 


THEORY  OF  STRUCTURES. 


J3.  In  a  steel  S"Xli"  cyp-bar  it  was  found  that  the  line  of  artion  of  a 
stress  of  100,000  lbs.  was  .25  in.  from  the  centre  Calculate  the  extra  stress 
produred  by  this.  •l"'-.  l'J20  Ibs./sq.  in. 

24.  An  iron  bar  is  rectangular  in  section,  its  width  iK-inj?  '6  ins,  and  its  tluck- 
ncss  1  in.  The  tensile  strength  of  the  metal  is  50,000  Ibs./sq.  in.  Find  the 
total  tensile  force  which  will  break  the  bar,  the  line  of  action  of  the  pull  being 
i  in.  distant  from  the  axis  of  the  bar.  Aii.9.  S5,714S  lbs. 

25.  .\  tension  bar  8  ins.  wide  and  IJ  in.s.  thiek,  is  slightly  cur\'ed  in  the 
plane  of  its  width,  so  that  the  mean  line  of  the  stress  passt-s  2  ins.  from  the  axis 
at  the  middle  of  the  bar  Calculate  the  maximum  and  minimum  stresses  io 
the  material.     Total  load  on  bar,  25  tons.  .ln.<i.  6J  and  IJ  tons/sq.  in. 

26.  A  tension  bar  of  T  section  has  a  2"Xj"  flange  and  a  4"Xi"  web. 
The  line  of  action  of  the  pull  /'  on  the  bar  is  1  in.  from  the  centre  of  gravity 
of  the  .section.  Find  the  maximum  and  minimum  stresses  developed  in  the 
metal.  Am.  IViP  and  -  ,",,»tP  or  I^IP  and  A',  A 

27.  A  horizontal  beam  of  weight  P  rests  at  the  ends  in  recesses.  A  weight 
W  sus|x>nded  from  the  centre  of  the  beam  by  a  string  .3  ft.  in  length  makes 
(A)  revolutions  jmt  minute.  Find  the  minimum  value  of  Pso  that  the  Iwam  may 
not  rise  out  of  the  rece.sses.  Atus.    iV«P. 

28.  Find  the  skin  .stress  due  to  bending  in  a  coupling-rod  from  the  follow- 
ing data:  Radius  of  crank,  10  ins.;  length  of  rod,  4  ft.;  diameter  of  rod, 
3  ins.;  number  of  revolutions  per  minute,  120.  Am.  1092  Ibs./sq  in.  1 

2Q.  The  coupling-rod  of  a  locomotive  is  10  ft.  in  length,  the  crank-nidius 
i.s  1  ft.  in  length,  and  the  driving-wheels  are  fi  ft.  in  diameter.  What  must 
be  the  depth  of  the  ro<l  if  the  stress  in  t  is  not  to  exceed  5  tons  \h}T  square 
iiKJi  at  70  miles  an  hour?  A  tis.  1 1 .2S  ins. 

30.  .\  locomotive  has  two  pairs  of  f)-ft.  driving-whe«>ls  and  S-ft.  coupling- 
rods  (>  ins.  deep,  of  wrought  iron,  capable  of  withstanding  a  working  load  of 
20,(HK)  lbs.  |)er  .scjuare  inch.  The  ir.inks  are  12  ins.  in  length.  Find  the  speed 
i'orrcs|M)nding  to  the  greatest  centrifugal  force.        Aii.i.  \)0X>  miles  jx>r  hour. 

31.  .\  shaft  5J  ins.  deep-  5  ins.  wideXitS  ins.  long  has  one  end  absolutely 
li\cd,  while  at  the  other  a  wheel  turns  at  the  rate  of  270  revolutions  per  minute; 
a  weight  of  20  lbs.  is  concentrated  in  the  rim,  its  centre  of  gravity  lieing  2J 
ft.  from  th(>  axis  of  the  shaft.  Find  the  maximum  stress  in  the  material  of 
the  sliaft.  and  also  find  the  maximum  deviation  of  the  sluift  from  (he  .straight, 
/■;iM'iiig27,<HH).(HK)lbs.  Ans.  4StK)  lbs.  jht  sq.  in.;  .;U43I  in. 

32.  An  iron  bar  12  ft.  long  and  <i  ins.  deep  is  held  freely  at  the  ends  .'ind 
rnpidlv  rutated.  every  |Miint  in  its  axis  docribing  a  circle  of  24  ins.  radius. 
iMiid  the  number  of  revolutions  jier  minute  for  which  the  maximum  intensity 
of  stress  is  lO.tKK)  lbs.  's<i.  in.  Ann.  1210. 

n.  \  2-111,  wrought-iron  bar  10  ft.  lotig  is  held  at  the  ends  and  is  whirled 
■•ihi'  It  :'.  Mirallcl  axis  at  the  rate  of  .'"lO  revolutions  jK'r  miniite.  If  the  distance 
Ih-Iui  1:1  liie  a\is  of  the  bar  and  the  axis  of  rotation  is  10  ft.,  fnid  the  niaxinium 
Htrt'.ss   to   wliiih   the   material   is  siihjceted,  Aiis.   17,I4H..")  il>s.  sq.  i:i. 

34.  A  steel  coupling-rod  for  a  loeoniotive  with  t»-ft.  tirivers  and  a  crank- 


EXAyrPLES. 


olQ' 


radius  of  1  ft.  is  10  ft.  long  and  has  an  I  cross-section  with  a  4"xj"  wcl> 
jiiid  equal  2"Xj"  flanges.  Find  the  maximum  intensity  of  stress  for  a  speed 
of  SO  miles  per  hour.  Auk.  19,410  Ibs./sq.  in. 

35.  The  coupling-rod  of  a  locomotive  is  of  uniform  I  .section  and  of  the 
fallowing  dimensions:  Dei)th=fi  ins.,  width  =.3  ins.,  thickness  of  \\eb  =  li  ins., 
thickness  of  flanges  =  1  in.,  length  between  centres  =91  ins.  The  driving- 
wheels  have  a  diameter  of  82  ins.  and  the  crank-throw  is  14  ins.  Assuming 
tiiat  the  rod  weighs  .3  Ib./cu.  in.,  find  the  ma.ximum  stress  due  to  bending 
when  the  engine  run.s  at  70  miles  per  hour.  Ans.  824.5  Ibs./sq.  in. 

36.  Find  the  bending  stress  in  the  middle  section  of  a  coupling-rod  of 
rectangular  section  from  the  following  data:  Radius  of  coupling-crank,  11  ins.; 
length  of  coupling-rod,  8  ft.;   depth  of  coupling-rod,  4.5  ins.;   width,  2  ins.; 
'revolutions  per  minute,  200.  Arts.  .'5797  Ibs./sq.  in. 

If  the  rod  has  an  I  section  with  a  3"  X  1"  web,  what  will  the  Ijending  stress 
be?  .4/(.v.  4537  lbs. /s<i.  in. 

7.  A  unif'irmly  loaded  beam,  20  ft.  1'  ng,  is  supported  <  11  two  pn  ps  at 
the  same  level.  The  overhang  at  each  end  is  the  s:inic.  Find  the  position 
it  the  props  sii  that  the  ends  (  f  the  beam  may  lic  h^  rizi  ntal. 

A  UK.  Overhang  =())}  ft. 

38.  Determine  the  i.s-seeles  section  i^f  maximum  strength  which  can  be 
(lit  out  iif  a  circular  section  of  given  diameter,  and  c<  mpare  the  strengths  of 
tiie  two  sections.  Ana.  W  =  Sxdiam.;  175N^to  1782. 

39.  Show  that  the  moments  of  resistance  of  a  circular  section  and  of  the 
strongest  rectangular  section  that  can  be  cut  out  of  the  same  are  in  the  ratio 
of  9n\  3  to  112,  and  that  the  areas  of  the  sections  are  in  the  ratio  of  33  to 
14  V  2. 

40.  t'how  that  the  moments  of  resistance  of  an  isosceles  triangular  section 
and  of  the  strongest  rectangular  section  that  can  !«  cut  out  of  the  same  ar& 
iti  the  ratio  of  27  to  16,  and  th.it  the  areas  of  the  two  sections  are  in  the  ratio 
of  9  to  4. 

41.  Determine  the  dimensions  of  the  strongest  section  in  the  form  of  (a) 
a  rectangle  with  vertical  sides,  (b)  an  i.sosceles  triangle  with  horizontal  base, 
ili.if  Clin  l)e  cut  out  of  an  elliptical  section  having  a  vertical  major  axis  of 
ii'iigth  '2p  and  a  minor  a.\is  of  length  '2q. 

Ans.  (a)  <lcplh  if  rect.-pv'j;  width  1  f  rect.  "^nT 

(/))  depth  <  f  triangle -Ip;  base  of  triangle  ^^N''20. 

42.  Determine  the  isosceles  section  of  maximum  strength  wiiich  can  lx>  rut 
mil  of  a  circular  section  of  given  diameter,  and  compare  the  strengths  of  the 
two  sections.  .1«,i.  //  =  ;X(liam.;  M'yvfy.  \JH2. 

43.  A  round  and  a  H<iuare  Imam  of  ecpial  length  and  e(]ually  loaded  artr 
to  Im!  of  equal  strength.  Find  the  ratio  of  the  diameter  to  the  side  of  tiki 
"■'I""'"''-  AtiH.  i,'M.i  3.3. 

44-  Compare  the  relative  strengths  of  two  rectangular  Imams  of  equal 
)•  ngth,  the  breadth  (fc)  ^nd  depth  (<f}  of  one  being  the  depth  (/<)  and  tm-adth 
fiO  of  the  other.  Ann.  d\  h. 

45-  Compare  the  uniformly  diatribute<i  loadg  which  ran  be  borne  by  two 


550 

Iwams 
II  lillll'; 
of  rosi 
46. 
tion  a  I 

lllC   Il't 

47. 
inatcri; 
4;  (^) 
bciiisr  I 

48. 
inertia 


THEORY  OF  STRUCTURE.^. 

of  ic'c-tannular  .section,  the  several  linear  dimensions  of  the  one  being 
<  the  correspoiKlinf;  (liinensions  of  the  other.  Also  compare  the  moments 
stance  of  corr('s])on(linf:  sections.  -'"•^■-  "'!  «  • 

('onii)aro  the  moments  of  resistance  to  IxMidii.g  of  a  rectangular  sec- 
1(1  of  the  rluimlioidal  and  isosceles  sections  which  can  be  inscrilH>d  in 
tanple.  the  base  of  the  triangle  beinf;  the  lower  eilsi"  of  the  rectangle. 

All.''.  4:1  ;1  or  4:1  :2. 
Compare  the  relative  strengths  of  two  beams  of  the  .same  length  and 
il  («()  when  the  .sections  are  siiiiiliir  and  h:.ve  areas  in  the  ratio  of  1  to 
when  one  section  is  a  circle  and  the  other  a  scpiare,  a  side  of  the  latter 
■ijnal  to  the  «liameter  of  the  former.  .1»/.-'.  ('/)  1  to  S:  (b)  of)  to  33. 
At  every  p'int  (f  a  !)eam  the  1$.M.  is  prop'rti  nal  to  the  moment  of 
Sii  'W  that  the  curv(>  of  the  neutral  a   is  is  a  jiarabola. 


49.  There  are  two  beams  of  the  same  >cdioiial  area,  the  one  .1  iM'ing  cir- 
cular and  of  ladiu-  /■,  I  he  oilier  II  being  square.  .\  hollow  square  of  side  r 
is  cut  through  ilic  middle  of  .1.  \Miat  must  be  the  radius  of  a  hollow  :dui:d 
throuch  the  middle  ot  />'.  so  tli:it  liolh  sections  may  have  the  same  moment 
of  resistance?     The  siiles  of  the  .s<|Uares  are  vertical.  Ans.  rX.71. 

50.  A  triangular  knihM'dge  of  a  weighing-machine  overhangs  IJ  inches 
and  supports  a  lo:id  of  '-'  tons  (.assume  evenly  distributed).  Taking  the  trian- 
gle to  1h'  e(iuihiteial,  liiid  the  re(iuisite  size  for  a  te.isile  stress  at  the  apex  of 
10  tons  per  s.piare  inch.  Ans.  1.687  ins. 

51.  .\  circular  link  of  a  chain  having  an  internal  diameter  of  3  ins.  is 
made  of  iron  having  a  diameter  of  1  in.  and  one  side  is  cut  through.  Deter- 
mine the  maxiniuni  anil  mininium  stresses  in  the  link  when  the  chain  is 
loaded  with  Um  lbs.  .t«v.  •.>l,f.:}()  Ibs./sq.  in.  and  10,0'.>1  Ib.s./sq.  in. 

5J.  A  bar  of  Ian  li  ('.  fi.  long  ■  _'  ins.  s<iuare,  resting  on  sui)|)orts  at  the 
two  eiiils,  f:iils  iDidcr  a  loai!  of  ol.'i  lbs.  at  the  centre.  Find  the  breathh  of 
each  of  two  cantilevers,  of  »  ft.  length  by  10  ins.  depth,  made  of  this  ma- 
terial, which  are  to  carry  a  cistern  of  water  weigliing  2  tons,  .")  being  a  factor 
of  safety,  and  the  load  uniformly  distributed.  Am.  2.07  ins. 

53.  A  rectangular  Ix-am  of  pine  8.'.)  ins.  deep  by  !i  ins.  wide,  of  uniform 
Beetioti  throughout,  is  supported  horizontally  on  two  walls  15  ft.  ajiart. 
What  weight  will  the  beam  safely  carry  at  .">  ft.  from  one  of  the  walls,  the 
breaking  load  biing  four  times  the  safe  load?  How  much  must  the  depth 
be  increased,  the  iireadll;  remaining  the  same,  if  the  load  is  shifted  to  the 
middle  of  the  Uam?  (Assume  that  a  pine  bar  1.")  ins.  long  l)y  1  in.  by  1  in., 
aupiMirted  at  both  ends,  fails  under  a  h)ad  of  3iin  lbs.  at  the  centre.) 

.\iis.  3:UL'  lbs.;  new  depth  =9.44  ins. 

54.  What  must  In-  the  ihi.kiiess  of  .'i  mil.l-steel  tulx^  of  10  ins.  e\tern;il 
diameter  to  cirr^  ;  lo:id  of  KI.OtKI  lbs.,  placed  centrally  iM'tweeil  two  su|)- 
ports  10  ft.  .•i|i!!'i.  the  -ale  working  unit  .-tre.ss  Ix'ing  10,000  Ibs./sq.  in.? 

Auk.  .3N  ill. 

55.  A  bloc!^  "1  lie  3  ins.  wide  and  4  ms.  <leep  has  Its  en<ls  restuig  uimn 
8uptH)rts  30  iii>.  iiiKiri  i.nd  carries  a  uniformly  distributed  lo.ad  of  4S00  Hn. 
An  increase  o.  pressure  to  the  extent  of  llL'.'i  Ibs./sq.  in.  h)wer8  the  freezing- 


EXAMPLES. 


551 


l«iint  1°  F.  Assuming  that  the  ordinary  theory  of  flexure  holds  good,  find 
the  tomiMTature  of  the  ice.  Ans.  30°  F. 

56.  A  ijcam  supported  at  the  ends  ran  just  bear  its  own  weight  W  together 

with  a  single  weight  7-  at  the  centre.     What  load  may  be  i)lared  at  the  cen- 

tri'  (if  a  l)oam  whose  transverse  section  is  similar  but  m'  as  great,  its  length 
111  iiig  /(  times  as  great?  If  the  beam  could  support  only  its  own  weighi ,  what 
wciiild  Ik'  the  relation  l)etween  m  and  n? 


57.  .\  wooden  l)eam  112  ins.  deep,  fi  ins.  wide,  and  12  ft.  long  is  emliedded 
ill  a  wall  at  one  end.  What  weight  will  the  beam  carry  at  the  outer  end  if 
the  breaking  weight  of  a  fx>am  1  ft.  long,  1  in.  broad,  and  1  in.  deep,  .sup- 
I'Mitcd  at  the  ends  and  loaded  at  the  centre,  is  .")(K)  ll)s?  Ans.  9000  lbs. 

58.  .\  sjjruce-tiee  (M)  ft.  high  and  14  ins.  in  diameter  at  the  ground  level 
i-  subjected  to  a  horizontal  wind  pressure  which  may  Ije  a.ssumed  to  l)e  uni- 
fc.niily  distributed  over  the  upix-r  .30  ft.  of  the  tree.  Find  the  intensity  of 
this  pressure  if  the  maximum  fibre  stress  at  the  ba.sc  is  lOSO  Ibs./.sq.  in. 

Atis.  ISlbs./lin.  ft. 

59.  Find  the  stress  in  tons/.sq.  in.  at  the  skin  and  also  at  a  point  2  ins, 
■■(iin  the  neutral  axis  in  a  piece  of  10"XS"  oak,  (a)  with  the  10-in.  side  ver- 
liid,  (/()  with  the  S-in.  side  vertical.  The  oak  rests  upon  supports  .3  ft. 
:i|i:irt  and  carries  a  load  of  4900  lbs.  at  its  middle  point.  Also  compare  (c) 
till'  strength  of  the  lx?am  with  its  strength  when  a  diagonal  is  horizontal. 

Ans.  (a)  330 J,  132 A;  (h)  413^,20633;  (r)  4:  vil  or  .J:  v/4l. 

60.  In  the  ease  of  a  tram-rail,  the  area  of  the  modulus  figure  is  8.24  sq. 
ill-;..  ;md  the  distance  between  the  two  centres  of  gravity  is  S.S.')  ins.;  the 
iH'iir:d  axis  is  situated  at  a  distaiu'e  of  3.1  ins.  from  the  skin  of  the  bottom 
Ihiii-c.     Find  the  /  and  Z.  An.<i.  70.9;  22.87. 

61.  Show  that  the  rmxhilu.t  of  rupture  of  any  material  is  18  times  the  load 
hIihIi  will  break  a  beam  12  ins.  long,  1  in.  deep,  and  1  in.  wide  when  applied 
111  the  centre. 

6j.  Find  the  limiting  length  of  a  wTought-iron  cylindrical  beam  4  ins. 

iianietcr,  the  modulus  of  rupture  lieing  42,0(X)  lbs.     What  uniformly  dis- 

lutcd  load  will  break  a  cylindrical  lx>am  of  the  same  material  20  ft.  long 

I   I  ins  in  diameter?  Ans.  04.8  ft.;  SSOO  lbs. 

63.  .\  red-piiK  iK'am  IS  ft.  long  has  to  support  a  weight  of  10,000  lbs.  at 
iciitre.  The  section  is  rectangular  antl  the  depth  is  twice  the  breadth. 
I  ni'l  the  transverse  dimensions,  the  modulus  of  rupture  Ix^ing  S.")00  lbs.  and 
lit  liciiig  a  factor  of  safety.     (Neglect  the  weight  of  the  beam.) 

Ans.  /.=0.S4in8.:  rf  =  10.f>8ins. 

'>4-  A  round  oak  cantilever  10  ft.  long  is  just  broken  by  a  load  of  fiOO  lbs. 
>ii-|K'ii(|cd  from  the  free  end.  Find  its  diiinieter,  the  modulus  of  rupture  l)eing 
liiiiOi)  lbs.     (Neglei't  the  weight  of  the  iM-am.)  Ans.  4. IS.")  ins. 

'15.  Determine  the  breaking  weight  at  the  centre  of  a  cast-iron  Ix-am  of  6  ft. 
-1  III  and  4  iiis.  square,  the  ctjc'fficielil  111  rupture  Ix'ing  3i).JKH)  lbs. 

Ans.  2fi,(>6<'.f  lbs. 

06.  The  flooring  of  a  corn  warehouse  is  supported  U|)on  yellow-pine  joiata 


552 


TllEOnr  OF  STIiUCTih'ES. 


20  ft.  in  the  rlear,  8  ins.  wide,  10  ins.  deep,  and  spared  3  ft.  centre  to  centre. 
Find  the  hpip;ht  to  which  corn  weiphing  48J  Ib.s./cu.  ft.  may  be  heaped  upon 
the  floor,  10  being  a  factor  of  safctj'  and  18,000  lbs.  the  modulu."  of  rupture. 

Ans.  2.75  ft. 

67.  \  yollow-pino  beam  14  ins.  wide,  15  ins.  deep,  and  resting  upon  sup- 
ports 120  ins.  a])art  broke  down  under  a  uniformly  dis*ril'.tcd  load  of  00.97 
tons.     F"ind  the  coefficient  of  rupture.  Ans.  2731.456. 

68.  Find  the  brc.-iking  weight  at  the  centre  of  a  Canadian  ash  beam  2i 
ins.  wide,  3J  ins.  deep,  and  of  45  ins.  span,  the  coefficient  of  rupture  being  7250. 

Ans.  4934,',  lbs. 

69.  A  timber  beam  6  ins.  deep,  3  ins.  wide,  90  ins.  between  supports,  and 
weighing  .")()  Ibs./cu.  ft.  broke  down  under  a  weight  of  10.00)  lbs.  at  the  centre. 
Find  the  modulus  of  rupture.  Ans.  8911J. 

70.  A  wrought-iron  bar  2  ins.  wide,  4  ins.  deep,  and  144  ins.  between  supports 
carries  a  uniformly  distributed  load  IF  in  addition  to  its  own  weight.  Find 
W,  4  being  a  factor  of  safety  and  50,000  lbs.  the  modulus  of  rupture. 

Ans.  52355  lbs. 

71.  Find  the  length  of  a  beam  of  Canadian  ash  0  ins.  square  which  would 
break  under  its  own  weight  when  supported  at  the  ends.  The  modulus  of 
rupture  =7000  lbs.,  and  the  weight  of  the  timlK>r  =30  Ibs./cu.  ft. 

.4ns.  274.95 

72.  The  teeth  of  a  cast-iron  wheel  are  3i  ins.  long,  2}  ins.  deep,  and  7  ins. 
wide.  What  is  the  breaking  weight  of  a  tooth,  the  modulus  of  rupture  being 
5(KK)  lbs?  Ans.  50,625  lbs. 

73.  A  WTOUght-iron  bar  4  ins.  deep,  J  in.  wide,  and  rigidly  fixed  at  one  end 
gave  way  at  32  ins.  from  the  load  when  loaded  with  1508  lbs.  at  the  free  end. 
Find  the  modulus  of  rupture.  Ans.  4181J. 

74.  .\  cast -iron  l)eam  is  12  in-<.  wide,  rests  upon  supports  18  ft.  apart,  and 
carries  a  12-in.  brick  wall  which  is  121  ft-  ''>  height  and  weighs  112  Ibs./cu.  ft. 
Taking  fi3,0tX)  as  the  modulus  of  rupture  for  a  uniformly  distributed  load  and 
5  as  a  factor  of  safety,  find  tlie  depth  of  the  beam,  (a)  neglecting  its  weight, 
(6)  taking  its  weight  into  ac<'ount. 

.Also  (r)  determine  the  depth  of  a  cedar  Ijeam  which  might  be  substituted 
for  the  cast-iron  Ix-am,  taking  11,200  lbs.  as  the  modulus  of  rupture  for  the 
cedar.  .4«.s-.  (a)  0  ins.;  (b)  OJ  ins.;  (r)  14.23  ins. 

75.  \  cast-iron  girder  27i  ins.  deep  rests  u(>on  supports  20  ft.  apart.  Its 
bottom  flange  h.as  an  area  of  48  sq.  ins.  and  is  3  ins.  thick.  Find  the  breaking 
weight  at  the  centre,  the  ultimate  tensile  strength  of  the  iron  being  l."),000 
Ibs./sq.  in.     (Neglect  the  effect  of  the  web.)  Ans.  253,84()i',  il>s. 

76.  Is  it  safe  for  a  man  weighing  KU)  lbs.  to  stand  at  the  eentre  of  a  spruce 
(jiank  10  ft.  long,  2  ins.  wide,  and  2  ins.  thick,  supjiorted  by  vertical  ropes  at 
the  ends?    The  .safe  working  strength  of  the  timl)er  is  1200  Ibs./sq.  in. 

Ans.  No;  tlie  maximum  safe  weight  at  the  centre  is  53i  lbs. 

77.  A  timber  beam  weighing  42  Ibs./cu.  ft.  rests  on  piers  10  ft.  apart;  it 
i«  R  itis  thick.  What  depth  mti^t  it  l)c  made  if  the  deflection  in  the  centre 
under  its  own  weight  and  a  load  of  1  ewt./ft.  run  is  not  to  exceed  i  of  an  inchf 
(Assume  £-700  tons/sq.  in.)  Ana.  9.1  ins. 


EXAMPLES. 


553 


78.  A  piece  of  timber  10  ft.  long,  12  ins.  deep,  8  ins.  wide,  and  having  a  work- 
iiip  strength  of  1000  Ibs./sq.  in.  carries  a  load,  including  its  own  weight,  of 
w  Ibs./Iin.  ft.    Find  the  value  of  w,  (a)  when  the  timber  acts  as  a  cantilever 
(h)  when  it  acts  as  a  beam  supported  at  the  ends.     Find  (c)  stress  in  material 
3  ins.  from  neutral  axis  tt  fixed  end  of  cantilever  and  at  middle  of  beam. 

.4ns.  (a)  320  lbs.;    (6)   1280  lbs.;    (c)  .500  Ibs./.sq.  in. 

79.  A  cast-iron  beam  of  square  cross-section  1  in.  deep XI  in.  wide  is 
tested  on  a  span  of  30  ins.  The  breaking  load  in  the  centre  is  7i  cwt.  Cal- 
culate (d)  the  maximum  intensity  of  tensile  stress,  assuming  the  beam  formula 
to  hold  up  to  the  breaking-point,  {h)  the  probable  deflection  in  the  centre 
Vou  may  assume  £■=  12,000,000  Ibs./sq.  in. 

AriR.  (a)  45,360  Ibs./sq.  in. ;    (6)  .810.5  in. 

80.  A  yellow-pine  beam  14  ins.  wide,  1.")  ins.  deep,  and  resting  upon  suf)port3 
129  ins.  apart  was  just  able  to  bear  a  weifiht  of  34  tons  at  the  centre.  AVhi;t 
weight  will  a  beam  of  the  same  material  of  45  ins.  span  and  5  ins.  square  bear? 

Ans.  4.133  tons. 

81.  A  horizontal  circular  tube  of  steel  is  7  ft.  in  diameter,  \  in.  thick,  100 
ft.  long  sup[K)rted  at  the  ends,  its  total  load  distributed  uniformly  all  over 
iH'ing  30  tons;  what  are  the  greatest  stresses  in  the  metal?  The  tube  is  filled 
with  compressed  air;  what  must  its  pressure  be  if  there  is  just  no  compressive 
stress  in  the  metal?  State  what  is  now  the  nature  of  the  stress  inthcmet;il 
at  the  place  where  it  is  gn^atcst.  Ans.  6i  tons/sq.  in.;  38.6  Ibs./sq.  in. 

82.  Ah  oak  beam  of  circular  section  and  22  ft.  long  is  strained  to  the  elastic 
limit  (2  t<>n.s/sq.  in.)  by  a  uniformly  distributed  load  of  2,4  tons.  Find  the 
diameter  of  the  beam.  What  load  2  ft.  from  one  end  would  strain  the  material 
to  the  same  limit?  Ans.  7  ins.;  3.088  tons. 

83.  \  uniform  beam  of  weight  W,  crossing  a  given  span  can  bear  a  uniformly 
distributed  load  W,.  \Vh.at  load  may  be  placed  upon  the  same  Ijeam  if  it 
(Tosses  the  span  in  n  equal  lengths  supported  at  the  joints  by  piers  whose 
wiihhs  may  be  disregarded?  Ans.  11' {W,+W ,)—][',. 

84.  A  cast-iron  water-main  .30  ins.  inside  diameter  and  32  ins.  outside  is 
uiisui)iK)rted  for  a  length  of  12  ft.     Find  the  stress  in  the  metal  due  to  bending. 

Ans.  95.8  Ibs./sq.  in. 

85.  A  wrought-iron  bar  1  i  ins.  wide  and  20  ft.  long  is  fixed  at  one  end  and 
carries  a  loud  of  .")00  lbs.  at  the  free  end.  Find  the  depth  of  the  bar,  so  that 
the  stress  may  nowijcre  exceed  10,000  Ibs./sq.  in. 

.Ins.  0.928  ins.;  if  weight  of  bar  is  included,  the  depth  (/  is  given  by 
</'-4.8<i-48=0,  and  (/=9.73  ins. 

86.  Determine  the  diameter  of  a  solid  round  wrought-iron  beam  resting 
upon  supjwrts  60  ins.  apart  and  aljout  to  give  way  under  a  load  of  30  tons  at 
1  )  ins.  from  one  end.  Take  .'i  as  a  factor  of  sjifety  and  8960  Ibs./sq.  in.  as  the 
s.ife  working  intensity  of  stress. 

Alls.  5.47  ins.;  if  weight  of  hcam  is  taken  into  account,  </  is  given  ]y 
2019.W4  +  196;d'  -  12.320fi»  -0. 

87.  A  yellow-pine  beam  14  ins.  wide  and  15  ins.  deep  was  i.laeed  U|K)n  supporN 
10  ft.  9  ins.  apart  and  deflected  I  in.  under  a  load  of  20  tons  at  the  centre. 
Find  E,  neglecting  the  weight  of  the  beam.    What  were  the  intcnsitieh  of 


I 


I   I 


I 


554 


rilEORY  OF  STRUCTURES. 


the  normal  and  tangential  stresses  at  2  ft.  from  a  support  and  2i  ins.  from 
neutral  plane  upon  a  plane  inclined  at  30°  to  the  axis  of  the  beam? 

.4ns.  i;.'7L',ll->  lbs.;  132.83  and  218.91  lbs. 

88.  A  Ix-am  of  uniform  section  of  depth  (/,  with  equal  flanges  and  of  span  /, 

is  built  into  walls  so  that  its  ends  are  horizontal  and  at  the  same  level.     One 

of  the  walls  settles  a  distance  »  without  disturbing  the  horizontality  of  the 

ends  of  the  Warn.     Show  that,  due  to  settling,  the  maximum  stress  indueed  is 

— :;— ,  h  Ix'ing  \oung  s  moduhis. 

8g.  .V  stress  of  1  lb. /s(i.  in.  itroduces  a  strain  of  ^nnltxii  '"  a  beam  12  ins. 
square  and  20  ft.  l)elween  sMjijions.  Find  the  radius  of  curvature  and  the 
central  dellection  under  a  load  of  2(K)0  lbs.  at  the  middle  jxiint. 

Alls.  2400  ft.;  }  in. 

go.  .\  bar  of  wrought  iron  3  ins.  broad  and  1}  ins.  thi<k  is  supported  in  a 
horizoiit.Ml  jiosition  at  two  points  2}  ft.  apart  What  dedeetion  at  the  middle 
will  be  caused  by  [tlacing  therea  load  of  l.")cwts.?   £=3x10'.    Aiis.  Mm  in. 

91.  .\  2"X4"  beam  20  ft.  long  has  both  ends  fixed  and  is  loaded  at  the 
centre  with  a  weight  of  4(K)  lbs.  Find  tlie  deflection  at  the  centre.  How 
much  work  is  done  in  bending  the  beam?     (Take  E  =2S,(K)0,000  Ibs./sq.  in.) 

Alts,  il  in.;  77t  in.-lbs. 

92.  .\  steel  girder  20  ft.  long  between  supjjorts  is  loaded  with  .3000  lbs. 
per  foot  run.  ("alculate  what  its  moment  of  inertia  must  be,  so  that  the 
deflection  at  the  centre  is  ,;'„„  of  the  span,  the  value  of /?  in  pounds  and  inches 
being  taken  as  30,(KM),000.  .\lso  calculate  the  slo]X!  at  the  ends  of  such  a 
Ix.ani.  -4ws.  1800;  9.10  minutes. 

93.  .\  beam  S  :is.  wide  and  weighing  .")0  lbs.  /cu.  ft.  rests  upon  supjiorts 
30  ft.  apart.  Find  its  depth  so  that  it  may  deflect  J  in.  under  its  own  weight. 
(E  =  1 ,2(K),(KK)  lbs.)  .4«.s-.  9.18.J  ins. 

94.  .\  rectangular  girder  of  given  length  I  and  breadth  h  rests  upon 
two  supi)orts  and  carries  a  weight  P  at  the  centre.  Find  its  depth  so  that 
the  elongation  of  the  lowest  fibres  may  be  rluir  of  the  original  length. 


Aii.t. 


^ 


lOOW 

hE    ■ 


95.  A  l)eam  AB,  60  ft.  long,  is  fixed  horizontally  at  .4  and  hinged  at  a 
point  C  40  ft.  from  .4,  and  the  other  end  li  is  supported  on  a  pier  at  the  same 
level  as  .1.  The  beam  carries  a  uniformly  ilistributed  load  of  2000  Ibs./ft.  run 
over  the  whole  of  i*s  length.  Determine  the  shearing  force  and  bending  moment 
at  every  point.  .\lso  find  the  deflection  and  slo])c  at  C. 
An.t.  Reaction  at  .1  =  ">0,  at  B  =  10  tons;  B.M.  at  .4  =  1200  ft.-tons. 

,  ,       ,^     2088000    .      ,,,.      ;S02000     .  ^    ,.         ,  ,,    92lfi0000 
Slojns  at  C  for  AC  =  — ^,—  ;  for  BL  -      ^,,—  ;  deflection  of  t  = 


EI 


EI 


EI 


96.  Deduce  expressions  for  the  slofw  and  deflection  at  the  end  of  a  canti- 
lever i>f  reel  angular  .section  having  a  length  of  tU)  ins.,  a  breadth  of  2  im.,  and 
a  depth  of  3  ins.,  when  loaded  uniformly  with  100  lbs.  /ft.  run.    (/!,'  =30,(KM),0<K).) 

An."!,  isa;  I'o  in. 


» 


EXAMPLES. 


555 


97.  An  angle-iron  3"x3"xA"  was  placed  upon  supports  12  ft.  9  ins. 
apart  and  doflected  IJ  ins.  under  a  load  of  8  cwts.  uniformly  distributed  and 
•J  cwts.  at  the  centre.     Find  E  and  the  position  of  the  neutral  axis. 

Auk.  £:  =  lf),070,Gn  lbs.;  neutral  axis  ;?!;  in.  from  up[«r  face. 

98.  A  yellow-pine  beam  14  ins.  wide,  1.")  ins.  deep,  and  weighing  32  Ibs./cu.  ft. 
was  placed  ufwn  supports  10  ft.  6  ins.  apart.  Under  uniformly  distributed 
loads  of  59,734  lbs.  and  of  127,606  lbs.  the  central  deflect  ions  "were  respec- 
tively .18  in.  and  .29  in.  fijid  the  mean  value  of  E.  Also  determine  the 
additional  weight  at  the  centre  which  will  increase  the  first  deflection  by 
,V,  of  an  inch.  Arts.  2,5G0,9(J0  lbs.;  24,197  lbs. 

99.  For  the  load  of  59,734  lbs.,  find  the  maximum  intensities  of  thrust, 
tension,  and  shear  at  a  point  half-way  lietween  the  neutral  axis  and  the  out- 
side skin  in  a  tran.sverse  section  at  one  of  the  points  of  trisection  of  the  beam. 
.Mso  find  the  ineUnations  of  the  planes  of  principal  -stress  at  the  point. 

Ans.  800.01,  3.55,  401.78  lbs.;  0=3°  48J'. 

100.  If  the  shape  of  the  section  and  the  load  are  given,  and  the  linear  dimen- 
sions of  section  are  chosen  so  as  to  make  the  greatest  stress  a  given  quantity, 
slio'w  that  the  deflection  at  centre  is  proportional  to  (length)!. 

loi.  ,\  steel  rectangular  girder  2  ins.  wide,  4  ins.  deep  is  placed  upon  sup- 
ports 20  ft.  apart.  If  E  is  35,000,000  lbs.,  find  the  weight  which  if  placed 
at  the  centre  will  cause  the  beam  to  deflect  1  in.  Aiis.  1296A  lbs. 

102.  A  cylindrical  beam  of  2  ins.  diameter,  60  ins.  in  length,  and  weighing 
}  Ib./cu.  in.  deflects  ^^  in.  under  a  weight  of  3000  lbs.  at  the  centre.     Find  E. 

Ann.  77=21,688,210  lbs. 

103.  A  3"X3"xJ"  angle-iron,  with  both  ends  fixed  and  a  dear  span  of 
■JO  ft.,  carries  a  uniformly  distributed  load  of  .')00  lbs.,  which  causes  it  to  deflect 
.:?009  in.  Find  E.  What  single  load  at  the  centre  will  produi-e  the  same 
di'tlection?     Find  the  work  done  due  to  bending  in  each  case. 

Ans.  18,116,000  lbs.;  2.Wlbs.;  52.12  in.-tons;  48.86  in.-tons. 

104.  Show  that  the  work  done  in  deflecting,  within  the  elastic  limits,  a 

uniform  rectangular  bar,  sui)[K)rtcd  at  the  ends  and  loaded  in  the  middle, 

1  /' 
IS  equal  to  the  volume  of  the  bar  multiplied  by  —  ~-,  /  being  the  maximum 

Is  E 

stress  in  the  bar. 

105.  A  girder  fixed  at  both  ends  carries  (2n  +  l)  weights  W  concentrated 
at  |)()ints  dividing  the  length  of  the  girder  into  2n+2  equal  divisions.  Find 
1  !i(^  total  central  deflection.  w  + 1  ]yp 

-'"•^■-  mm- 

106.  The  deflection  of  a  uniformly  loaded  horizontal  beam  supported  at 
tiic  ends  is  not  to  excetxl  1  in.  in  50  ft.  of  span,  and  the  stress  in  the  material 
is  not  to  exceed  400  Ibs./sq.  in.  Find  the  ratio  of  span  to  depth,  E  Jx-ing 
1  ,.'00,000  Ibs./sq.  in.,  and  the  neutral  axis  being  at  half  the  depth  of  the  beam. 

Ann.  24. 

107.  A  round  wrougiit-iron  bar  I  ft.  long  and  d  ins.  in  diameter  can  just 
r;Irry  its  own  weight.  Find  /  in  terms  of  d,  (a)  the  allowable  deflection  being 
1  in./per  100  ft.  of  span,  E  being  30,000,000  lbs.;    (6)  the  allowable  stre.ss 


55C 


THEORY  OF  STRUCTURES. 


Ans. 


being  SOfiO  Ibs./sq.  in.;    (r)  the  stiffness  given  by  (a)  and  the  strength  given 
by  (6)  being  of  equal  importance. 

Ans.   (a)  l=<^'-2rm'\  (b)  l  =  \  -"iTk/;  (c)/-?SM. 
io8.  A  square  steel  bar  /  ft.  ioiif;  and  having  a  side  of  length  d  ins.  can  just 
carry  its  own  weight;  its  stiffness  is  ts'ijo  and  the  maximum  allowable  work- 
ing stress  is  7  tons/sq.  in.     Find  I  in  terms  of  </,  E  being  l.'{,000  tons. 

Z(inft.)  ^13 
rf(in  in.)~21' 

log.  A  beam  is  supported  at  the  ends  and  bends  under  its  own  weiglit. 
Show  that  the  upward  force  at  the  centre  which  will  exactly  ne  .trail ze  the 
bending  action  is  equal  to  |  or  i  of  the  weight  of  the  beam  (»')>  according 
as  the  ends  are  free  or  fixed.     Find  the  neutralizing  forces  at  the  quarter  spans. 
Ans.  Ends  jrec:  iVi"'  at  each  or  iVjit'  at  one  of  the  points  of  division. 
Ends  fixed:   /„«'  at  each  or  Jii'  at  one  of  the  points  of  division. 
no.  A  horizontal  girder  .1/i  of  length  /,  and  fixed  a    A  and  at  B,  carries 
two  weights  P  and  Q,  concentrated  at  jwints  C  and  I)  respectively.     If  AC  =X 
and  BD  =  Y,  show  that  the  bending  moment  at  C  is  greater  or  less  than  the 
bending  moment  at  D  according  as  PX\3l-'2X)  ^QY'(3l-2y). 


tlHJ! 


111.  Show  that  the  work  done  in  bending  a  horizontal  beam  is  the  same 
whether  it  has  two  ends  fixed,  or  one  end  fixed  and  one  resting  ui)on  its  sup- 
port, or  two  ends  resting  upon  supports,  if  the  load  intensities  in  the  sevt  1 
cases  are  in  the  ratio  of  2v  (i  to  3  to  '2. 

112.  X  horizontal  girder  AC  is  fixed  at  .1,  rests  upon  the  support  at  C, 
carries  a  uniformly  distributed  load  of  intensity  ir,  and  is  hinged  at  B,  divid- 
ing the  girder  into  segments  AB=a  and  BC  =b.  Find  (1)  the  reactions  at  .1 
and  C,  (2)  the  moment  of  fixture,  (3)  the  deflection  at  B. 

.Ins.  (I)«.(a4);    uf;    (2)f(a+/,);    (3)|j';j4). 

113.  A  horizontal  beam  with  both  ends  absolutely  fixed  is  loaded  with  a 
weight  W  at  a  point  dividing  the  sjjan  into  two  segments  a  and  6.     Show 

that  the  deflection  at  the  point  is  ^jfA   ^rz)  1  ^"tl  fi"d  the  work  done  in 


bending  the  beam. 


W'/  ab  \» 

Ans.  ;7F,,( 1 1  . 

iiEl\a+b/ 


114.  A  girder  with  both  ends  fixed  carries  two  equal  loads  W  at  points 
dividing  the  girder  into  segments  a,  b,  c.  Determine  the  reactions  and  bend- 
ing moments  at  the  supports. 

Ans.  R,  =  W --,  .  .    ,. ; 

'  (a +0+0'  ' 

2a'  +  CwV)  +  3ab'  +  h'  +Cm'c+  6abc  +  3l'c 


R^  =  W 

ia  +  b+c)* 

.     _  „,2a'r  +  2abc  +  bc^  +  ab'. 


(a  -rh  -re 


ii/,=r 


2ac'  +  2abc  +  a'b+b*c 


li 


EXAMPLES. 


557 


115.  A  bridge  n  ft.  in  the  rlear  is  formed  of  two  cantilevers  which  meet 
in  the  centre  of  the  span  end  are  connected  by  a  bolt  capable  of  transmitting 
:i  verti(;al  pressure  from  the  one  to  the  other.  A  weight  W  is  placed  at  a 
liistance  6  from  one  of  the  abutments.  Find  the  pressure  transmitted  from 
one  cantilever  to  the  other,  and  draw  the  curve  of  bending  moments  for  the 
loaded  cantilever.  .        ^  /         b'       b'\  /   h'       hK 

^ns.«.=ir(l-3-,+2^);     /e.  =  r  (.3I, - 2^.)  . 

116.  Two  weights  P  and  Q  (<P)  are  carried  by  a  horizontal  girder  of 
length  /  resting  upon  supports  at  the  ends,  the  distance  between  the  weights 
Ix'iiig  a.  Place  the  weights  so  as  to  throw  a  maximum  bending  moment  on 
the  girder  and  find  the  value  of  this  moment.  Also  find  the  corresponding 
work  of  flexure.  ,        ^  Pl  +  0(l-ri\ 

Ans.  Distance  of  P  from  support  =~^^  ^^    . 


Max.  B.M. . 


\Pl+Q(l-a)\' 
MP  +  Q)l     • 


117.  The  section  of  a  cantilever  of  length  /  is  an  ellipse,  the  major  axis 
(rcrliral)  being  twice  the  minor  axis.  Find  the  deHection  at  the  end  under 
:i  smgle  weight  W,  /  being  the  coefficient  of  working  strength  and  E  the 
coefficient  of  elasticity.  ,  297    fp  ,  j 

V3.500i?'U7 
I  j8.  A  bar  of  wrought  ii  on  4  ft.  long,  4  ins.deep,  and  1 J  ins.  thick  is  fixed  atone 
•■n.l  and  loaded  with  2000  lbs.  at  the  free  end.     Find  the  maximum  slope  and 
t  he  maximum  deflection,  E  being  28,000,000  lbs.  Am.  .010423 ;  .334  in. 

119.  A  horizontal  girder  AB,  of  length  /,  and  fixed  at  A  and  B,  carries  a 
load  P  at  a  point  C.    If  AC  -a,  find  the  maximum  deflection, 

Ans    -^(^-">'°' 
3£7(3i-2o)'- 

ijo.  A  horizontal  girder  AB,  of  length  I,  is  fixed  at  A,  and  rests  upon  its 
.-support  at  «.  It  carries  p  weight  P  at  a  jwint  C,  and  AC=a.  Find  the 
position  of  the  most  deflected  point,  and  show  that  the  bending  moment  at 
<'  IS  greatest  when  a=/X.634. 

.Ins.  Distance  of  most  deflected  point  from  point  of  fixture -zf-^^^V 

VSl-aJ  ■ 

"I.  A  piece  of  greenheart  142  ins.  between  supports,  9  ins.  deep,  and  5  ins. 
\Mdc  was  tested  by  being  loaded  at  two  points,  distant  23  ins.  from  ihe  centre, 
"iili  equal  weights.  Under  weights  at  each  point  of  4480  lbs.,  11,200  lbs  ' 
•||"1  17,920  lbs.  the  central  deflections  were  .13  in.,  .37  in.,  .67  in.,  respectively 
I  Hid  the  mean  coefficient  of  elasticity.  The  beam  broke  under  a  load  of 
•i:'.:{()S  lbs.  at  each  point.     Find  the  coefficient  of  bending  strength. 

-^n.  8,.-,08,0001bs./sq.  in.;  23,017  Ibs./sq.  in. 

J2J.  A  beam  of  span  I  is  uniformly  loaded.     Compare  its  strength  and 

Mifltiess  (<i)  when  merely  resting  npon  s.ipix,rts  .it  the  ends;    (i)  when  fixed 

•It  one  end  and  resting  upon  a  support  at  the  other;    (r)  when  fixed  at  both 

f  "'Is.    In  case  (c)  two  hinges  are  introduced  at  points  distant  xfrom  the  centre 


558 


THEORY  OF  STRUCTURES. 


2v  2 


show  that  the  slrcitfith  of  the  beam  is  economized  to  the  best  effect  when 
x=       ,  and  that  the  stiff iifss  is  a  maximum  when  x  —  ^  vcrv  nearly. 

Am.  3:3:2;  .5:2.08:1. 
A  stecl-pinte  beam  of  uniform  section  and  30  ft.  span  has  both  ends 
h.\ed  and  is  freely  hirificd  at  the  points  of  trispction.  Determine 
the  n<::tral  Jixis  (<;)  for  a  uniformly  distributed  load  of  (MKK)  lbs.; 
(/))  for  a  sirifxlc  load  of  lO.tXK)  lbs  concentrated,  ;i/\s^,  7i  ft.  and, 
aecimd,  I.")  ft.  from  a  support. 

124.  .\  channel  of  the  dimensions  shown  l)y  Fisr.  .")42  rests 
upon  sup|M)rts  20  ft.  apart.  If  the  ^t^(^-s  dcvclniK'd  in  the 
material  is  not  to  exceed  IKXX)  lbs. /»if|.  in.,  find  the  allowable 
unil'ornily  distributed  load  when  .1.1  is  (a)  vertical,  (li)  horizont;d.  Also 
find  (f)  the  central  dellection  iii  each  ca.se.    (£^  =28X10"). 

Ans.  (a)  2.j,400  lbs.;  (b)  6970  lbs.;  (r)  ^^  in.,  -^  in. 

^^1  loo 

125.  Find  the  moment  of  resistance  to  bending  of  a  steel  I  Ix-aiii,  eac'* 
flange  consisting  of  a  pair  of  3"X3"Xi"  angle-irons  riveted  to  a  12''Xj" 
web,  the  coefficient  of  strength  l)eing  5  tons/sq.  in.  What  load  will  the  beam 
cfirry  at  5  ft.  from  on(>  end,  its  span  being  20  ft.?  Finil  the  central  detiection, 
and  also  the  d"(lection  at  the  loaded  |)oint,  A"  being  l.^.OOO  tons. 

Att<.  287. H.")  in. -tons;  (5.4  tons,  disregarding  weight  of  Iwam,  orfi.l3  tons, 
taking  weight  of  beam  into  account;  .4  in.  at  centre  and  .3  in.  at  loaded  point. 

126.  A  plate  girder  of  (i4  it.  span  and  8  ft.  deep  carries  a  dead  load  of  2 
tons/lin.  ft.  .\t  any  wction  the  two  flanges  are  of  equal  area,  and  their  joint 
area  is  equal  to  that  of  the  web.  Find  the  sectional  area  at  the  centre  of 
girder,  so  that  the  intensity  of  stress  in  the  metal  may  not  exceetl  3  tons/.sq.  in. 
The  dellection  of  the  girder  is  I  in.  at  the  centre.  Find  E  and  the  radius  of 
curvature.  Aus.  128  sq.  in.;  l.'),360  ft.;  2.">,804,800  lbs. 

127.  Taking  the  oK-fficient  of  (linrt  elasticity  at  1.5,000  ton.s,  the  roeffi- 
ciciif  of  hilcral  elasticity  at  ()(),0(M)  tons,  and  the  limit  of  elasticity  at  10  ton.s 
deterinine  the  greatest  deviation  from  the  straight  line  of  a  uTought-iron 
girder  of  breadth  h  and  depth  </.  b* 

■^"'^-  245SW- 

128.  An  I  iH'am  7J  ins.  deep,  with  a  flange  3i"Xi"  and  web  fij"xr'. 
was  placed  on  centres  (»  ft.  apart  and  tested  by  loading  at  the  centre.  The 
following  readings  were  obtained: 


Lnail  in 
I'nunda. 

IVflcctinn  in 
Inches. 

I.i)ti<l  in 

PlIllllll.H. 

r)<'fi<'<  1  ii  ,11  iu 

InrlirN. 

I.'iikI  itt 
I'minil" 

Deflwliim  ia 
Inrhe-i. 

0 

0 

Ki.tMN) 

.110 

2S.(HH) 

0  4(10 

4,<KI0 

.028 

20.0(K) 

.l.iS 

.'iO.tMN) 

1.10 

N.IKK) 

.()."> 

24, (XK) 

.l(i7 

32,000 

2. 2.5 

12,<)(K) 

.082 

27,(KX) 

.310 

I    .: 


i)('iciiMlne  (he  mixliitus  nf  clji.iiinty  i/'.'i  iM^fon-  the  yield-fx)int  is  renched, 
and  the  maximum  stresii  on  the  beam  when  the  hiad  is  10,(XX)  lbs. 

Ans.  28, 1 7 0,000  11  IS.;  27  1 1 ".8  lbs. 


EXAMPLES. 


559 


ijg.  A  0"X9"  beam  168  ins.  centre  to  centre  of  supports  was  tested  by 
roiiccKruting  gradually  increasing  loads  and  the  following  observations  were 
made: 


Lout]  in 

Deflection  in 
Inches. 

Load  in 
Poumla. 

neflwiion  in 
Inches. 

Load  in 
Poundii. 

Deflection  in 
Indies. 

1,000 

.11 

9,000 

1  32 

19,000 

Fracture 

3,000 

.41 

12,000 

1.79 

6,000 

.86 

15,000 

2.28 

Di.srcgarding  weight  of  beam,  find  E  and  skin  stress  et  fracture.  Hence 
indicate  the  material  of  the  beam.  Am.  1.104,500  lbs.;  6,568  lbs. 

130.  A  simplex  steel  beam  with  equal  flange  areas  rests  upon  supports 
(K)  ins.  centre  to  centre,  and  is  tested  by  being  loaded  in  the  centre.  The  follow- 
ing o'jsj-rvntions  were  made: 


Ld.'kI  at  (^entre 

Central  Deflec- 

Load at  Centre 

Central  Deflec- 

Ixiad at  Centre 

Central  Deflee 

in  Pounds. 

tion  in  Inches. 

in  Pounds. 

tion  in  Inches. 

in  Pounds. 

tioninlnchea 

1,000 

Oil 

5,000 

.059 

8,000 

.094 

2,000 

.023 

6,000 

.070 

9,000 

.107 

3,000 

.035 

7,000 

.082     • 

10,000 

.119 

4,00^ 

.047 

Determine  the  coeFicient  of  transverse  elasticity  and  also  find  the  maxi- 
mum tensile  and  fibre  stresses  under  a  load  of  10,000  lbs.  (/-13.0;  depth  of 
iM-ain  -5  ms.)  Ans.  27,574,000  lbs.:  27  .'.74  Ib-i. 

131  A  gas-pipe  resting  upon  supports  45  ins.  apart  has  an  external  diam- 
eter t)l  .'/«■  inch  and  an  internal  diameter  of  srven  tenths  of  an  inch.  The  pipe 
wa.-f  loaded  at  the  centre  and  the  following  observations  made: 


Load  in 

Deflection  in 

Load  in 

Deflection  in 

Load  in 

Deflection  in 

Pounds. 

Incbea. 

Pounds. 

Inches. 

Pounds. 

Inches. 

0 

0 

40 

.062 

80 

.127 

10 

.015 

50 

.078 

90 

.142 

20 

.030 

60 

.094 

100 

.158 

30 

.046 

70 

110 

Calculate  the  modulus  of  elasti.-ity.  Atis.  31,800,tXX)  lbs. 

13a.  The  following  observations  were  made  in  the  transverse  test  of  a 
l)<)Ugla.s-lir  beam  I4.S.5  ins.  deepXO  ins.  wideXl.'iO  ins.  l)etween  supimrts: 


lal  ill  Onire 

DnflrK'tiim  in 

Load  at  (Vntre 

I>efle**tion  in 

Iauh\  at  (Vntre    Doflc -lion  in 

ill  I'ound". 

Inchen. 

in  Pounds. 

Inches. 

in  Poun4 

s.                lliclicrt. 

1  ,(K)0 

1.525 

16,000 

2  ItW 

32,(MK) 

2  s.V) 

2,000 

1.570 

IS.CMK) 

2  2.'iO 

34,000 

2  070 

4,000 

1  fi.W 

20.000 

2  3.30 

36,(XK) 

3.0r).5 

6,(X)0 

1.740 

22,000 

2  420 

3S,(X)0 

3  180 

S,000 

1.820 

24,0(X) 

2  510 

40,(HM) 

3  2«K) 

10. (KK) 

1.900 

26.0<K) 

2  .590 

41,700 

3    Failed  on 

12,000 

1  090 

2S.000 

2.680 

*  tension  side 

14,000 

2.060 

30,000 

2 .  7M 

42,000 

( Shcurctt  ion- 
(  gitudinally 

560 


THEORY  OF  STRUCTURES. 


Calculate  the  modulus  of  elasticity,  and  the  maximum  stress  developed 
at  the  iwint  of  failure.  -4«,s.   1,0-'0,()00  lbs.;  7091  lbs. 

133.  From  tho  transverse  test  of  a  white-pine  beam  9"X9"X14'  between 
supports  the  followinR  observations  were  made: 


Load  at  Centre 

Deflertion 

Load  at  Centre 

Deflertion 

Load  at  Ontre 

Deflection 

ill  P.iuiiils, 

ill  Inohes. 

ill  Piiunda. 

in  Inches. 

in  Founds. 

in  Incbm. 

0 

l.jl 

S,000 

2.07 

16,000 

3  97 

2,000 

1.76 

10,000 

2 .  9S 

18,000 

4.48 

4,000 

2  00 

12,(KK) 

3  30 

19,000 

4.69 

6,000 

2  37 

14,000 

3  62 

This  beam  commenced  to  fail  on  the  compression  aide  when  the  load  was 
17,000  lbs.  Calculate  the  modulus  of  elasticity  and  the  maximum  stress 
developed  before  failure  took  place.  Atis.  1,118,000  lbs.;  5877  lbs. 

134.  A  reinforced  concrete  beam  is  to  support  at  ni<u-length  a  concen- 
trated load  of  10,(X)0  lbs.,  the  distance  centre  to  r^ntre  of  supports  being 
12  ft.  .\ssuming  E  to  1m«  *J9,000,000  li)s.  for  the  steel  reinforcing  bars  and 
3,000,000  lb.s.  for  concrete  in  compression,  also  that  the  compressive  and  tensile 
strengths  of  the  concrete  are  21KK)  and  200  lbs.  sq.  in.  rcsi)ectively,  determine 
the  section  of  the  In-am,  the  elastic  limit  of  the  steel  l)eing  .^O.OOO  lbs.  'sq.  in. 

Ans.  Sectional  area  in  sijuare  inches  of  concrete  =99.2  and  of  8tecl=.63.5. 

135.  Two  cantilever  girders,  the  one  4  ft.  and  the  other  0  ft.  long,  are 
placed  at  right  angles  to  each  other,  with  their  ends  meeting,  and  support 
at  their  intersection  a  common  load.  What  mast  be  the  ratio  of  moments 
of  inertia  in  order  that  the  cantilevers  may  carry  equal  loads?    .4hs.  8:27. 

136.  In  a  transverse  test  of  a  slate  beam  3.96"  X 3.97" X  G()"  between 
the  supfwrts,  and  with  the  bed  horizontal,  the  following  observations  were 
made: 


Central  l.<>a>l 
in  I'nundf. 

0 

1,000 
2,000 


Deflection 
in  Inohen. 

0 
.020 

.038 


Ontral  Loail 
in  Poundn. 

3,(MN) 
4,(NM) 
5,000 


Deflertion 
in  Inrhen. 

0.-)7 
.075 
095 


Central  Load 
in  Poiindn. 

6,tK)0 

7,000 


Deflection 
in  Inchea. 

.110 
Failed 


Find  E  and  the  coefficient  of  lx;nding  strength. 

Alls.   1.-., 470.000  lbs.;  11,104  lbs. 
137.  In   a  transverse  test    of  a  slate  l)eam   3.99" X 3.98" X 66"  between 
supports,  and  with  its  bed  vertical,  the  following  obs»>rvations  were  made: 


Central  Ltia*! 
in  l''»iiiiln. 

0 

1,(MM> 

2,(K)0 

3,000 


Deflpf'tinn 
ill  liit'lie^. 

0 

017 
035 
052 


Ontral  Lo»<J 
in  P<iiincl!". 

4,(MM) 

5.(NM) 

6,(»(K) 


DeflertiMH 
in  Ini'he", 

.070 

.0S9 

lOtJ 


(VntTHl  l.uail 
in  roumlN. 

7,000 

H.tMK) 
S.-'JOO 


Defleetino 
in  Inrtiw, 

122 

140 

Failed 


Find  E  and  the  coefficient  of  bending  strength. 

Ann.  1626X10*  lbs.'  13,315  11) 


EXAMPLES. 


561 


138.  In  a  transverse  test  of  an  oak  beam  2.94  ins.  wide X 11 .4  ins.  deepX  144 
ins.  between  supports  the  following  okservations  were  made: 


t'cntral  Ixiad 

Deflection 

Central  Loud 

Deflection 

Central  Load 

Deflection 

in  Pdunila. 

in  Inches. 

in  Pound!!. 

.n  Inches. 

in  Pound.^. 

in  luchcs. 

2,000 

.17 

7,000 

.64 

12,000 

1.120 

3,000 

.26 

8,000 

.73 

13,000 

1 .230 

4,000 

.35 

9,000 

.82 

14,000 

I.TjO 

5,000 

.45 

10,000 

.92 

15,000 

1..500 

6,000 

.55 

11,000 

1.015 

15,500 

Fjtiled  on 
tension  side 

Determine  E  and  the  coefficient  of  landing  strength. 

Am.  1,800,()(X)  ll).s.,.^.|.  in.:  8703  Ibs./.sq.  in. 

139.  A  wrought-iron  beam  of  rectangular  section  and  20  ft.  span  is  16  ins. 
deep,  4  ins.  wide,  and  is  loaded  with  a  proof  load  at  the  centre.  If  the  proof 
strength  is  7  tons/sq.  in.,  find  the  proof  deflection  and  the  resilience,  E  being 
r.',(K)0  tons  (1  ton -2240  lbs.).  .\>,s.  .029  ft.;   650  ft.-lbs. 

140.  In  a  rolled-steel  beam  (.symmetrical  about  the  neutral  axis)  the 
iniiinent  of  inertia  of  the  section  is  72  in.-uiiits.  The  beam  is  8  ins.  deep, 
is  l;ii(l  across  an  opening  of  10  ft.,  and  carries  a  distributed  load  of  9  tons. 
liiid  the  maximum  fibre  stress,  also  the  central  deflection,  taking  E  at  13,000 
tons/sq.  in.  ,l„.s.  7i  t<>'>s/«q- in.;  .216  in. 

141.  -\  girder  30  m.  long  has  both  ends  fixed  and  carries  a  uniformly  dis- 
tributed load  of  5800  k./lineal  metre.    Find  the  maximum  deflection  and  the 


Work  of  flexure. 


Ant. 


783  X 10" 


mm. 


567675X10" 


km. 


64A7      '  El 

14a.  A  steel  beam  of  circular  section  is  to  cross  a  span  of  15  ft.  and  to 
carry  a  load  of  10  tons  at  5  ft.  from  one  end.  Find  its  diameter,  the  stiffness 
Ining  such  that  the  ratio  of  maximum  dcfkction  to  s|)an  is  .0012.").  (A' =  13,000 
'""••*•*  Am.  9.3  ins. 

143-  Determine  the  dimensions  of  a  beam  of  rectangular  section  which 
ininht  Ix-  Mubstitut«d  for  the  round  Ix-ain  in  tiie  preceding  question,  the  stiff- 
in  ss  remaining  the  same  and  the  coeflficient  of  working  strength  being  7i 
'"'"*  •■"I-  in-  Anx.  1.8  ins.  X  13.4  ins.  deep. 

144-  Two  equal  weighu  are  placed  s.nnmetrically  at  the  points  of  tri- 
.'.linn  of  ii  beam  of  uniform  sj-ction  .supported  at  the  ends.    These  weights 

n  I  hen  removed  mid  other  two  equal  weights  are  placed  at  the  quarter  spans. 
IiihI  the  r.itio  of  the  two  m-ts  of  weights  M^  that  the  maximum  intensity  of 
-  I  -s  may  Im-  the  sjitne  in  each  case.  Also  show  that  the  8(ifjncHHo{  the  beam 
1-  the  same  in  each  case.  ,4„,,_  3  („  4 

145-  A  cast-iron  JK-ain  of  an  inverted  T  section  rests  uixm  »up|K)rts  22  ft. 
:>l  lit :  the  web  is  1  in.  thick  and  20  ins.  dwp;  the  flange  is  1.2  ins.  thick  and 
I  -  MIS.  wide ;  the  Ix-am  can  ies  a  unifon.ily  distributed  load  of  99,000  lbs.    Find 


Ii 


maximum  deflection.  E  being  17,920,0(K)  lbs, 

Anx.  .822  in.  (/ -1608.65). 
146.  Find  the  maximum  deflection  of  a  cnsf-inm  cnntjlovor  2  irw.  wideXJ! 
deep X 120  ins.  long  under  iU  own  weight,  E  being  17,920,000  Ibe. 

Ant.  H  in. 


'   if 


562 


THEORT  OF  STRUCTURES. 


147.  An  elastic  beam  rests  upon  supports  at  its  ends,  and  a  weight  placed 
at  a  iwint  .1  protluces  a  certain  deflection  (d)  at  a  point  B.  Show  that  if 
the  weight  is  transferred  to  B  the  same  deflection  (d)  is  produced  at  A. 

148.  A  lx>am  of  20-ft.  span  carries  a  load  which  varies  uniformly  in  intensity 
•om  0  at  .  ne  end  to  100  lbs.  at  the  other  end.  Find  the  work  done  in  bend- 
i..  the  iK-am.  •  (-'-"W' 


Ans. 


■  in. -lbs. 


14Q.  A  pitch-pine  beam  14  ins.  witle,  15  ins.  deep,  and  wei{.'Iiing  45  Ibs./cu. 
f'.  is  placed  upon  supports  10  ft.  9  ins.  apart  and  carries  a  load  of  20 
t.ms  at  tlic  centre.  Find  the  deflection  and  radius  of  curvature,  E  being 
1  .-.'TO.OOO  lbs.  V/hat  stiffness  does  this  give?  What  amount  of  uniformly 
distributed  load  will  produce  the  same  deflection? 

.Ins.  .361  in.;  3843  ins.;  .0028,  .32  tons. 

150.  A  beam  is  supported  horizontally  on  two  posts,  one  under  each 
end;  C  is  a  ix)int  of  the  beam  one  fourth  of  its  length  from  one  of  the  points 
of  siipiwrt.  Compare  the  curvature  at  C,  supijosing  the  beam  to  be  uniformly 
loaded,  with  what  it  would  be  if  the  lieam  were  without  weight  and  the  load 
conien'trated  at  the  middle  iwint,  the  total  load  in  both  cases  being  the  same. 

Ans.  3  to  4. 

151.  A  straight  bar  of  wrought  iron  l"Xl"  section  is  loaded  as  a  tie  bar 
with  5  tons.  It  is  found  that  the  jwrtion  between  two  points  on  it  4  ft.  apart 
elongates  .010  m.  What  is  the  value  of  El  If  the  bar  be  subject  to  a  bend- 
ing moment  of  1800  in.-lbs.,  what  would  be  the  radius  of  curvature?  Fmd 
also  the  greatest  stress  and  deflection  if  the  bar  be  supjKjrted  at  points  4  ft. 
apart  and  centrallv  loaded  with  120  lbs. 

Am.  12,()32  tons;  <»7.4f>fi  ft.;  8640  Ibs./sq.  in.;   .1313  m. 

152.  A  cantilever  of  n-ctangular  section  and  constant  breadth  is  loaded 
at  the  free  end.  If  the  curvature  is  constant  show  that  the  greatest  stress 
developed  in  any  secti<'i  is  proportional  to  the  depth  of  the  section  and  also 
to  the  cube  root  of  the  distance  of  the  section  from  the  loaded  end. 

153.  A  carriage-spring  is  made  uj)  of  six  plates  each  i  in.  thick  X  3  ins.  wide 
and  the  top  plate  is  •«•)  ins.  in  length;  determine  the  initial  curvature  and 
dip  of  this  plate.  Also  find  the  overlap,  the  deflecti(m,  and  the  load  which 
will  produce  a  deflection  of  .M  in.,  taking  A' -=30,000,000  Ibs./sq.  in.  and 
assuming  a  pMof  stress  of  30,0(X)  Ibs./sq.  in. 

.4h.'!.  2:rf)  ins.;    .048  ins.;  Sins.;  .0002592  ins.;   2470  lbs. 

154.  A  weight  of  2.">00  lbs.  is  to  Im-  the  proof  load  of  a  carriage-spring  made 
of  i"y.r'  steel  plates.  If  the  proof  stress  is  30,000  lbs.,  how  many  plates 
will  Ix-  required,  the  top  plate  lM>ing  30  ins.  in  length?  Ann.  <•. 

155.  Finil  the  load  which  will  deflect  2  ins.  a  carriage-spring  made  of  ten 
Btritis  each  |"X2J",  the  lengtii  of  the  upjH'r  plate  being  40  ins.  and  the  over- 

.,  .t-     ..  ..in'UvO  ,4 WK,  3206  lbs. 

156.  A  cantilever  of  length  /,  siiecific  weight  w,  and  square  in  section, 
a  «de  of  the  section  being  2fc  at  the  fixed  and  2a  at  the  free  end,  bends  under 


EXAMPLES. 


56$ 


its  own  weight.  Find  the  slope  and  deflection  of  the  neutral  axis  at  the  free 
ctid.  HAice,  also,  deduce  corresponding  results  when  the  cantilever  is  a  regular 
pyramid.  ^  ^    (6-(-a)t(f     {b  +  '2a)wl\     td'       wl* 

4Eb'     '   ~    8Eb'  "  '  4Eb''  8£6»" 
157-  If  the  section  of  the  cantilever  in  the  preceding  example,  instead  of 
Ixing  square,  is  a  regular  figure  with  any  number  of  equal  sides,  show  that 
tlie  neutral  axis  is  a  parabola  with  its  vertex  4t  the  point  of  fixture. 

158.  Find  the  slope  and  deflettion  at  the  fi'ec  end  of  the  following  canti- 
levers when  bending  under  their  own  ^-'ht,  I  being  the  length,  26  the  depth 
;it  the  fixed  end,  w  the  specific  weight,  .   ,d  E  the  coefficient  of  elasticity: 

(u)  Of  constant  thickness  t  and  win  profile  in  the  form  of  a  trajiezoid 
with  the  non-parallel  sides  equal  and  of  depth  2a  at  the  free  end. 

(b)  Of  circular  section  and  with  profile  in  the  form  of  an  isosceles -triangle. 

(c)  Of  constant  thickness  atid  with  profile  in  the  form  of  a  parabola  sym- 
metrical with  respect  to  the  axis  and  having  its  vertex  at  the  free  end. 

(</)  When  the  depth  2a  in  (a)  is  nil,  i.e.,  when  the  profile  is  an  isosceles 
triangle. 

U)  Due  to  a  uniformly  distributed  load  of  intensity  p  over  the  canti- 
lever (f). 

I/)  Due  to  a  weight  W  at  the  free  end  of  (a). 

{g)  Due  to  a  uniformly  distributed  load  of  intensity  p  over  the  cantilever 
(o).     Hence  also  deduce  the  deflection  and  sloiie  when  the  depth  2a  is  niL 
»■<*  .    _  3uf'       /    a'  a      6»  - nb'  +Ha'b-2a' 

■2Elb-a)'ib-a"^'b^~ 
1  "•/•  ,  ^   2wP      Awl' 


Ans.  (a) 


(6)t^,;t 


id) 


2Eb'' 

IwP 

3  £6''  6  b'E' 

•2Eb'''  4Eb''' 


66' 


)• 


5Eh''  15£6'' 
I'       3   pi' 


^'  2  EbH'  10 E^t 


(/)? 


Wl' 


Wl' 


4  Eatb''  2  Et(b-a)'  ^  b 


'    i   l,-n' 


3h-a 
2b' 


(a)  7 


p? 


4  Et(h 


P        I.      b     (3b-a)(b-a)\ 
^'^''a 26' )' 


4W(6-a)n*^'''"«6-' 


(2b'  +  !iab-a')(b-a) 


a) 
90°. 


)  = 


'•JKb-ayl" "*6  '  26" 

4  Etb" 

159.  A  vertical  row  of  water-tight  sheet-piling  12  ft.  high  is  supported 
!)>  a  series  of  uprights  placed  0  ft.  centre  to  centre  and  securely  fixetl  at  the 
li.se.  Kind  the  greatest  deviation  of  an  upright  from  the  vertical  when  the 
«Mi<r  rises  to  the  top  of  the  piling.  What  will  the  maximum  deviation  be 
wiicn  the  water  is  (>  ft,  from  the  top? 

An»    "'•![!.  .3110400.      «*  ,.     ,M^  «*<;,■.      ,,    218700 
"*"'•  30EI Br  '   30W^*-^)  +T^1F/'^-f)*- 


24£/' 


El 


564 


THEORY  OF  STRUCTURES. 


160.  Show  that  the  curved  profile  of  a  rantilevcr  of  unifonn  strength 
designed  to  carry  a  load  \V  at  the  free  end  is  theoretically  a  cubudl  parabola. 
Also  show  that  by  taking  the  tangc.ts  to  the  profile  at  the  fixed  end  as  the 
boundaries  of  the  cantilever  a  cantilever  of  approximately  uniform  strength 
is  obtained  having  a  depth  at  the  free  end  equal  to  two  thirds  of  the  depth 
at  the  fixed  end  (breadth  to  be  proportional  to  depth). 

161.  Design  a  wheel-spoke  33  ins.  in  length  to  be  of  approximately  uniform 
strength,  the  intensity  of  stress  Ix'ing  1000  Ibs./sq.  in.;  the  load  at  the  end 
of  the  siwke  is  a  force  of  1000  lbs.  applied  tangentially  to  the  wheel's  periphery, 
and  the  section  of  the  siwke  is  to  be  (a)  circular,  (b)  elliptical,  the  ratio  of 
the  depth  to  the  breadth  being  2J. 

.fins,  (rt)  Depth  at  hub  =6.9.')2  ins.,  at  p<!riphery  =4.634  ins. 
(b)        "      "     "     =9.43.^)  "       "         "  =6.29     " 

Breadth  at  hub  =3.774 =2.516" 

162.  A  beam  of  17  ft.  span  is  loaded  with  7,  7,  11,  and  11  tons  at  points 

1    0   11,  and  15  ft.  from  one  end.     Determine  the  depths  at  these  points,  the 

beam  Iwing  of  uniform  breadth  and  of  approximately  uniform  strength;  the 

coefficient  of  working  strength  =2  tons/sq.  in.,  the  depth  of  the  section  of 

maximum  resistance  to  bending  =  16  ins. 

,     11358     ,,     277X16' 
Ans.  0=  ,,,„u  ,  rf,  ■ 


1088 

and  rf.'  ■■ 


1262 
670X16' 
1262    ■ 


,,    1067X16'. 

rf,' .  .^^^ — ;  di'  - 1 6' ; 


1262 


163.  Design  a  timber  cantilever  of  approximately  uniform  strength  from 
the  following  data:  Length  =12  ft.;  square  section;  load  at  free  end-1  ton; 
cfK'fficicnt  of  working  strength  =}  ton/sq.  in.  What  must  be  the  dimensions 
at  the  fixed  and  free  ends  so  that  the  cantilever  might  carry  an  additional 
uniforndy  di.stributcd  load  of  2  tons? 

yln.-i.  Side=l.").l  ins.  at  fixed  end  and -10  ins.  at  free  end;  side -19.1  ins. 
at  fixed  end  and -J(  19.1  ins.)  at  free  end. 

164.  IVsign  a  cantilever  10  ft.  long,  of  approximately  uniform  strength, 
to  carry  a  load  of  4000  lbs.  at  Hie  free  end,  the  coefficient  of  strength  being 
2000  Ibs./sq.  in.,  and  the  section  in)  a  rectangle  of  constant  breadthand  12  ins. 
deep  at  the  fixed  end;    ('<)  a  sfpmre. 

How  will  tlic  results  he  modified  if  i.  i"  to  carry  an  additional  uniformly 
distributed  load  of  4S(K)  ll)s.? 

Am.—  l-'irt<t.  (<i)  ^  =  10  ins.,  rf  atJVee  end-6  in.i.;    (b)  8ide-Vl440 
at    ;ixed  end  and-'<y42(>J  at  free  end.  _ 

Secimil.  (a)  6-16  ins.,  rf  at  free  end  "5..S  ins.;  (fc)  side-^  :.'304 
at  fixed  end  and  --^aOtH  at  free  end. 

165.  Design  a  cantilever  10  ft.  long,  of  constant  breadth  and  of  appmr- 
imaiely  uintorm  strength,  to  carry  a  uiillorMily  disirihuted  load  of  50()fi  lbs- 
on  the  half  of  the  length  next  the  free  end,  the  intensity  of  stress  l>eing  2000 
less  sq.  in.,  and  the  section  a  rectangle  12  ins.  deep  at  the  fixed  end.    What 


EXAMPLES. 


565 


must  the  dimensions  be  if  1000  lbs.  are  concentrated  at  30  ins.  from  fixed  end? 
Ans.  ^=9|  ins.;   d  at  centre  =6.928  ins.;    at  free  end=0. 

6  =  10  ins.;  depth  =9.48  ins.  at  7i  ft.  from  free  end,  =6.708 
ins.  at  centre,  and  =0  at  free  end. 
i66.  A  gallery  30  ft.  long  and  10  ft.  wide  is  supported  by  four  9"X5" 
rantilpvers  spaced  so  as  to  bear  equal  portions  of  the  superincumbent  weight. 
Wliat  load  per  square  foot  will  the  gallery  bear,  the  coefficient  of  working 
sticDgth  being  700  Ibs./sq.  in.?  Find  the  depth  of  cast-iron  cantilevers  3  ins. 
wide  which  may  be  substituted  for  the  above,  the  coefficient  of  working  strength 
bring  2000  Ibs./sq.  in.  How  should  the  depth  vary  if  the  cantilevers  are 
to  l)e  of  uniform  strength? 

.Ins.  lOi  lbs.;  4</'=189;  variation  of  depth  for  cast-iron  -cantilever  is 
pivcn  by  6400(i' =  12Ix',  x  being  distance  from  free  end. 

167.  A  span  of  60  ft.  is  cros.sed  by  a  l)eam  hinged  at  the  points  of  trisec- 
ticiii  and  fixed  at  the  ends;  the  beam  ha.s  a  constant  breadth  of  3  ins.  and  is 
to  Ik-  of  unijorm  strength;  the  inten.sity  of  stress  is  3  tons/sq.  in.  Determine 
till'  dimensions  of  the  beam  when  a  load  of  A  ton  per  lineal  foot  covers  (o)  the 
whole  span;    (6)  the  centre  span. 

Ans.  (a)  Depth  at  support  =4v'l2  ins.,  at  centre  ='v/20. 
(6)       "       "        "       =  v^80  "      "      "      -n/20. 

168.  The  weight  of  200  Ibs./sq.  ft.  upon  a  platform  60  ft.  long  and  10  ft. 
wide  is  equally  borne  by  six  cast-iron  girders  of  rectangular  section,  triangular 
ill  profile,  10  ft.  long  and  3  ins.  wide.  Find  the  depth  at  the  fixed  end,  taking 
2  tons/sq.  in.  as  the  coefficient  of  strength. 

If  £  =  17,000,000  lbs.,  find  the  deflection  at  the  free  end. 

Ans.  24j  ins.;  .276  in. 

169.  Find  the  limiting  length  of  a  cedar  cantilever  of  rectangular  section 
in  which  the  length -40 X depth,  w-36  Ibs./cu.  ft.,  and  /-1800  Ibs./sq.  in. 

Ans.  60  ft. 

170.  A  steel  cantilever  2  ins.  square  has  an  elastic  strength  of  1,5  tons/sq. 
ill.     What  mu.st  its  limiting  length  be  so  that  there  may  be  no  setf 

An.'i.  23.4  ft. 

171.  Find  the  limiting  length  (-64xdppth)  of  a  WTought-iron  beam  of 
I  iniilar  section,  the  clastic  strength  l)eing  8  tons/sq.  in.  What  will  this  length 
I"  if  ;i  b<'am  of  I  section  having  equal  flange  areas  and  a  web  area  equal  to 
till  joint  area  of  the  flanges  is  substituted  for  the  circular  section? 

.4ns.  84  ft.;    224  ft. 

1 7 J.  A  rectangular  cast-iron  beam  having  its  length,  depth,  and  breadth 
in  iIk'  ratio  of  60  to  4  to  1  rests  u\mn  supftorts  at  the  two  ends.  Find  the 
iliiiii  ii-;ions  of  the  beam  so  that  the  intensity  of  stress  under  its  own  weieht 
III  IV  iiowhoreexceed 4.500  Ibs./sq.  in.     Ans.  /"128  ft.;  (i=8|"jft.;  h---'2i%  ft. 

173.  The  effective  length  of  the  Conway  tubidnr  bridge  is  412  ft.;  the 
'  ■  live  depths  of  a  tube  at  the  centre  and  quarter  spans  are  23.7  ft.  and 
- -'  -I  ft.  res|)ectively ;  the  sectional  areas  of  the  top  and  bottom  flanges  are 
"tively  (j4r»  f..j.  iiis.  aiut  530  sq.  ins.  at  the  centre  and  .566  sq.  ins.  an  1  461 
^<{.  ins.  at  the  quarter  spans;  the  corresponding  sectional  areas  of  the  web 
ar.  j.-,7  »,|.  ina.  and  241  sq.  ins.     Assume  the  total  load  upon  a  tube  to  be  equiv- 


r)66 


THEORY  OF  STRUCTURES. 


alent  to  3  tons/lin.  ft.,  and  that  the  continuity  of  the  web  compensates  for 
the  weakening  of  the  tension  flange  by  the  rivet-holes.  Find  (a)  the  flange 
stresses  and  the  deflection  at  the  centre  and  quarter  spans,  E  being  24,000,000 
lbs.  What  (6)  will  be  the  increase  in  the  central  flange  stresses  under  a  uni- 
formly distributed  hve  load  of  J  ton/lin.  ft.? 

Ans.  (a)  4..59  tons/sq.  in., 3.04  tons/sq.  in.;  8..5(>  ins. 


1.3(57 


1.15S      " 


6.72  " 


{!>)  Stresses  and  deflections  are  increased  in  ratio  of  3  to  4. 
174-  Design  ii  wooden  cantilever  12  ft.  long,  of  circular  section  and  uniform 
filrciigth,  to  carry  a  uniformly  distributed  load  of  2  tons,  the  coefficient  of 
working  strength  being  1  ton/sq.  in.     Also,  find  the  deflection  of  the  free 
t>nd. 

Aux.  Taking  fixed  end  as  origin  and  z  being  radius  in  inches  at 
di.stance  x  ft.  from  origin,  then   1  la' =  14(12 -x)'. 


Deflection  at  end  =- 


[37.9 . 
E 


ins. 


175.  A  beam  ^B  of  span  /,  carrying  a  uniformly  distributed  load  of  inten- 
sity u;  rests  ui)on  a  supjwrt  at  Ii  and  is  imiH>rfectly  fixed  at  A,  so  that  the 

1   wP 
n(<utral  axis  at  A  has  a  slojw  of  ;jj^^.     The  end  B  is  lower  than  .1  by  an 

1   1/7* 
amount  —  g^~.     Fmd  the  reactions.     How  much  must  B  be  lowered  so  that 

tlio  whole  of  the  weight  may  be  borne  at  .4?    Find  the  work  done  in  bending 

tlic  beam.  21        11  7  irl*  53     ir'P 

^^ns.  ^^ul. -.1;    --;    -^2^_. 

176.  Two  angle-irons,  each  2"X2"Xi",  were  placed  upon  supports  12  ft. 
9  ills,  apart,  the  transverse  outside  distance  between  the  bars  iM-iiig  9}  ins., 
and  were  prevented  from  turning  inwards  by  a  thin  plate  uiM)n  the  upjier 
faces.  The  bars  were  tested  under  uniformly  distributed  loads,  and  each 
was  found  to  have  dellected  2,^^  ins.  when  the  load  over  the  two  was  1008  lbs. 
Find  E  and  the  iMwition  of  the  neutral  axis. 

Hotii  bars  failed  t.igethcr  when  the  total  load  consisted  of  lOJ  cwts.  (cwt.  = 
112  lbs.)  uniformly  distributed  and  3  cwts.  at  the  centre.  Find  the  maximum 
stress  in  the  metal. 

-t/i.s-.  17,22(l,l.«)lbs.;  j;  in.  from  upper  face:  20,323  and  39,.-)77  Ib-s./sq.  in. 

177.  A  horizontal  girder  of  uniform  strength,  of  rectangular  section  and 
of  length  /  rests  upon  supports  at  its  ends  and  carries  a  uniformly  distributed 
load  of  intensity  w,  which  develops  the  mnif  maximum  ulrcxs  j  at  every  cross- 
.section  of  the  girder.  If  the  depth  (d)  is  constant  and  the  breadth  variable, 
find  the  maximum  deflection.  ,    //» 

4  Ed 

J 78.  A  semi-girder  of  unijorm  Mrcngth,  of  length  /,  and  of  rectangular 
seel  ion  carries  a  weight  W  at  the  free  end  which  produces  a  maximum  stress 
/  at  every  cross-se.tioii  of  the  beam.  Prove  that  the  maximum  deflecfion  i* 
4   ijl)'>  /   l>  \i     ,  ,      , 

3  ~E~  VOUJ    ^^   '^"  *"''  "rcadth  (h)  is  constant  and  the  depth  variable,  and 


EXAMPLES. 


567 


fli.it  it  is  f wire  as  great  as  it  would  Ik  if  the  section  were  uniform  throughout 
Jiiul  equal  to  that  at  the  supjjort.  What  would  be  the  maximum  deflection 
if  the  semi-girder  were  subjected  to  a  uniformly  distributed  load  of  w  lbs. 
per  unit  of  length? 

Ans. 


E  ^  5.„- 


179.  The  vertical  soctions  of  a  cantilever  ABC,  of  length  /,  and  fixed  at 
the  (11(1  BC  are  circular.  The  profile  is  a  cubical  parabola  with  its  vertex 
at  1.  Show  that,  under  its  own  weight,  the  slope  at  the  free  end  of  the  neutral 
axis  i.s  ten  sevenths  of  the  maximum  deflection  divided  by  the  length. 

180.  A  horizontal  beam  of  length  /  carries  a  load  P  concentrated  at  a 
(listaiKc  a  from  each  end.  Find  the  maximum  deflection  due  to  P  (a)  if  the 
two  ends  rest  upon  the  supports,  (b)  if  both  ends  are  fixed. 

'  '  EI  \8     6  /  '    ^"^  EI  \8     6  /  • 

181.  What  shoild  be  the  diameter  of  a  screw-blade  for  a  cast-iron  pile  of 
-'4  IMS.  diameter,  .so  that  the  stress  at  the  root  of  the  blade,  which  is  2  ins.  thick 
iiKiy  not  exceed  1  ton/sq.  in.  under  a  uniform  soil  pressure  of  2  tons/sq,  in.7 

.  ^Ins.  25.63  ins. 

182.  If  a  bar  of  cast  iron  1  in.  square  and  1  in.  long,  when  secured  at 
oiM.  end  breaks  transversely  with  a  load  of  6000  lbs.  susjwnded  at  the  free 
<ii(l,  what  would  be  the  safe  working  jjre.ssure,  employing  a  factor  of  safety 
of  10,  k-tween  the  two  teeth  which  are  in  (•f)ntact  in  a  pair  of  spur-wheels 
wimse  width  of  tooth  is  6  ins.,  the  depth  of  the  tooth,  measured  from  the  point 
to  I  ho  root,  being  2  ins.,  and  the  thickness  at  the  root  of  the  tooth  1}  ins. 
(Assume  that  one  tooth  takes  the  whole  load.)  Ans.  4050  lbs. 

183.  Show  that  the  moment  of  resistance  of  the  equal  flanged 
rollcil  joist  with  the  web  vertical  is  to  the  in.mu-nt  of  resist- 
.-tnce  when  the  web  is  horizontal  in  the  ratio  of  uinhi' -2bh,')  • 

I'  \{h:~h,)a'+h,{a-2by\.    (Ex.  a  =3  ins.,  fc  =  1.2  ins., /,,  =  5  ins      ^ 

'-^!*"'t^   .         ,  ,  An..  5.7  to  1.    '   i^T^ 

184.  A  piece  of  grccnheart  140  ins.  between  supjiorts,  9  ins.  wide,  and 
^  Ills,  deep  was  sui-cessivcly  siibjecte.l  to  loads  of  4,  8,  and  10  tons  at 
nie.rntre,  the  corresponding  deflections  l)eiiig  .32  in.,  .64  in.,  and  1.28  ins. 
I   II. !  /;  ftiul  the  total   work   done  in  Iwiuliiig  the  l)eam. 

\Mi;it  were  the  (orrcspondiiig  inch-stresses  at  three  fourths  of  the  depth 

An».  E  ^  1S61  tons;  13.44  in.-tons;  J|  tons,  U  tons,  V^  tons. 

185.  The  square  of  the  radius  of  gyration  of  the  equal-flanged  section 
-I  a  wrought-iron  gir.ler  of  depth  d  is  .V/';  the  area  of  the  .section  =  5rf'-  the 
M'MM  =oO  ft.  In  addition  to  its  own  weight  it  carries  a  uniformly  distributed 
1  .111  of  IV,  Ibs./lin.  ft.;  the  maximum  intensity  of  stress=10  000  Ibs./sq.  in 

1  irul  the  depth.     Also  determine  the  stiffness,  E  being  25,000,000  lbs. 

186.  The  section  of  a  beam  is  of  the  form  and  dimensions"  show^^  by 'the 
(igure,  and  the  coefficient  of  strength  is  10  tons/sq.  in.     Find  the  moment 


^ 


568 


THEORY  OF  STIiiCTURES. 


L-4-'-^ 

1 ^ 

♦  ■ 

1 

1  'f 
5 

1 

Fig.  ,')-14. 


of  resistance  and  the  greatest  load  which  may  be  placed  at  the  centre 
of  such  a  In^ain  of  10  ft.  span.        Arts.  W2}  in.-tons;  3.42  tons. 

187.  The  up|)er  chord  of  a  Howe  truss  is  24  ins.  wide  X 12 
ins.  decj>  and  is  made  up  of  four  12"XG"  timbers;  the  lower 
chord  is  24  ins.  widcXKJ  ins.  deep  and  is  made  up  of  four 
IC'XO"  timlx'rs:  the  distance  between  the  inner  faces  of  the 
chords  is  24  ft.  Find  the  moment  of  resistance  to  bending,  taking  800  Ibs./sf]. 
in.  as  the  coefficient  of  tensile  strength  and  neglecting  the  effect  of  the  web. 

Ans.  Neutral  axis  is  137?  ins.  from  bottom  face  of  lower  chord; 
moment  of  resistance  =S7, HI, (HO  in. -lbs. 

188.  A  girder  supported  at  the  ends  is  30  ft.  in  the  clear  and  carries  two 
stationary  loads,  viz.,  7  tons  concentrated  at  6  ft.  and  12  tons  at  IS  ft.  from 
the  left  supi)ort.  Find  the  position  and  amount  of  the  maximum  deflection, 
and  also  the  work  of  flexure.  The  girder  is  built  up  of  plates  and  angle-irons 
and  is  24  ins.  deep.  If  the  moment  of  resistance  due  to  the  web  is  neglected, 
and  if  the  intensity  of  the  longitudinal  stress  ii-  not  to  exceed  5  tons/sq.  in., 
what  should  be  the  flange  sectional  area  corresponding  to  the  maximum  bend- 
ing moment?  Determine  the  work  of  flexure  and  the  necessary  flange 
sectional  area  at  the  centre  if  the  girder  is  subjected  to  a  uniformly  dis- 
tributed load  of  40  tons  instead  of  the  isolated  loads. 


Ans.  Max.  def. 


-(x-G)» 


4536     ) 


EI  X  being  15.22  ft. 


\Vork=^"^.^^-^  ft.-tons,  10.32  sq.  ins.;   work  =  ^-^^  ft.-tons, 

sectional  area  =  15  sq.  ins. 
i8g.  A  beam  AB  of  span  /  carrying  a  uniformly  distributed  load  of  inten- 
sity u-  is  fixed  at  A  and  merely  supjwrted  at  B.    The  end  B  is  lowered  by  an 

u-i*       ,-    ,    . 
amount  —7:7.     rmd  the  reactions.     How  much  must  B  be  lowered  so  that 

the  whole  of  the  weight  may  be  borne  at  .4? 

Solve  the    example    supp<jsing  the    fixture   at   A   to   be  imperfect,   the 


neutral  axis  making  with  the  horizontal  an  angle  whose  tangent  is  tt; 


1^  wl^ 

48  El . 


Ans.  {Ini  at  .1,  ,\wl  at  B;  [  |^;      JuZ,   iul;    ^^  |^. 

190.  The  section  of  a  girder  of  24  ft.  span  supported  at  the  two  ends  is 
shown   by   Fig.  545.     The   allowable   working  stress   |ier  s(|uare 
incli  is  4  tons.     Determine   the   load   which  may   be    uniformly 
distributed  over  such  a  iK-am,  and  find  the  maximum  defle<'tion. 
iA'=.,^  iij'  11..-.)  .l„.s-.  10..-.3tons;  ..384  in. 

191.  A  steel  bar  of  uniform  rcctaiiirular  section,  10  ft.  Ions;, 
has  to  supi)ort  a  lojid  of  2  tun-  at  llic  centre  and  to  satisfy 
the    ''onditi'in   thtit     when   ihi 


«• 

\ 

1 
1 

f 
1 

1     . 

i 

the 


1!'!!    1!l".' 


St  ^f'^ 


d    i*   <1   t 


the  deflection  at  the  centre   is   .2  i 

the  bar,  being  given  that  A' =  13,500  tons/sq.  in.       .4 


()t)K/sfj.   in. 


Flo.  .'545. 


Ddermine  the   ireadth  and  depth  of 
-i'i  ins.  ;    5i  ins. 


BXAMPLBS. 


569 


1Q2.  Two  horizontal  beams,  crossing  at  right  angles  at  their  middle  points 
a:v  ngKlly  connected  at  the  crossing  and  have  their  ends  resting  upon  sup- 
I-Ms.    ^how  that  if  a  load  P  is  placed  at  the  crossing  the  portion  borne  by 

a    mrl..  iK-am  is  /' ^  ( 1 +£j,  „  and  m  being  the  ratios  of  the  lengths  and 
III  nji'iitsof  inertia  of  the  beams. 

193.  A  liorizontal  bracket  of  length  a  is   attached  to  the  upper  end  of  a 
v.i Ileal  pilar  of  length  I  which  has  its  lower  end  built  in.     When  carrying  a 
lo.„l  II   at  the  extremity  of  the  bracket  the  pillar  bends  slightly.     Show  that 
(ill.,  t,.  flexure  of  the  pillar,  the  bending  moment  at  its  base  is  increased  what- 
ev(  r  the  length  of  the  bracket,  in  the  ratio 

\w, 

in  ME  is  Young's  modulus  and  /  the  moment  of  inertia  of  the  cross-section 
of  the  pillar  about  line  through  its  centre  perpendicular  to  the  plane  of  bending. 

194.  A  rolled  beam  with  equal  flanges  and  a  web  whose  section  is  equal 
to  th,.  joint  section  of  the  flanges  has  a  span  of  24  ft.  and  carries  a  weight  of 
S  tons  at  the  centre.  If  the  stiffness  is  .001  and  if  the  coefficient  of  strength 
per  s,,uare  inch  is  5  tons,  find  the  depth  of  the  beam  and  the  web  and  flange 
«..tional  areas.     (£;- 15,000  tons.)      Ans.  16  ins.;    10.8  and  5.4  sq    ins 

195.  In  a  rolled  joist  the  sum  of  the  two  flange  areas  ahd  the  web  area 
IS  a  .onstant  quantity.  Find  the  proportion  between  them  which  will  eive 
a  joist  of  maximum  strength,  the  thickness  of  the  web  being  fixed  by  practical 
nuisKJerations.  a„,    r'i„„„„  „  ■.       l 

,„,     .  u^  •        u  \  .  ^"s-  flange  area  =f  web  area. 

.96.  A  wrought-iron  beam  of  I  section,  20  ft.  between  supports,  carries 
a  uniformly  distributed  load  of  4000  lbs.  and  deflects  .1   in.;    the  ;ffective 

■'"'.;  T:\  ^  =3«'T'^  ^^'■'  ^'^  ^^^^-Joint  area  of  the  equal  flanges 
I.n.i  the  total  sectional  area.  Also  fi.id  the  width  of  a  rectangular  section 
>s  ins.  deep  which  might  be  substituted  for  the  above. 

.    .  J     ^  ■^"»-  ■'-240;  area=22J  sq.  ins.;  width=5i  ins. 

197.  A  girder  has  a  moment  of  resistance  of  550,000  f  t.-lbs.,  a  depth  of  3.2  ft 
and  a  web  area  of  15  sq.  ins.     Determine  the  area  of  the  flanges,  the  coefRcieut 
of  strength  for  compression  and  tension  being  14,000  lbs./s(|.  in. 

198.  A  plate  girder  30  ft.  long  and  3.6  ft.  deep  carrier'a  load^''of"TO00 
II'-  li...  ft  Calculate  the  area  of  the  flange  so  that  the  flange  stress  .shall 
"nt  .  xcoed  12,,'-.00  lbs./sq.  in.  If  two  6"X4"xf'  angles  (area  of  section  4  5 
i-'\.  HIS.  each)  and  one  cover-plate  be  used  for  each  flange,  find  the  length  of 
tho -over-plate.  Ans.  1.1  sq.  ins.;  18.96  ft. 

100.  A  street-car  weighing  40,000  lbs.,  the  weight  being  concentrated  on  2 
axir  s  s  ft.  apart,  is  to  be  carried  across  a  span  of  20  ft.  by  a  number  of  timber 
strmsors  16  ins.  deep.  Find  the  total  width  of  the  stringers  so  that  the  fibre 
ST.  -,  may  not  e.xceed  1000  lbs./sq.  in.  ,\ns   36  ins 

^00.  Determine  the  moment  of  inertia  of  a  built-up  beam  composed  of 
t";;  .■■iiial  flanges  each  consisting  of  a  7i"Xi"  plate  connected  with  a 
-1      .     wh,  by  four  angle-irons,  each  3"  X  3"  Xi".    If  the  working  strew 


it 


570 


THEORY  OF  STRICTURES. 


is  (■>  tons/sq.  in.,  find  the  moment  of  resistance  of  the  section.     If  the  span 
is  I'O  ft.,  what  uniformly  distributed  load  will  the  beam  safely  carry? 

Ans.    30")8:  1408  in.-tons;  48.9  tona. 

201.  .\  cast-iron  l)eam  of  rectangular  section,  1-  ins.  dee|i,  ti  ins.  wide,  and 
If)  ft.  long,  carries,  in  addition  to  its  own  weight,  a  single  load  I';  the  coeffi- 
cient of  working  strength  is  20(X)  lbs./s(i.  in.  Find  the  value  of  P  when  it  is 
placed  («)  at  the  middle  [M>int ;    ('<)  at  '2\  ft.  from  one  end. 

.1/i.s-.  (a)  4'.'00lbs.;   {h)  9.i775  lbs. 

202.  The  to])  and  bottiim  flanges  of  a  rolled  section  of  wrought  iron  are 
8" X  I",  and  the  web  is  of  the  same  thickness.  The  heiglit  over  all  is  12  ins. 
What  is  the  bending  moment  when  the  greatest  tensile  stress  is  10,000 
Ibs./sq.  in.?  Au)<.  647,.500  in.-lbs. 

203.  .\  rolled  iron  beam  of  I  section  is  1  in.  thick  throughout  and  rests 
upon  supports  10  ft.  apart.  The  lianges  are  8  ins.  wide  an<I  the  depth  over 
all  is  8  ins.  What  weight  will  the  joist  carry  at  the  centre  if  the  safe  working 
stress  is  4  tons^sq.  in.?  .l».s.  7.18  tons. 

204.  Calculate  ap|)roximately  the  .safe  central  load  for  a  single-web  riveted 
girder  G  ft.  decj);  llangcs  IS  ins.  wide,  2i  ins.  thick.  The  flange  is  attached 
to  the  web  by  two  4"  X  \"  angles.  Neglect  the  strength  of  the  web.  and  assume 
that  the  section  of  each  flange  is  reduced  by  two  rivet-holes  J  in.  in  diameter 
passing  through  the  flange  and  angles.  (Span  of  girder,  oO  ft.;  stress  in 
flanges,  5  tons/sq.  in.)  .l«.s.  110  tons. 

205.  \  trough  section.  Fig.  o46,  is  used  for  the  flooring  of  a  bridge;  each 
section  has  to  support  a  uniforndy  distributed  load  of  l.oO  Ibs./sq.  ft.,  and 

a  concentrated  central  load  of  4  tons.     Find  the  span  for 

which  such  a  section   may  be  safely  used.     Skin  stress 

=  5   tons/sq.   in.;    j)itch    of    corrugation,    2   ft.;  depth, 

1  ft.;    width  of  flange  =8  ins.;    thickness  =  J  in. 

Ans.  22.8  ft. 

206.  \  lattice  girder  of  80  ft.  span  and  8  ft.  deep  is 

designed  to  carry  a  dead  load  of  00  tons  and  a  live  load 

of  120  tons  uniformly  distributed;  at  the  centre  the  nd 

sectional  area  of  the  bottom  flange  is  4.1  sq.  ins.  and  the  gross  .sectional  area 

of  the  top  flange  oOi  sq.  ins.     Find  the  position  of  the  neutral  axis  and  the 

maximum  flange  intensities  of  stress.  If  the  live  load  travelsat  60 miles  an  hour, 

what  will  be  the  increased  pressure  due  to  centrifugal  force?  (1  ton  =2240  lbs.) 

Ans.  3.o4G  ft.  frt)m  toj);    11,200  Ibs./sq.  in.;   8920.35  Ibs./sq.  in.; 

.594000 

EI 

207.  Two  tracks  6  ft.  apart  cross  the  Torksey  Bridge,  and  are  supported 

l)\-  single-webbed  plate  cross-ginh^rs  2.')  ft.  long  and  14  ins.  deep.     If  the  whole 

of  the  weight  upon  a  pair  of  drivers,  viz.,  10  tons,  is  directly  transmitted  to 

one  of  these  cross-girders  find  the  maximum  deflection  of  the  girder  and  the 

work  of  flexure  when  the  ends  (o)  are  fixed  to  the  main  girders,  (/i)  merely 

'•est  on  these  girders. 

499.32  141fi.36.'j 

Ans.  (a)   -.,■  •  at  14.205  ft.  from  end;   —fn —  ft.-tons. 


Fig.  546. 


lbs. 


El 


El 


EXAMPLES. 


571 


208.  Find  the  uniformly  distributed  load  which  can  be  borne  by  a  rolled 
T-iron  beam,  G"x4"Xi",  10  ft.  lonp,  fixed  at  one  end  and  free  at  the  other 
tliocoofficientof  strength  being  10,000  Ibs./sq.  in.  .4ns.  438  lbs.    ' 

209.  One  of  the  tubes  of  the  Hritannia  bridge  has  an  efTective  length  of 
170  ft.,  depth  of  27i  ft.,  and  deHects  12  ins.  at  the  centre  under  a  uniformly 
.iistnbuted  load  of  l.JS?  tons.  Find  E  and  the  central  flange  stresses,  the 
sjM'tioiial  areas  of  the  top  flange,  bottom  flange,  and  web  being  648  sq.'ins., 
■iS,")  sq.  ins.,  and  302  sq.  ins.,  respectively. 

Ans.  A' =22,910,496  lb.s.;  /,  =5.37  tons /sq.  in.;  /.=4.S1  tons/sq.  in. 

210.  Compare  the  resistance  to  bending  of  a  wrought-iron  I  section  when 
the  l)cam  is  placed  like  this,  I,  and  like  this,  -.     The  flanges  of  the  beam 
air  (•)  ms.  wide  and  1  in.  thick,  and  the  web  is  }  in.  thick  and  measures  8  in- 
iutween  the  flanges.  ^^s.  4.57  to  1. 

211.  Comp.ire  the  strength  to  resist  l)ending  of  a  rolled  joist  with  an  8"X  1" 
w(.b  and  two  eijual  Hanges  each  4"x  1",  when  placed  with  its  axis  vertical  and 
with  Its  axis  horizontal.        Ans.  Ratio  of  moments  of  resistance  =  7.25  to  1. 

212.  The  deflections  of  a  licam  at  three  points  .1,  B,  and  C,  are  2,  2J,  and 
4  mm.  resi)ectivcly,  under  a  load  of  10  tons  at  a  point  D.  If' the  load'  at  I) 
1^  now  removed,  and  if  loads  of  S,  12,  and  6  tons  are  concentrated  at  the  points 
.1,  li,  and  C  respectively,  find  the  deflection  at  D.  Ans.  7  mm 

213.  A  railway  girder  is  101.2  ft.  long,  22.25  ft.  deep,  and  weighs  3764 
ll)s./lm.  foot.  Find  the  maximum  shearing  force  and  flange  stresses  at  25  ft. 
I'nim  one  end  when  a  live  load  of  2500  Ibs./lin.  ft.  crosses  the  girder. 

-1».>i.  168,078.3  lbs.;  268,155.5  lbs. 

214.  A  floor  with  superimpos<'d  load  weighs  160  Ibs./sq.  ft.  and  is  carried 
l>y  tubular  girders  17  ft.  centre  to  centre  and  42  ft.  bctwen  bearings.  Find 
the  (l.'pth  of  the  girders  (neglecting  efTect  of  web),  the  safe  inch-stress  it\  the 
inr)  il  being  9000  lbs.  and  the  sectional  area  of  the  tension  flange  at  the  centre 
•'-'  '^'l-  '"•■'•  A».^.  24.99  ins. 

215.  A  hollow  tube  of  wrought  iron  of  3  ins.  outside  and  2J  ins.  inside  diam- 
eter IS  20  ft.  long ;  find  its  weight.  What  is  its  deflection  under  its  own  weight' 
What  further  weight  on  fts  middle  will  it  carry  safely  if  /=4J  tons/sq.  in.? 

.4ns.  21  lbs.;  .42  in.;  144Albs.;  158.7  lbs. 

216.  Through  what  height  must  a  weight  of  400  k.  fall  upon  the  middle 
of  a  joist,  200  cm.  long  X 24  cm.  deep,  without  developing  a  greater  stress 
than  1600A:/cm.^  80  per  cent  of  the  work  done  being  used  up  in  producing 
flexure  and  taking  /  =4288  aii.l  £•  =  2  X 10  A/cm.=.  Arts.  3.75  cm. 

217.  A  timber  stringer  6"xi6"  rests  on  two  supports  20  ft.  apart  and 
-  ames  two  concentrated  loads  of  16,000  lbs.  each  at  a  distance  of  2  ft.  from 
each  support.  Find  the  greatest  intensity  of  shearing  stress  along  the  fibre 
of  the  timl)er.  4„,    3^00  j^g  /,^  {„ 

If  the  ultimate  resistance  to  compression  is  8000  Ibs./sq.  in.  and  the  resist- 
ancc  to  shear  along  the  fibre  600  Ibs./sq.  in.,  indicate  the  probable  mod<  of 
failure  as  the  load  is  increased.  .4  «.s.  Intensity  of  shear  250  Ibs./sq.  in. 

218.  A  cast-iron  beam  has  a  cruciform  section  with  equal  ribs  2  ins.  thick 


.ill 


THEORY  OF  STRUCTURES. 


aiul  4  ins.  long.  If  the  intensity  of  longitudinal  shear  at  the  neutral  axis  is 
1  tnii/s(i.  in.,  find  the  total  shear  which  the  section  can  bear,  and  also  find 
till-  moment  of  resistance,  the  least  coefficient  of  working  tensile  and  com- 
pressive stress  btMiig  1  ton/s(|.  in.  Ann.  'td.'M  tons;   34.4  tons. 

219.  A  cast-iron  beam  of  an  inverted  T  section  has  a  uniform  depth  of  20 
i.is.  and  is  22  ft.  between  supports:  the  flange  is  12  ins.  wide  and  1.2  ins.  thick; 
the  wel)  is  1  in.  thick;  the  load  uixmi  the  beam  is  I'jOO  Ibs./lin.  ft.;  E  = 
17.000,000  lbs.  Fitid  the  deflection  at  the  centre,  ae  moment  of  resistance 
to  l)eiiding,  the  maximum  tensile  and  compressive  intensities  of  stress,  and 
the  position  of  the  neutral  axis.     Why  is  the  flange  placed  downwards? 

Ans.  .122  in.;    320,700  in.-lbs.;    1402  lbs.  and  3274  Ibs./sq.  in.; 

r).0()2  ins.  from  flange  siirface. 

220.  The  section  of  a  lK>am  is  in  the  furm  of  an  i.sosceles  triangle  with  its 
base  horizontal.  Show  that  the  moment  of  resistance  to  bending  of  the 
stronjiiwt  tr:ipc/,oi(lal  In-am  that  can  be  cut  from  it  i"  very  nearly  iVo/W', 
6  being  the  width  of  the  base  and  d  the  depth  of  the  triangle. 

aai.  Find  the  sectional  area  of  a  wrought-iron  beam  of  T  section  which 
may  be  substituted  for  the  cast-iron  beam  in  Example  21'J,  the  depth  being 
the  same  and  the  coefficients  of  sfength  uer  S()uare  inch  being  3  tons  in  com- 
prr-ssion  and  ."»  tons  111  tension.  Why  should  the  flange  be  uppermost?  What 
should  the  lutal  sectional  area  be  if  the  flange  and  web  are  of  eijual  area? 

Ana.  1U.14  sq.  in. 

:«22.  A  cist -iron  beam  of  T  section,  with  a  1.02"  X  1.02"  upper  flange  and 
a  .5.07"X.")7"  web,  rests  ujMin  supports  102  ins.  apart  and  carries  a  load  of 
8300  lbs.  at  the  middle  jMant.     Find  the  tensile  fibre  stress. 

.4».s.  33,527  lbs. /sq.  in. 

223.  A  iR'am  of  1.')  ft.  span  c.irries  loads  of  2,  4,  1,  and  3  t(ma  at  dis- 
tances of  2,  .">,  7,  and  II  ft.,  re<{H'ctively,  from  the  left  supjxjrt.  Determine 
the  deflection  graphically,  taking  /  =  242  and  £'=12,000  tons  per  sq.  in. 

Ans.  0.3  iu. 

224.  Fiml  tlu-  mo  ent  of  resistance  of  the  section  (Fig.  547)  (a)  about 
the  axis  .1.1,  ('()  alniut  the  axis  /J/i,  each  passing  through  the 
centre  of  gravity  ('  of  the  si'ction,  the  allowable  extreme  fibre 
stress  iM'ing  Ifi.OOO  Ibs./scj.  in.  The  section  consists  of  one 
l<t"x3''eover.  /»■«  l.">-in.X3.)-lb.  channels.  and<Hy4"Xi"  bars. 

Arix.    Oi)     2,S72,320     in.-lbs.;     (6)  . 

2,47fi,.")HC.  in.-lbs. 

225.  Fig.  .')4S  shows  a  chord  s«'ction 
designed  fora  .">(i0-ft.  double-track  s|)an. 
The  unsupfiorled  length  is  3.')  ft.  Find  ((/i  the  moment 
of  resistaiK'c  iilxnil  a  verliiiil  a\i>;  {In  the  ilttlection 
and  iniixiniu:!  fibre  stress  due  to  its  own  weight.  The 
seeti'in  consists  of  oi,<-  4S"  .  J"  cover,  lour  top  angles 
I'.ich  4"  'I":,  J",  fitjlit  42"XU"  web  plates  in  pairs,  and  jour  lH)tt»m   angles* 

eacli  7"  ^'35"-  1.:". 

Ans.  {'!)  43,121,3(0  in, -lbs.;  {h)  fibre  Hfn>«8-.'>I2  lbs./s«.  in.;  <leflcclion 
...014  in.,  E  being  30,(K)0,(Ktu  lb8./s«|.  in. 


flST^- 


c,     t 

I 


Fifi    .517 


I'k.  .5 is 


-il 

ii 


EXAMPLES. 


573 


jj6.  Taking  ft,  fc  as  the  tentile  and  compressive  intensities  of  stress,  find 
111'  moment  of  resistance  to  bending  of  a  section  consisting  of  a  20t"X7t" 
lop  (langc,  an  80<"X10/"  bottom  flange,  and  a  trapezoidal  web  4<  ins.  thick 
Ml  tlio  top,  8/  ins.  thick  at  the  bottom,  and  I20t  ins.  deep.  Also,  com- 
pare the  maximum  and  average  intensities  of  shear. 

Anfi.  84,HCAIU*  or  36,419/r/'- 
2J7.  A  ca.st  iron  channel-beam  having  a  web  12  ins.  wide  and  two  sides  7  ins. 
deep,  the  metal  being  everywhere  1  in.  thick,  -osses  a  .sp-'n  of  14  ft.  If  the 
tensile  intensity  of  stress  is  1  ton/sq.  in.,  what  uniformly  distributed  load  will  liie 
l>eam  carry  (o)  with  the  web  at  the  bottom ;  (/))  with  the  web  at  the  top?  Find 
('■)  the  maximum  compressive  intensity  of  stress  to  which  the  metal  is  sub- 
jeited,  and  (d)  compare  the  maximum  and  average  intensities  of  shear.  Also 
If)  what  should  be  the  area  of  a  rectangular  section  to  bear  the  same  total 
shear?  Ann.  /  =  110J;  (a)  Ut  tons;  (h)  m  tons;  (e)  V.Vaq.  ins. 

jj8.  a  trough-shaped  steel  Ix'am  of  the  dimensions  shown  by  Fig.  rA9  has 
to  carry  a  uniformly  distributed  load  of  4')  Ibs./cu.  ft.  If  the  length  of 
the  beam  between  the  supports  is  10  ft.,  find  to 
whut  height  the  load  may  Ik-  raised  so  that  the  ^^^^/V-x' 
stress  developed  may  not  exceed  10,0()0  Ibs./sq.  in.  ;-„-.^,^. 
--jirsi,  assuming  both  ends  free;  second,  assuming 
l>oih  ends  fixed.  .4ns.  126.3  ft.;  189.6  ft.  Fio.  549. 


J29.  In  the  follow ng  cases  determine  the  position  of  the  neutral  axis, 
the  moment  of  n>si8tance.  the  shearing  strength,  the  ratio  of  the  maximum 
to  the  average  intensity  of  shear,  the  cot>fficients  of  strength  jier  .square  inch 
l>.'inc  4i  tons  for  tension  and  compression  and  3J  tons  for  shear: 
(<i)  A  rectangle  2  ins.  wide  and  6  ins.  deep. 

Ans.  .\t  centre;  .M  in. -tons:  28  tons;  3  to  2. 
('')  \  square  with  a  diagonal  vertical,  the  length  of  a  side  iH-iiig  4  in.-tons. 

Ann.  M  centre;  24 vli" in.-tons;  49|  tons;  9  to  8. 
''•)  A  circular  section  4  ins.  in  diameter. 

Anx.  At  centre;  28.2  in.-tons;  .33  tons;  4  to  3. 
(</)  A  regular  hexagonal  sj-eticni  with  a  diameter  (a)  vertical,  (h)  hori- 
zontal, n  being  a  side  of  the  hexagon. 

45        -    I  .'j 
AnK.  (a)  .\t  centre;  ^r.n'VS;  -^a'\%  7  to  5. 
tlJ  4 


4.')  ,    2()2.")V'3 


(/>)  At  centre;  --.a';  ■^' 


o';  1.258. 


16     '      629 

(i)  A  triangular  seetion  6  ins.  deep,  with  a  bajw  6  ins.  wide,  the  sides 
I'lniiequal.  A)ii<.   I  ins.  from  vertex;  4()..">  in.-tims;  42  tons;  3  to  2. 

/I  A  ijouiilc-fee  »c>ction  comi)o.sed  of  a  30"x|"  web  and  four  angle-irons 

•  >'ii.v'x;H"xi". 

.IfM.  At  centre;  1  J2J.6  in.-tons;  34..')3I  tons;  3.1.1."i  to  1. 
((/)  A  section  having  a  semicircular  top  flange  of  8  ins.  e^iternal  diameter 


574 


THEORY  OF  STRUCTURES. 


and  1  in.  thick,  a  web  14  ins.  deep  and  1  in.  thirk,and  a  bottom  flange  8  ins. 
wide  and  1  in.  thick.  .1//^.  G00.:>7  in.-tons;  44.27  ton.s;  .3833. 

(/()  .\  section  havinjt  a  semi-elliptic  top  flange  2  ins.  thick,  the  internal 
major  and  minor  axes  Ix'ing  S  ins.  [horizontalln)  and  4  ins.  (vertically),  respect- 
ively, a  bottom  flange  8  in.s.  wide  and  2  ins.  thick,  and  a  web  10  ins.  di>ep 
and  2  ins.  thick.  Ans.  S4.")  in.-tons;  73.20  tons;  34.23. 

(i)  A  section  having  a  semi-elliptic  top  flange  2  ins.  thick,   the  external 

major  and  minor  axes  being  10  ins.  (Iwrizon- 

lo: »;      ffiii,^)  .,,,,1  c  i„jj    (vrrlimlly) ,  resi)ectivrjy,  a 

trai)cz(ii(lal  web  S  ins.  deep  having  a  width 
of  3  ins.  at  the  top  and  G  ins.  at  the  bot- 
tom, and  a  bottom  semicircular  flange  of  10 
ins.  diameter. 

l».s.  Il41i  in.-tons;  12.0  tons;  .382. 
{'j)  The  sections  shown  by  Figs.  .").")0  and 
.").")1.     What  is  the  ratio  of   the  maximum 
tensile  and  compressive  stresses  in  eaih  section? 

-Wi.v.    1S71  ill.  tons;  rjO;   M;      212  in.-tons;  137;  .89:  75  to  37. 
(/,)  A  traiH'zoidal  scc'.ion.  tlic  top  -idc,  bottom  side,  and  dc.in  h  (inches) 
Ix'ing  in  the  ratio  of  I  to  2  to  4.  Aus.  Ih  from  top  .side;  ,V(i'i' in.-tons. 

(/)  A  section  in  the  form  of  a  rhi'tnbus  of  depth  'Ir  and  with  a  horizotital 
diagonal  of  length  2'>.  .      Anx.  |bc';  '{Ik;  9  to  S. 

(m)  An  angle-iron  2" X 2" Xl". 

Anx.  Neutral  axis  divides  depth  into  s<'gments  of  \\  iii.  and  H  in.; 
;;•';  in.-ton-*:  2.3  tons:  4107  to  l")8t). 
(h)  \  hollow  circular  s<'cti(m  of  external  railius  ('  and  internal  radius  (  '. 
<)()(■« -t-^*    33      V*-V" 4rH-cr+f" 


Fid.  5.JI. 


-lux. 


2S 


(' 


t  C +(•(■' +C'"    3 

in)  A  cruciform  .section  made  up  of  a  flat  steel  bar  10  ins.  by  \  in.  and  four 
steel  angles,  each  4"x4"Xj",  all  riveted  together.  (.Neglect  weakening 
effect  of  rivet -holes.)  .!«.>.■.  7().42.">  in.-tons;   31.7  tons;    2.20S. 

J30.  \  iH-aiu  of  triangular  section  12  ins.  deep  and  with  its  base  horizontal 
can  \wt\T  a  total  shear  of  UK)  tons.  If  the  safe  maximum  intensity  of  shear 
is  4  tons/sq.  in.,  liml  the  width  of  the  ba.se.  Ana.  GJ  ins. 

aji.  In  a  rolle<l  Ix-ani  with  e<iual  flanges,  the  area  of  the  web  is  i)ro{K)r- 
tional  to  the  /(th  |M)wer  of  the  depth.  Find  the  most  e<'onomical  distribu- 
tion of  metal  Ix'tween  the  flanges  and  web,  and  the  moimMit  of  resistance  to 
U'tidiiig  of  the  section  thus  designed.  Also  find  the  ratio  of  the  average  to 
the  maxiiniiin  intensity  of  shear. 

.\ii!<.    .Vrca    of    each    flange:    web    area:  :2h- 1  :ft;    mnmrnt    <ij    rrsistanrc 

~--^'—IAh,  I  Ining  the  coefhcii-nt  of  strength,  A  the  total  wctionnl  an-n, 

2  »i  +  1 

.      ,      ,      ,    ,          ,                 ,    .            (N  +  l)(4w  +  l) 
and  h  the  deptli  of  thi'  wi-l);    rnlu)  o/  shnir.s'^ -,^- . 

aji.  Assuming  that  the  web  atul  flanges  i:(  a  rollcfl  Ix'am  are  rectaiigulftr 
in  sction,  determine  the  ratio  of  the  maximum  to  the  Hveragi"  intensity  of 


EXAMPLES. 


575 


shciir  ill  a  section  from  the  following  data:  the  total  depth  is  ;j- times  the  breadth 

of  each  flange,  n  times  the  thickness  of  each  flange,  and  2n  times  the  thickness 
111"  the  web.  Show  also  that  this  ratio  is  4^  or  V,  according  as  the  area  of 
I  lie  web  is  equal  to  the  joint  area  of  the  two  flanges  or  is  equal  to  the  area  of 
taih  llange.  How  much  of  the  shearing  force  is  borne  by  the  web?  How 
MUK'h  by  the  flange?  .  .       3(n'  +  12n-12)(n +6) 


Ans.  ratio  = 


70'- 


2(/i'-f-18«'-3f);i+24)' 

233.  A  built-up  beam  is  composed  of  two  equal  flanges,  eaoh  consisting 
of  a  (')J"Xi"  plate  connected  to  a  24"  Xi"  ^'pI'.  "P^n  ''  ''»s-  '"  ♦he  middle, 
liy  nii'aiis  of  four  e<jual  angle-irons,  each  3"X3"X  i".  Dotcrniine  the  moment 
of  resistance,  the  maximum  shearing  strength,  the  ratio  of  the  maximum 
to  the  average  intensity  of  longitudinal  shear,  and  the  intensity  of  longitudinal 
shear  12  ins.  from  the  neutral  axis,  G  tons/scj.  in.  Ix'ing  coefficient  of  strength. 

Am.   1391i  in.-tons;  09.3  tons;  2.3:  .146  ton/sq.  in. 

234.  Find  the  moment  of  resistance  to  l)ending,  the  resistance  to  shear, 
.■inij  the  ratio  of  maximum  to  the  average  intensity  of  a  shear  in  the  case  of 
a  section  consisting  of  two  equal  flange^  each  composed  of  a  pair  of  .">"X3i"X  |" 
aticle-irons  rveted  to  a  31i"Xi"  ?b,  the  .5-in.  sides  of  the  angles  being 
iioi'izoiital  and  4J  toasAsq.  in.  Iwing  the  coefficient  of  strength. 

Ans.  1501.06  in.-tons;  46.21  tons;  3.0.')8. 

235-  The  floor-l)eam  for  a  single-track  bridge  is  1.5  ft.  between  bearings, 
.111(1  each  of  its  flanges  is  comiwsed  of  a  pair  of  2}"X2f"Xi''  angle-iroiis 
liveted  to  a  30" Xj"  web.  The  uniformly  distributed  load  (including  weight 
of  iH.iin)  upon  the  beam  is  4200  lbs.,  and  a  weight  of  1600  lbs.  is  concentrated 
at  eaeli  '  *■  the  rail-crossings,  2}  ft.  from  the  centre.  Find  (a)  the  maximum 
tlaMi;e  stress,  {h)  the  ratio  of  the  maiimum  and  average  intensities  of  shear; 
.1  the  stifTness,  E  being  27,000,000  lbs. 

Ans.  (n)  1103.7  lbs.;  (h)  2.039;  (r)  .000O..63. 

236.  .\  lattice-girder  of  100  ft.  sjian  carries  SO  tons  uniformly  distributed; 
tlic  (lirder  's  10  ft.  deep  and  the  .safe  working  stress  is  4  tons/sq.  in.  If  the 
niltii  of  the  flange  must  be  20  ins.  lo  carry  th?  load  exclusive  of  the  weight 
oi  the  ginler,  what  must  Ik;  the  wi(,th  of  the  flange  when  the  weight  of  the 
girder  is  taken  into  account?  Ans.  23  ins. 

237-  A  plate  (tinier  of  double-tee  section  and  of  80  ft.  span  is  8  ft.  dwp 
■iiicj  c.irrics  a  uniformly  di.stributeil  load  of  80  t(ms.  If  the  width  of  the  flange 
imi-i  lie  12  ins.  to  carry  the  load  exclusive  of  the  weight  of  the  girder,  what 
iiiii^l  the  width  Im-  when  this  weight  is  taken  into  account?       Ans.   13J  ins. 

238.  Design  the  central  s«>ction  of  a  plate  girder  of  4.")  ft.  span  and  5  ft. 
iIk  p  to  carry  a  dead  load  of  .500  Wis./h.  run,  a  live  load  of  3200  Ibs./ft.  run 
Hi.i  ail  inii)act  load  of  2400  Ibs./ft.  run,  and  also  determine  the  length  «  f  the 
'I  I'll-"'  I'lates.  Ans.  23.6  ft. 

239.  .V  tubular  girder  rents  Ufxin  sup|w>rt«  .36  ft.  apart.  At  6  ft.  from 
oil"  iiiil  the  flannes  nm  each  27  ins.  wide  and  2|  ins.  thick,  the  not  area  of  the 
'■  iiioii  llange  Ix'iiig  (K)  ins.,  while  the  web  consists  of  two  A-in.  plates,  30  ins. 
dip  and  IN  me.  apart.     Neglecting  the  effect  of  the  angle-imns  uniting  the 


676 


THEORY  OF  STRUCTURES. 


wol)  plates  to  the  flanges,  determine  the  moment  of  resistance.  The  girder 
has  to  carry  a  uniformly  distributed  dead  load  of  ofi  tons,  a  uniformly  dis- 
trihutod  live  load  of  .54  tons,  and  a  local  load  at  the  given  section  of  100  tons. 
What  are  the  corresponding  llange  stresses  |)er  square  inch?  How  many 
;-iii.  rivets  are  required  at  a  given  si-ction  to  unite  the  angle-irons  to  the 
flanges? 

Aitx.  23S.13Xc.K<fliciont  of  strength;  3..31S(>  tons;  3.S06tons;  six,  f,  being 
4  tons/sq.  in. 

J40.  A  Ix-am  of  rectangular  .section,  of  breadth  h  and  depth  J,  is  actt>d 
upon  by  a  coujile  in  a  plane  inclined  at  4.">°  to  the  axis  of  the  section.  Com- 
pare the  moment  of  resistance  to  bending  with  that  about  either  axis. 

bV-2     dV2 
'1»«-   b+d'T^- 

241.  If  the  plane  of  Ix-nding  does  not  coincide  with  the  plane  of  symmetry 
of  a  bcaiUj  show  that  the  neutral  axis  is  parallel  to  a  line  joining  the  centres 
of  two  <'ircles  into  which  the  Ix'am  would  l)e  iM-nt  by  two  comiM)nent  couples 
whose  axes  are  the  principal  axes  of  inertia  of  the  section,  each  couple  being 
suftposfd  to  act  alone. 

242.  If  a  spirn'  spring  is  fastened  to  the  barrel  so  that  there  is  no  change 
of  direction  relatively  to  the  barrel,  show  that  the  tendency  to  unwind  is  directly 
jiroportional  to  the  amount  of  winding  up.     (Condition  of  jHTfect  isochronisra.) 

243.  .\  metal  l)eam  is  subjected  to  the  action  of  a  In-tiding  moment  steadily 
appiii'd  hcijoiiil  the  elastic  limit.  .\s,suming  that  the  metal  acts  as  if  it  were 
{x-rfecily  |)lastic,  i.e.,  so  that  the  stress  throughout  a  transverse  section  is 
viiijorm,  compare  the  moment  of  resistance  to  In'mling  of  a  section  of  the 
Ix-ain  with  the  moment  on  the  assumption  that  the  met;d  continued  to  fulfil 
the  onliiiary  laws  of  elasticity,  (n)  the  section  in-ing  a  rectangle;  (h)  the  si'c- 
tioii  tM'iiiR  a  circle.  Am.  (<i)  3  to  2,  (h)  .V)  to  33. 

244.  The  neutral  axis  of  a  symmetrically  loaded  girder  whose  moment 
of  inertia  is  constant  as.sumes  the  form  of  an  elliptic  or  circular  arc.  Show 
that  the  biiding  moment  at  any  |)oint  of  the  deflected  girder  is  inversely 
pro|M)rtionai  to  the  cuIk-  of  the  vertii'al  distance  In'tween  the  |M)int  and  the 
centre  of  the  ellipse  or  circle. 

245.  The  llange  ( f  a  girder  eonsi.sts  of  a  j>air  of  angle-irons  and  of  a  ])late 
which  extends  over  the  niiddli'  iH)rtion  of  the  girder  for  a  certain  required 
distance.  Show  that  the  greatest  economy  of  material  is  securt>d  when  the 
I  iiL'ili  of  the  plate  is  two  thirds  of  the  span  and  the  sectional  areas  of  the 
plat,'  and  angle-irons  arc  as  4  to  .'>,  the  girder  Ix-ing  uniformly  loaded. 

246.  The  llange  of  a  uniforndy  loaded  girilor  is  to  consist  of  two  plates, 
eaih  of  which  extends  over  the  middle  |«)rtioii  of  the  girder  for  a  certain  rt- 
<linrcd  distance,  and  of  a  pair  of  angle-irons.  Show  that  the  gn-atest  economy 
of  material  is  realizeil  when  the  lengths  of  the  plates  and  aiigic-irons  arc  m 
the  ratio  of  1-' :  1H:'.':J.  and  when  the  areas  of  the  plates  aic  in  the  ratio  of  I:'). 
What  should  l)e  the  relative  Icngthn  of  the  plates  if  they  are  of  equal  are.-i.' 

.li/,i.  nx-.'-nv'iF+l). 


EXA^TPLES. 


577 


247.  Determine  the  moment  of  resistance  to  bending  of  a  section  of  a 
l»nn  >n  whu-h  the  top  flanRe  is  compos,.,!  of  two  340-mm.X12-mm  plates 
:.n.l  ..>r  .  40-mm  X  10-mm.  plat,.,  and  the  l.,ttom  flange  of  „„.  340-mm  v 
nn„„.  plate  and  ,.«r  84()-„.m.XS-mm.  plate,  the  flanges  being  riveted  to  a 

tV""""  r  '''""- ''3-""^«"«  "f  f"""-  100-mm.XlOO-mm.XS-mm.  anHe- 
irons.  The  c.oeffi,.K.„t  of  strength=G  k./mm.'.  What  will  the  nunnonC o( 
n.,>tance  be  if  Ihrre  400-mm.Xlo-mm.  plates  are  substitute,!  f,.r  the  ton 
iLu.fr..  and  of,e  400-mm.X  lo-m,„.  plate  is  substituted  for  the  b,.tt.,m  flanged 

,,»    p.  ^    .  ,,  ,  ^«s-  02.92  k.ni.;  10(1.3.-)  k.m. 

248.  Iloor-beams  4.4  m.  In-twc-n  l)earings  an,!  s,.ace,!  2..-.1S  m  centre 
...  .-..Mtre  have  a  section  compo.sed  of  two  eq.ial  flanges,  ea,.h  consisting  „f  two 
s..-.Mrn.X.S.,-mm.X12-mm.  angle-irons  rivetc.l  to  a  4!K)-n>m  ■  7  ,„„,  VT 
A  ucght  ,.f  15()  k.  (,!u,.  to  longitu.linals)  an,!  a  weight  of  l.-,0  k.  (,lu,.  „;  rails' 
"  •  .  '.<■•'  300  k.  m  all,  are  ,.on,.entrat,.,l  at  tl„-  raiI-,Tossings,  and  the  tics 

mv  ...Is,,  t,.  ,.arry  a  un,f,.rn.ly  distributed  loa.l  of  400  k.  ,!ue  to  weight  of 
ln--!.,.am,  4(K)0  k.  ,lue  to  weight  of  platform,  and  4000  k./sq.  „,.  „f  puf,f,rm 
>1  ..■  ..y"-""Aload.  Find  the  moment  of  resista.u-e  to  ben,!ing  and  the  n.axT 
""""  "•"!«"  '"ttM..s.t,es  of  stress.  An...  4.9  k.m. ;  iTTSr.  k  /mm  " 

249.  h;a.h  ,.f  the  flanK,.s  ,.f  a  ginler  is  a  :J.->()-mm.X  lO-mm.  plate  an,!  is 
ny.N.l  to  a  l.S-m.XS-mm.  w.-b  by  m.-ans  of  two  l(X)-mm.  i  lOO-mm  X 
l-nuM  angle-rons  ■  Vt,.nnine  the  moment  of  resistan.-e  to  iK-n-ling,  the  coeffi- 
.-..  .  .strength  bc,-,„  fi  k./s„.  n„n..  („)  ,li.sr,.gar.li,.g  ,:,e  weakening  eC 
';t  nvctmg;  (h)  assum.ng  that  the  flang,.  plat,.s  are  rivete,!  to  the  anWcs  l.v 
-"-;'"•  ^'-t^-  .     ^  -1".^.  (.V)   10,S.C.0104  k.m.;  (h)  lOO.S-m. 

^50.  Ih,.  ,.ross-tie  ..r  a  single-track  bridge  is  4.1  m.  Ix^tween  Ix-arings 
..^  ..Hige  of  the  rails  Ix-ing  1..51  m.;  each  of  the  flang,.s  is  <.,.mpo.s,.,!  ..f^,' 
1  s-:nn,.  <S-„.m.  plat,-  rivet,-,!  to  a  .".O-mm.  X  S-mm.  web  bv  n„.ans  ,.f  two 
.0-nm..xa)-mm.X!)-mm.  angle-irons;  a  l„n,l  of  2<Mi  k.  (weight  of  rails  et,-  > 
1-  .-Hr,.,.,rat..,l  at  ea,-h  rail-.n,ssir.g.  What  uniformly  distribute,!  i,.a,!  will 
'  i-  .s.f,.ly  Ijear  the  metal's  ,.oefr..-ient  of  strength  lK..ng  «  k./.s,,.  mm  ' 
Ih.    I.;a.!  a,.tually  distribufd  over  the  tie  is  1!),7.S2  k.     Fin,!  the  maxinuini 

""'""V:^  ''''''\       .  -'"■^■-  -'-».!<>-'  k.;  4.04  k./s,,.  mm 

^V    l),.s,gn  a  longitu,linal  of  .4.-;  m.  ,lepth  whi.h  is  to  !k.  supmrt,.,!., 

in^rv.Mls  „f   {..}  m.  and  to  carry  at  its  mi.hlle  ,H,int  a  w,.ight  of  7(J00  k      1.. 

""I.iiicnt  (.f  strength  U'ing  .">  k.  s,).  mm.  ''  '    " 

1  •■>.  /  --^W.S7.^,  and  the  /  of  a  .s,.,.,!,,,,  with  tw..  equal  flanges,  eH,.h  com- 

«  '  1 .  !^''ivT5.ir,  """"■  ""'■"""•  ""•''•-''■""''  "^■''^«»  t"  "  ■»-^»-"'n'-  X  8-mm. 

-•Si.  A  longitu.linal  2..54,S  m.  lK.tw,.en  In-arings  ,.onsists  of  two  equal  flanges 

'-  I   .nn,,»,s,.,l   „f  two    70-mm.x7()-mm.xO-n.m.    angle-irons   riv,., ,,   i 

■•'Mnn,.  ..-mm    w,.b  ,.late.     Fin,!   the   flang,.   intens.tv   „f   stress   under   a 
!"      'iiiini  l„ad  <if  7000  k.  at  tli,.  .elitre.  " 

•'"■"•  /".IK)01H!)2S4.-)0S:  stre,ss~:,.r.  k  /mm  ' 
-'M.  A  ,ross-f.e  resting  uiM.n  sup|«.rts  at  the  en,ls  and  2.2(i  m    iH-tw.vn 
I"  -li.gs  ,.  ,.,.m,K..sed  of  two  ,.,,ual  flanges,  ..onsistins  of  tw<,  7(U„,t„  v7,Kn,m 
■  •-",„>.  angle-irons  riveted  at  the  to,,  to  «  4.'iO-mm.x7-mm.  web  plate  and 


578 


THEORY  OF  STRUCTURES 


at  the  bottom  to  a  300-mm.X7-mm.  web  plate,  the  interval  between  the 
web  plates,  which  is  oix-n,  being  2.53  m.;  the  tie  is  designed  to  carry 
a  uniformly  distributed  load  of  G7«  k./lineal  metre  of  its  length,  and 
aWo  a  load"  of  n,(144.S  k.  at  each  of  the  i^ints  distant  ..'j;.-)  m.  from  the 
bearing.     Find  the   position  of   the  neutral  axis  and  the  tniixunum  flange 

A„s.  l.olf)  m.  from  top  ilange;  /  =  .0231.ft4.')fi4198;  maximum  B.M. 
-=4815.8101  k.m.;  maximum  tensile  stress  =.37  k./mm.';  ma.ximum  com- 
pressive stress  =  .314  k./nmi.'. 

254.  Find  the  maximum  concentrated  load  on  a  cross-tie  for  a  smgle  track 
due  to  a  six-wheel  locomotive,  the  wheels  l)eing  2.3  m.  centre  to  oentre,  the 
ties  l)eing  3.2  m.  centre  to  centre  and  the  weight  on  each  wheel  being  7000  k. 

Alts.  10,937.5k. 

255.  The  floor-beams  for  a  double-track  bridge  are  8.3  m.  between  bearings 
and  are  spaced  2..J8  m.  centre  to  centre.  The  distance,  centre  to  centre, 
between  track  rails  is  1.5  m.,  and  between  in.side  rails  is  2  m.;  the  tie  has 
equal  flanges,  each  consisting  of  two  TO-mm.  X 90-mm.  X 0-mm.  angle-irons 
riveted  to  a  <X)0-mm.X7-mm.  web;  the  maximum  live  load  uiKin  the  tie  is 
that  due  to  a  weight  of  3000  k.  tipon  each  of  the  six  wheels  of  two  locomotives, 
the  wheels  boing  2.4  m.  centre  to  centre.  If  the  coefficient  of  working  strength 
is  5*  k  /so   mm.,  what  uniformly  distributed  load  will  the  tie  carry? 

'    ^    ^  .4  ns.  2298  k. 

256.  Determine  the  safe  value  of  the  moment  of  inertia  (/)  of  a  cross- 
tie  for  a  double-track  bridge,  the  length  of  the  tie  between  l«>aring8  b'ing 
7  624  m.,  its  depth  .0  m.  the  gauge  of  the  rails  1.5  m.,  the  distance  between 
inside  rails  2  m.  The  uniformly  ilistributed  load  uiwn  a  tie  consists  of  mi 
k  '/s...  m.  due  to  platform,  etc.,  and  of  IS(K)  k.  due  to  weight  of  tie;  the  ties 
are  3  584  m.  centre  to  .entre,  the  live  loa.l  is  that  due  to  a  weight  of  7000  k. 
upon  each  of  the  centre  wheels  of  a  six-wheel  locomotive  an.l  a  weight  of 
ryjOO  k.  uix)n  each  of  the  front  and  rear  wlu-els,  the  wheels  being  2.4  m.  centre 
to  centre;   the  safe  coefficient  of  stn-ngth  =<)  k./sq.  mm. 

257.  The  cros«-ties  of  n  single-track  bridge  consist  of  two  equal  flanges, 
Pnch  coini-oscd  of  two  7(Mnnr.  >  70-n.m.  ,<<Mnm.  angle-irons  riveted  to  a 
6.-i()-n.ni.  X  7-min.  web:  the  tics  an  ».l  m.  long,  and  each  carries  1!),14(.  k. 
(viz  384  k.  I'>r  tl<s.  27<12  k.  I<>r  phtjorm,  ami  1(1,0(X)  k.  jor  proof  loiul)  uiii- 
f<.rI,'i^•  di-<tril).it<Ml  and  (135  k.  (./»«  /-  lonuilmllwiU,  mils,  ,7c.)  concentrated 
at  e.:ch  rail-crossing,  i.e.,  at  7.".5  mm.  from  the  middle  point.  .\s,suming  that 
the  cross-ties  an'  merely  supportcl  at  the  ends,  find  the  maximum  imensity 
of  stress. 

.\„.    .5.7724  k./mm.';   -     .0018423.      The   fixture   of    the   en.Ls   approxi- 

inalclv  d<nil)les  tin-  strength. 

258.  Till'  loniriindiiiMls  of  the  bridge  in  the  last  ("iample  con^s'st  of  two 
pairs  nf  T(Mn!n.  ^T(l-iiitii.  ■  '.(-mm.  anulc-irons  riveted  to  a  4-ni.x7-mni.  wch; 
the  cross-ties  an-  3.2  in.  centre  to  centre.  Ih-termine  the  maximum  intensity 
of  stress  due  to  a  loa.l  of  7000  k.  c.ncent rated  on  the  longitudinal  half-w.'iy 


EXAMPLES. 


579 


lictwcen  the  cross-ties,  assuming  that  it   is  an   independent  girder.    What 
ivduld  the  stress  be  if  the  ties  were  3  m.  centre  to  centre? 

Ans.  -=.29231706;  .192  k./mm.';  .18  k./inm.' 

250.  The  section  for  the  Estressol  bridge  cross-ties  is  the  same  as  that 

for  tlio  (irande  Baise  (Ex.  257)  bridge  ties;   the  load  at  each  rail-crossing  i.s 

jili.-)  k.,  and  the  uniformly  distributed  load  is  18,0(52  k.     Find  the  maximum 

iiitiiisity  of  stress  in  the  flanges,  assuming  that  the  ties  are  merely  supported 

^^  "''■  ^'"'l''  .4ns.  5.26  k./mm.'. 

260.  The  area  of  the  compression  flange  of  a  cast-iron  beam  is  17  sq.  ini .; 
the  thickness  of  the  web  is  a  certain  fraction  of  the  depth;  the  unit  stresses 
lire  ill  the  ratio  of  2  to  5.     Find  the  areas  of  web  and  tension  flange  which 


will  give  a  .section  of  maximum  strength. 


.4ns.  12  8q.  ins.;  3.2  sq.  ins. 


261.  Determine  suitable  dimensions  for  a  ca.st-iron  beam  20  ins.  deep 
at  a  section  subjected  to  a  bending  moment  of  1200  in.-tons,  the  coefficients 
of  strength  per  square  inch  being  2  tons  for  tension  and  8  tons  for  compression. 
Take  thickness  of  web=/j  in. 

.his.  Sectional  area  of  tension  flange  =.36  sq.  ins.;  of  compression  flange. 
=  2J  sq.  ins. 

262.  In  a  cast-iron  beam  the  area  of  the  web  is  one  half  the  area  of  the 
tc;  sion  flange,  the  depth  of  the  beam  is  9  ins.,  and  the  unit  stresses  are  2  tons 
[XT  s(juare  inch  in  tension  and  4  tons  per  square  inch  in  compression.  The 
maximum  moment  of  resistance  is  162  in.-tons.  Find  (a)  the  flange  and  web 
anas,  (6)  the  length  of  the  beam  so  that  its  stiffness  might  not  exceed  .001, 
(f)  the  net  weight  on  the  beam,  and  (d)  the  work  of  flexure,  E  being  7500* 
tons. 

Ans.  (n)  9,  3i,  and  4i  sq.  ins.;  (6)  9  ft.;  (r)  11.75  tons;  .41472  in.-tons. 

263.  \  cast-iron  girder  of  20  ft.  span  has  a  top  flange  of  4"Xli"  a  lx)ttom 
llar,.z..  of  12"X  U".  and  a  web  of  16"X  H".  Find  the  position  of  the  neutral 
axN.  If  the  maximum  tensile  stre.s,s  is  2000  Ibs./sq.  in.,  find  the  uniformly 
'iMiil.utcd  load  winch  the  girder  may  carry  and  the  maximum  compressive 
'""'''■'•  Ans.  lli^  ins.  from  top;  3190  lbs. 

264.  A  cast-iron  girder  with  a  4"X1"  upper  flange,  a  25"X1"  lower  flange 
hmI  a  2()"X1"  web  must  not  carry  a  greater  loa.l  than  will  develope  tensile 
■I'l.l  .  niiipressive  stres.^s  of  2000  and  mX)  lbs  sq.  in.  respectively.  D«.tcr- 
rnii.n  the  moment  of  resistance  of  the  section.  Ans.  933,.334  in. -lbs 

265.  A  ca-.-iron  girder  of  25  ft.  span  has  n  bottom  flange  of  36  sq.  ins. 
-'-iinii;,!  area.  Find  tlie  most  economic  arraniremciit  of  material  for  the 
V'  lp  in.l  toj)  flange  which  will  enable  the  Ik'.-uii  to  carrv  a  loa.l  of  IS.'HK)  llw 
.11  id  ft.  from  one  end,  the  tensile  and  conii)ressive  wi)rkitig  .strengths  Ix'inp 
-''"HI  and  Mm  Ibs./s.).  in.  res[)t.etively.  Assume  that  th..  thickness  of  the 
».  I,  i,  ;i  fraction  of  its  depth. 

l"v.   l).|,th-21l    ins.;    area   of   web=2S.S  sq.   ins.;    area  of  top  flaneo 
I  iii  s(|,  HIS. 

266     \   .louble-flunired  cast  .iron  girder  hn=  a  Hr-ctioi,a!  arr.i  of  j  sq    ins 
'li"  w.l,  ,s  I  in.  thick  an.l  21  ins.  .leep;   tli.-  moment  of  resistance  of  the  section 
I'  l(M),!t..0  ft.-ll)s.;   the  coefficients  of  .strength  are  2100  lbs.A.q.  j,,.  in  ten.sion 


Ml 


mf^ 


580 


THEORY  OF  STRUCTURES. 


and  5250  lbs.  in  compression.     Find  x,  the  position  of  the  neutral  axis,  and 
the  areas  of  the  two  flanges.  Ans.  a,  =2'J,Ve  sq.  ins.;  a^  =5Ji!  sq.  ins. 

267.  Find  the  safe  distributed  load  for  a  east-iron  Ix-am  of  the  following 
dimensions:  Top  lianne,  3"Xl";  bottom  llanfje,  8"Xl.r,";  web,  1.2.">  ins. 
thick;  total  depth,  10  ins.;  with  (n)  the  bottom  llange  in  tension,  {b)  when 
inverted;  span,  12  ft.;  skin  stress,  ;«KX)  lbs./s(i.  in.  Also  find  the  safe  central 
loa<ls  for  a  stress  of  IKKK)  lbs./s(|.  in.,  including  the  stress  due  to  the  weight 
of  the  beam.     The  beam  is  of  constant  cross-section. 

Am.  («)  0.414  and  2.98  tons;  (h)  3..-)42  and  1.414  tons. 

268.  Determine  the  thickness  of  the  metal  in  a  cast-iron  beam  of  12  ft. 
span  and  S  ins.  deeji  which  has  to  carry  a  uniformly  distributed  load  of  4000 
ll)s.,  the  section  beinu;  !(i)  a  hollow  s(|uare;  (!>)  a  circular  amndus.  The  coeffi- 
cient of  workinj:  stren-rth  =:5(MM)  Ibs./scj.  in.  -Mso  find  the  limiting  safe  span 
of  tiie  beam  under  its  own  wcisrht. 

Alls.  Ncfilecting  weight  of  beam,  (<i)  .111?  in.:  ('-)  .3.');},S  in.  Taking  weight 
of  beam  into  account,  (a)  .'M'J  in.;  (/»)  .727  in.  Limiting  span  =39.527  ft.  iu 
(it)  and  =32.09  ft.  in  ('-). 

200.  The  efTective  length  and  de|)th  of  a  cast-iron  girder  which  failed 
under  a  load  of  IS  tons  at  the  centre  were  57  ins.  and  51  ins.  resiK!ctively ;  the 
top  flange  was  2.33  ins.  by  .31  in.,  the  Ijottom  flange  0.07  ins.  by  .66  ins.,  and 
the  web  was  .200  in.  thick.  .\ssuming  that  the  ordi:iary  theory  of  flexure 
held  good,  what  were  the  maximum  intensities  of  stress  in  the  flanges  at  the 
point  of  ru|>ture?  Ans.  jt  -12.30  tons/sq.  in.;  /c='44.9  tons/sq.  in. 

270.  .\  double-flanged  cast-iron  girder  14  ins.  deep  and  20  ft.  l>efween 
supports  carries  a  uniforndy  distributed  load  of  20  tons.  Find  suitable  dimen- 
sions for  the  section,  the  tensile  and  compressive  inch-stre.sses  Ijeing  2  tons 
and  5  tons  resiH-ctively.  .Mso  find  the  stijjnrss  of  the  b<'am,  E  being  SOOO 
tons. 

.\iis.  Let  thickness  of  web  =1  in.;  rt,  =22J j  sq.  ins. ;  a,  =4,Vj  .sq.  ins. ;  stiff- 
ness -  1  :-  MO. 

271.  .\  cast-iron  beam  with  a  30"Xl"  web  is  subjected  to  a  lK;ndiiig 
monient  of  21(H)  in. -tons,  the  coefn'-iciit  of  strength  Iw-ing  2  tons  in  teiision 
and  S  tons  in  conipression.  Find  the  |)osition  of  the  neutral  axis  and  the 
areas  of  the  tension  and  compression  flanges. 

.\iis.  0  ins.  from  t  '>t  of  web;  1 }  sq.  ins.,  .')0  sq.  ins. 

272.  The  central  section  of  a  cast-iron  ginler  is  lOJ  ins.  dee]>:  its  web  area 
is  fur  timo  tiie  area  of  the  top  flange,  and  the  moment  of  resisfam-e  of  the 
section  is  3i>0,(KKJ  in.-l!»s.;  the  tensile  and  compressive  intensities  of  stress 
are  3000  and  7500  lbs./'.s(i.  in.  i 's|M'clively.  Find  the  span  and  load  .so 
that  the  girder  may  have  a  .v///J'«..s.s -.(M)l ,  E  iM-ing  17,IKK),0(K)  lbs. 

Ans.  (I,  !_',.,  sc|.  i'ls.i  rtj-'Ul  sq.  ins.;  Oj  +  H,=9i'3  sij.ins.;  (<pan  =  lt)3.2 
ins.;    uniformly  distributed  load     17,047  lbs. 

273.  Determine  suitable  diiiieii-idii-i  for  .'i  cast-iron  girder  of  20  ft.  spaa 
and  24  ins.  deep,  cjirrying  a  load  of  30,<K10  lbs.  ut  the  <'entre,  the  coefficieiiis 
of  -vorking  stnngdi  in  teti'^in!!  ■,\;]t\  insup.rcKuidti  fwing  n-sjiectively  20'K)  :i!i'i 


5000  Ibs./gq.  in. 


.1//.'-'.  (J,  =  'i;''  .sq.  ins.;  .r~'!*  K(|.  ins.;  (»,  =  V  sq- 


274.  The  dimensions  of  the  section  of  a  cast-iron  girder  are  the  following; 


EXAMPLES. 


581 


Top  flange,  G"X2";  lx>ttom  flange,  12"x3";  web,  16"x2".  Determine  the 
luisition  of  the  neutral  axis,  the  moment  of  resistance,  and  the  maximum  com- 
pnssive  streaa,  the  maximum  tensile  stress  being  6  tons/sq.  in. 

Any.  12.925  ins.  from  top. 

275-  Determine  the  dimensions  of  a  cast-iron  beam  at  a  section  whose 
moment  of  resistance  is  800  in.-tons  and  whose  depth  is  18  ins.,  taking 
2  tons/sq.  in.  as  the  maximum  tensile  intensity  of  stress. 

Anx.  a,  =VV-  sq.  ins.;  yl'  =  W  sq.  ins.;  o,-S?  sq.  ins. 

276.  A  4"X4"Xi"  inverted  T  section  is  used  as  a  beam  for  a  span  01 
I-'  ft.,  the  uniformly  distributed  load  on  the  flange  being  O.G  ton.  Find  the 
miiximum  tensile  and  compressive  skin  stresses. 

Ans.  4.343  tons  (comp.),  1.911  tons  (tens.). 

277-  A  sample  cast-iron  girder  for  the  Waterloo  Corn  Warehouses,  Liver- 
pool, 20  ft.  7}  ins.  in  length  and  21  ins.  in  depth  (total)  at  the  centre,  wr.s  p!-  •(  rl 
upon  supports  18  ft.  IJ  ins.  apart,  and  tested  under  a  uniformly  distribui;\l 
load.  The  top  flange  was  5"Xl}",  the  bottom  flange  was  18"X2  ',  and  t:. ; 
wtl)  was  H  ins.  thick.  The  girder  deflected  .15  in.,  .2  in.,  .25  in.,  and  .28  in. 
under  loads  (including  weight  of  girder)  of  03,7^3  lbs.,  88,571  lbs.,  107,468  Ib.s.i 
and  119,746  lbs.,  respectively,  and  broke  during  a  sharp  frost  under  a  load 
of  :i90,282  lbs.  Find  the  mean  coefliicient  of  elasticity  and  the  central  flange 
stresses  at  the  moment  of  rupture. 

Am.  /-3309.122;  x.^  17,279,567  lbs.;  20,121  lbs.,  47,168  lbs. 

278.  The  eflfective  length  and  central  depth  of  a  cast-iron  girder  resting 
upon  two  supports  were  resjwctively  11  ft.  7  ins.  and  10  ins.;  the  bottom  flange 
\va<  10  ins.  wide  and  U  ins.  thick ;  the  top  flange  was  2J  ins.  wide  and  J  in.  thick ; 
til'  tiiickness  of  the  web  was  %  in.  The  girder  was  tested  by  being  loaded  at 
points  :?J  ft.  from  each  end,  and  failed  when  the  load  at  each  point  w.is  17J 
tons.  What  were  the  total  central  flange  stresses  at  the  moment  of  rupture? 
What  was  the  central  deflection  when  the  load  at  each  point  was  7J  tons? 
(/•;  -1S,0(J0,000  lbs.,  and  the  weight  of  the  girder  =3.368  lbs.) 

Ans.  203,918  lbs.;  .388  in. 

27Q.  The  dimensions  «.  the  .section  of  a  rast-iron  girder  are  the  following- 
Top  flange,  4"xli";  bottom  flange,  12"xlf' ;  web,  16"xlJ".  Determine 
till    position  of  the  C.  of  G.  and  calculate  the  moment  of  inertia  of  the 

linn.  Find  also  the  moment  of  resistance,  the  greatest  i)ermissib!e  ten- 
y\v  -ircss  I)oing  21  tons/sq.  in.  If  the  gi-der  is  20  ft.  long,  and  is  supimrted  at 
it-  two  ends,  find  also  the  greatest  safe  load  which  it  will  carry  when  uniformly 
ilistrihutcd  along  its  length. 

Aii.'i.  7i  ins.  from  top;  22S0i  in.-units;  800  In.-tons;  26J  tons. 

i8o.  .\  continuous  beam  of  four  equal  si)ans  carries  a  uniformly  distributed 
"  •!  of  ,/•  intensity  jier  unit  of  length.  The  second  support  is"  depres.sed  a 
■•■i,,in  distance  ,1  Ih-1ow  the  horizontal,  and  the  reaction  at  the  second  s>ip- 
■  li  IS  twice  that  at  the  first.  Show  that  the  reactions  at  the  first,  second, 
'innl.  fourth,  and  fifth  .supjMjrts  are  in  the  ratio  of  the  numbers  15  30  36* 
!l.  nu,\  l.<;  find  ri.  With  this  same  v.thic  of  ,/  fiml  the  reactions  when' one 
<-!i'l  is  I'txal. 

1    U'l* 

■'"'•  '^^4sYl''    ^■*'*''"'''  2S-'>'-«',  ^-uiH.   lo:.,,-,  -.-Slr-r/,  where  --388. 


ii:i 


582 


THEORY  OF  STIiUCTVHES. 


281.  A  rontinuous  RinliT  of  tliroo  spans,  the  sido  spans  Iwirip  equal,  rarries 
a  uniformly  distrihulcd  loa<l  of  intensity  tr.  Find  the  error  in  the  H.M.  at 
the  intermediate  support,  if  an  error  J  is  made  in  the  lenjith  {n)  of  the  centre 
span,  (/»)  of  a  side  span.  If  the  three  .sj)ans  are  eaeh  of  length  /,  show  that 
this  error  is  .09»'-'/  in  the  first  case  and  Alu-Jl  in  the  second  case. 


282.  A  wrought -iron  pirder  of  I  section,  2  ft.  deep,  with  fianpes  of  equal 
area  and  having  their  joint  area  e(iual  to  that  of  the  web,  vi/..,  48  sq.  ins., 
carries  \  ton  per  liuval  foot,  is  100  ft.  loiifi,  consists  of  five  equal  spans,  and 
is  continuous  over  six  suppi>rts.  Find  the  reactions  wlu'ii  the  third  sui)ix)rt 
i.s  lowered  i  in.  How  much  must  this  support  he  lowered  so  that  the  reaction 
may  he  nil  at  (n)  the  first  support,  (b)  the  third;  (c)  the  fifth?  How  much 
must  the  support  Ik>  raised  so  that  the  reaction  may  l)e  nil  at  (d)  the  second, 

(c)  the  fourth,  and  (/)  the  sixth  supinm?     (A' =  l(v)00  tons.) 

Ans.  «,=•_'■■;;;  «,  =  l.-)55;  ]{,^:i{,;  /^  =  14i?;  If,=^,\;  R^='4-;  tone,  (a) 
IJins.;  (/.)   jriin.:  (<•;   2:1  ins.;    (,/)   l^'A  ins.;  (c)   UfTins.;  (/)  Oi  ins. 

283.  Each  of  the  main  girders  of  the  Torksey  bridge  is  continuous  and 
consists  of  two  e(|ual  spans,  each  l.'{0  ft.  long.  The  girders  are  douhle-weblx-d; 
the  thickness  of  each  web  |)late  is  J  in.  at  the  centre  and  |  in.  at  the  abut- 
ments and  centre  pier;  tiie  total  de])thof  the  girders  is  10  ft.,  and  the  depth  from 
centre  to  centre  of  the  flanges  is  'J  ft.  4J  ins.  Find  (a)  the  rea(ttions  at  the 
supports,  and  also  (b)  the  points  of  inflection,  when  200  tons  of  live  load  cover 
one  span,  tile  total  dead  load  ujum  ea<'h  span  being  ISO  tons  uniformly  dis- 
tributed. The  top  flange  is  cellular;  its  (/rDsa  sectional  area  at  the  centre 
of  each  span  is  ,'>]  sq.  ins.,  and  the  corresponding  nd  sectional  area  of  the  bottom 
flan:;!'  is  .")■")  s(i.  ins.    Determine  (c)  the  flange  stresses  in  tons  |)er  square  inch,  and 

(d)  the  position  of  the  neutral  axis.  (/ =372,.")(K).)  .-M.so  (c)  determine  the 
reactions  when,  fust,  li  and,  mrimd,  C  are  lowered  1  in.  (/{"  =  10,900  tons, 
and  dejith  of  \o\\  flange  =  1 1. SIS  ::i.s.) 

Aiix.  (n)  l.").j,  3r)0  and  o.')  tons;  (//)  ItXit's  and  7'.)J  ft.  from  end  supports; 
(( )  (j.7  and  7.3  in  loaded  span,  1.13  and  1.22  in  unloaded  span;  {d)  oS.3 
ins.  from  centre  line  of  top  flange;  (r),  firnt,  l.").5.41."),  349.17,  and  55.41.5 
tons;  s(riinit,  l.")4.793,  30.").414,  .j4.793  tons. 

284.  .\  continuous  girder  AliVDE  of  four  sjians,  each  of  length  /,  rests 
upon  sup|M)rts  at  .4,  li,  (',  I),  ami  K,  and  carrii  s  a  uniformly  distribut:>d  load 
of  intensity  w.  Hy  how  much  mast  the  supports  at  Ji,  (',  and  D  be  lowered, 
so  that  the  reactions  at  the  five  supjwrts  may  Ix'  equal? 

23  «7^    4  jW*    23  jW* 
AOEl  '  5  El  '  40  £,7  ■ 

285.  A  continuous  girder  of  two  spans  of  00  ft.  .and  (M)  ft.  respectively  is 
loaded  with  a  uniformly  distributed  load  of  2  tons  per  foot-run.  Determine  the 
reactions  at  the  piers  end  the  H.M.  over  the  c(>ntre  support  (a)  when  the  ends 
rest  ujwu  the  supports;  (^j  when  the  end  of  the  short  span  is  fi.\cd  horizon- 
tally. 

Ans.  (a)  33J,  193?,  and  72i  terns;  XhlTy  ft.-ton.s. 

(h)  301,  1".M)3J,  and  72'  tons;   -125  and  -1550  ft.-tons. 


Ann 


EXAMPLES. 


583 


286.  A  rolled  steel  joist,  40  ft.  in  length,  depth  10  ins.,  breadth  .5  ins.,  thick- 
ness throughout  i  in.,  is  continuous  over  throe  supports,  forming  two  spans 
of  JO  ft.  each.  What  uniforinly  distributed  load  would  produce  a  maximum 
.•(tnss  of  ")J  tons/s(i.  in.?  Slcetch  the  diagrams  of  l)ending  moments  and 
■■^•"■="'  f'"''''"-  Ans.  lO-ms  tons. 

287.  \  continuous  girder  of  three  sj)ans  carries  .1  !„ad  of  1  ton  per  lineal  foot. 
Tlic  two  side  spans  are  2S  and  84  ft.  in  length,  and  the  intermediate  span 
i-  ."it;  ft.  in  length.  Find  the  reactions  and  the  B.M.'s  at  the  supports  (a) 
uh.  11  the  two  ends'  rests  ujwn  supjwrts;  (6)  when  the  end  o'  the  28-ft.  span 
i>  lixcd  to  the  support. 

Ans.  (a)  React ons,  llj,  33i,  88J.  and  .34  tons;  B.M.'s,  -  70  and 
-672  ft.-tons. 
(b)  Reactions,   14^\,  30U,  S%\,  33^    tons;    li.M.'s,   -69*. 
-57H,  and  -674/7  ft.-tons. 

288.  .\  continuous  girder  fixed  at  one  end  has  three  equal  spans  each  of 
104  ft.  determine  the  B.M.'s  and  reactions  at  the  supports  when  a  load  of 
104  tons  is  uniformly  distributed,  (a)  on  the  span  next  the  fixture;  (6)  on  the 
iiiiiidic  span;-  (c)  on  the  remaining  span.  Hence  deduce  (J)  the  correspond- 
iiii;  results  when  the  whole  girder  carries  a  uniformly  distributed  load  of  1  ton 
jiir  lineal  foot. 

.\ti.s.  B.M.'s,  (a)    -1144,    -416,   +104  ft.-tons;    (6)    +312,   -024,    -520 
ft.-tons;  (c)  -104,  +208,  -728  ft.-tons;  (</)  -936,  -832,  -1144 
ft.-tons. 
Reactions,  (a)  59,  1,-6,  50  tons;  {h)   -9,  +62,  -.5,  +.56  tons;  (c)  3, 
-12,  +68,  +45  tons;   (</)  53,  100,  118,  41  tons. 

289.  A  continuous  girder  180  ft.  long  consi.sts  of  two  spans  of  100  ft.  and 
nf  SO  ft.  The  smaller  span  carries  a  uniformly  distributed  load  of  80  tons. 
liiid  the  force  required  to  hold  the  outer  end  of  the  uidoaded  span  upon  its 
support,  and  also  determine  the  remaining  reactions  and  the  B.M.  at  the 
intermediate  pier. 

Ans.  Required  force  =  -3|  tons;  35J  tons,  48  tons;  355S  in.-ton.s. 

2Q0.  A  continuous  girder  of  three  spans,  the  outside  spans  being  equal, 
is  uniforinly  loaded.  What  mu.st  be  the  ratio  of  the  lengths  of  the  centr>> 
iisd  :i  side  span  so  that  the  neutral  axis  may  be  horizontal  over  the  intermediate 
sii|i|i()rts?  What  should  the  ratio  be  if  the  ceiit^re  span  is  hinged  (a)  at  the 
iintrc;  (h)  at  the  points  of  trisection?     Ans.  \'3  :  v^;  (a)  l:\;(h)  3:  2v'2'. 

291.  A  girder  carrying  a  uniformly  distributed  load  is  continuous  over 
foni-  supports,  and  consists  of  a  centre  span  (Q  and  two  equal  side  spans  (/,). 
i mil  the  ratio  of  /,  to  l„  so  that  the  neutral  axis  at  the  intermediate  supports 
ni:iy  be  horizontal.  Also  find  the  value  of  the  ratio  when  a  hinge  is  intro- 
'liiied  Ui)  at  the  middle  point  of  the  centre  span;  (h)  at  the  jwints  of  tri- 
-vruoii  01'  the  centre  span;   (c)  at  the  middle  points  of  the  half  lengths  of  the 


lentre  span. 


I,' 


Ans.  j-^  -■ 
•a 


'3'  /.'""l  •/,'     9'  /,»~4- 

2g2.  A  continuous  girder  ABCD  is  fixed  at  A,  rests  upon  supports  at  B, 

<\  and  1),  and  carries  a  uniformly  distributed  load.     If  the  reactions  at  A, 

li.  C,  D  are  equal  show  that  the  ratio  of  the  length  of  CD  to  AB  must   be 


584 


THEORY  OF  STRUCTURES. 


prcuHT  than  unity  and  less  tlmn  J.    Also,  if  CD  is  equal  to  five  fourths  of 
AB,  show  that  .\«:BC':CI»:  :4:7:.). 

293.  A  pirder  consists  of  two  spans  AB,  BC,  each  of  length  /,  and  is  con- 
tinuous over  a  cpntrc  ])icr  B.  A  uniform  load  of  length  2a(<J)  and  of  in- 
tensity ir  travels  over  AB.  Find  the  rea<'tions  at  the  supports  for  any  given 
position  of  the  load,  and  show  that  the  bonding  moment  at  the  eentre  pier 

is  a  maximum  and  equal  to  — r^(  1  —  tv  )  when  the  centre  of  the  load  is  at 


at  distance 


from  ^1. 


294.  Show  that  a  uniformly  loaded  and  continuous  girder  of  two  equal 
spans     ith  both  ends  fixed  is  2.08  times  as  stiff  as  if  the  ends  were  free  and 

..J  rested  on  the  supports. 

295.  A  continuous  girder  of  two  spans  AB,  BC  rests  upon  supports  at 
.1,  B.  A  uniformly  distributed  load  EF  travels  over  the  girder.  G,  is  the 
centre  of  gravity  of  the  jwrtion  BE  upon  AB,  and  G'j  that  of  the  portion  BF 
upon  BC.     If  the  bending  moment  at  B  is  a  maximum,  show  that 

AEEBAG, 
CFFB    CG, 

296.  A  continuous  jnrder  rests  upon  three  supports  and  consists  of  two 
unequal  spans  AB{^li),  BC{=L).  A  uniform  load  of  intensity  w  travels 
over  AB,  and  at  a  given  instant  covers  .'  length  AD{=r)  of  the  span.  If 
Ri,  Ri  are  the  reactions  at  A  and  C  resi^ectivcly  show  that 


«,/,'  +  /W,'  =  UT(/,'-^r/.+^-J^'). 


Draw  a  diagram  showing  the  shearing  force  in  front  of  the  moving  load 
as  it  crosses  tht-  girder. 

If  the  live  load  may  cover  both  spans,  show  that  the  shearing  force  at 
any  point  D  is  a  maximum  when  .ID  and  BC  are  loaded  and  BI)  unloaded. 
Illustrate  this  force  graphically,  taking  into  account  the  dead  load  upon  the 
girder. 

297.  .\  girder  of  uniform  section  rests  on  two  supports  at  its  ends  on  the 
s;inic  level  and  is  loaded  in  such  a  manner  that  the  area  of  the  bending-momeiit 
(liaiirani  is  .t,  and  the  distance  of  the  centre  of  gravity  of  that  diagram  from 
tli"  middle  of  tlic  span  (measured  horizontally)  is  r,  the  span  of  the  l)eam  iM^iiig 
2/t.  If  the  beam,  instead  of  merely  In-ing  supported  at  each  end,  has  its  ends 
built  in  hori/.oiitally.  show  that,  with  the  same  loading  as  before,  the  bending 
moments  at  the  two  ends  are  given  by 


.1 


a±.V 


298.  A  continuous  girder  of  two  spans  AB,  BC  has  its  two  ends  .1  and 


EXAMPLES. 


585 


C  fixed  to  the  abutments.  The  load  upon  AB  is  a  weight  P  distant  p  from 
.1,  and  that  ujkjii  BC  a  weight  Q  distant  q  from  V.  The  length  of  AB^l,, 
(p|  I1C=L.  The  bending  moments  at  .1,  B,  C  are  .1/,,  .1/,,  M,,  respectively. 
i'lii>  areas  of  the  bending-moment  curves  for  the  spans  AB,  BC  assumed  to 
hr  independent  girders  are  -li,  A„  respectively.     Show  that 

MJ,  +M,{1,  +1,)  +3/3/,-  -2(.4.  +^J 
and  M,(l,  +1,)  =  -•2(A,p  +  A^). 

If  /,  ^/s  =/,  show  that  .1/,  is  a  maximum  if 

2l{Pr-Q,i)=3iPp'~Qq^. 

299.  A  continuous  girder  AC  consists  of  two  equal  spans  AB,  BC  of  15  m. 
raili.  Determine  the  bending  moments  at  the  supports,  the  maximum  inter- 
iTiidiate  bending  moments,  and  the  reactions  (a)  when  the  load  upon  each 
span  is  3000  k./m.;  (6)  when  the  load  per  metre  is  3000  k.  upon  AB  and 
liKM)  k.  u|)on  BC.  Consider  three  cases,  viz.,  (I)  when  both  ends  of  the 
(linier  are  free;  (II)  when  both  ends  are  fixed;  and  (III)  when  one  end  is 
tree  and  the  other  fixed. 

Ans.  Ca.se  I.  (a)  B.M.'s,  -84,375  k.m.,  47,460.9375  k.m. 
Reaction,  16,875  k.,  56,250  k. 
(6)  B.M.'s,  -.56,2.50 k.m., 58,.593.75 k.m.,-' 31.25  k.m. 
Reaction,  18,7.50  k.,  37,.500  k.,  3750  k. 
Case    II.  (a)  B..M.'s,  -56,2.50  k.m.,  28,125  k.m. 
Reactions,  22,.500  k. 
(6)  B.M.'s,  -65,625k.m.,  -37,500K.m.,  -9375k.m., 
33,398.4375  k.m.,  64,453,125  k.m. 
Reactions,  24,375  k.,  30,000  k.,  5625  k. 
Case  III.  (a)  B..M.'s,  -48,214?  k.m.,  -72,321?  k.m.,  24,537411 
k.m.,  52,088151  k.m. 
Reactions,  20,892?  k.,  51,428}  k.,  17,678}  k. 
(&)  B.M.'s,  -64,285?  k.m.,  -40,1781  k.ai.,  -32,.573,V, 
k.m.,  ll,623iVrk.m. 

Reactions,  24,107j  k.,  31,071?  k.,  4821?  k. 
300.  A  viaduct  over  the  Garonne  at  Bordeaux  consists  of  seven  spans, 
vi/..  two  end  spans,  each  of  57.375  m.,  and  five  intermediate  spans,  each  of 
77. 0(.  in.;  the  main  girders  are  continuous  from  end  to  end,  and  arc  each 
^iil  i,ii(  ted  to  a  dead  load  of  .30.50  k.  per  lineal  metre.  Under  a  live  load  of  405  k. 
!<i  metre  determine  the  absolute  maximum  bending  moment  at  the  third 
>ii|,|i.iit  from  one  end.  Also  find  the  corresponding  reactions,  the  jwints  of 
•  tioii,  and  the  maximum  deflection  in  the  first  and  second  spatis. 
1/.".  4,125,'"59  k.m.;  reactions,  46,470  k.,  .381,008  k.;  points  of  inflrrlinr, 
I  ni.  from  end  supiwrt  for  first  span;  7.88  m.  and  6374  m.  from  second 
iiirt  in  second  span. 

?oi.  A  {rirdcr  of  two  spans  each  of  100  ft.  is  continuous  over  the  three 
•  '■•vt'^  .1.  /;,  and  r.  Draw  the  .S.F.  and  H.M,  di.igrams  f.ir  .1  uniformly 
:iil)Utcd  lo.'id  of  100  tons  on  AB  and  .50  tons  on  BC  (a)  when  H..>'.  is  nil 
1  and  at  C,  (b)  when  A  is  fixed  horizontally.  Ans.  (a)  Reactions,  40| 
i.  I.'.J  tons;  B.M.'s,  9.37i  ft.-tons;  (6)  51H,  80A,  17?  tons;  B.M.'s,  -892*' 
7141  ft.-tons 


iiifl 


■  i\.\« 


686 


THEORY  OF  STRUCTURES. 


302.  A  horizontal  continuous  ffirdor  of  three  equal  spans,  resting  upon 
supports  at  .1,  li,  (.',  D,  carries  a  uniformly  distributed  load.  Show  that  the 
bending  moments  at  the  intermeilinle  supports  will  Ir-  unaffected  when  the 
supports  B,  C,  I)  are  dei)ressed  Ix-low  the  horizontal,  provided  that  the  amounts 
of  the  depressions  are  in  the  ratios  of  1  to  2  to  3. 

303.  The  bridge  over  the  Grande  Raise  consists  of  two  equal  spans  of 
V.).S  in.;  each  of  ilie  main  girders  is  continuous  and  rests  U|)on  abutments 
at  the  ends.  Fitid  the  position  of  the  jxiirits  of  inflection,  the  tx-nding  moment 
at  the  centre  support,  the  maximum  intermediate  IxMiding  moment,  and  the 
maximum  Hang"  stress  (</)  under  the  dead  load  of  1700  k.  |x<r  lineal  metre; 
(h)  under  the  same  dead  load  tog>'fher  with  an  additional  i)roof  load  of  2000  k. 
per  lineal  metre  on  one  spun.  The  depth  of  the  girder  =3.228  m.,  and 
/  =.003920232 144. 

Ans.  (a^  14.S.">  m.  from  the  abutments;    83,308.5  k.m.;    46,801;^ 
k.m.;    1.431.5  k./sq.  mm. 
(b)  1(>.18  m.  from  abutment  on  loaded  side;    11.138  m.  from 
abutment  on  unloaded  side;  132,313..3  k.m.;  121,l%.32 
k.m.;  2.273.')<»  k./sq.  mm. 

304.  A  continuous  girder  of  three  equal  spans  is  uniformly  loaded.     By 

how  much  must  the  two  iiuermediatc  supports  lie  depressed  to  jJHMluce  the 

7  W* 
same  reactions  at  all  the  supports?  ^"«-  ^g^y 

305.  A  h.)rizontal  girder  ABC  is  fixed  at  .1,  rests  upon  supports  at  B  and 
C,  and  carries  a  .iniformly  distributed  load  211'.  If  an  error  J  is  made  (1) 
in  .'/<,  (2)  ill  BC,  find  the  cons«'quent  errors  in  the  B..M.'s  at  A  and  B,  and 
if  AB-BC,  show  that  these  errors  are 


(1)    i§>»'-'- 


12 


1% 


r.W- 


306.  A  uniform  l-eanr«  is  su[)ported  by  four  equidistant  props,  of  which 
two  are  terminal.  Sho'v  that  the  two  points  of  inllection  in  the  middle  seg- 
ment are  in  the  .same  horizontal  {ilane  as  the  i)rops. 

307.  The  horizotit  il  girder  ABC  is  fixed  at  .1  and  rests  tipon  supiiorfs 
at  B  and  (',  AB  Uii.d  ecpial  to  BC.  If  the  depression  of  ('  is  11  times  that 
of  /*,  and  if  It,,  R„  B<  arc  the  reactions  at  A,  B,  C,  resijectivcly,  show  that 
H,  (48-1 7/1)  +  «i,(8  -  « )  -  IU>^  -  -:i" )  • 

308  A  continuous  girder  of  two  eciual  spans  (/)  is  uniformly  loaded.  Show 
that  the  ends  will  ju.st  touch  their  8Up|X)rt3  if  the  centre  support  is  raised  ^. 

300.  The  horizontal  girder  ABC  is  fixed  at  -4,  rests  upon  supports  at  B 
and  C,  and  carries  weights  It  ,  and  U  ,  concentrated  at  the  middle  |K>iiiin 
of  AB  and  BC  resjiectively.      Find  the  reactions  and  the  bending  momenta 


k'C 


EXA^fPLES. 


587 


1  .ind  R.     If  An  =  BC  and  if  n%=3ir„  show  tha;  the  momont  of  fixture 
/(//,  tli.'it  the  bonding  moment  at  li  is   —\\\\AB,  and  that  the  reactions 


at  .1,  ii,  atid  C  are  JH',,  'I'lr,,  and  JH',,  respectively. 

ildw  much  must  B  bo  lowered  so  that  the  reaction  at  B  may  bo  nil?  Find 
thf  corresfHinding  reactions  at  .1  and  C  How  much  must  C  now  be  lowered 
so  iliat  the  reactions  mav  bo  the  same  as  before? 

.4B'/25"ir, +43ir,x     -IB        ..,  „        AB 


.ift&. 


El  \  7()S 
4  X  depression  of  B. 
310.  .\  continuous  sjirder  consists  of  two  spans,  each  5f  "I.  in  leufrth;  t!i3 
clTiMtivc  depth  of  the  girder  is  S  ft.  If  one  of  the  end  l)oa:  si-,  cttlcs  to  the 
I'Miiii  of  1  in.,  find  the  .  .aximum  increase  in  the  flange  an(t  shearing  si -ess 
<,iiM'il  thereby,  and  show  by  a  diagram  the  change  in  the  distribution  of  the 
-iri-^cs  throughout  the  girder,  (.\ssume  the  section  of  the  girder  to  be  uni- 
l.iiiii,  and  take  i:-'-'.j,(X)(),00()  lbs.) 


Ans.  Increa.se  of  maximum  B.M.  =2V(i' 


V-'lGu-     7' 


"        "  .shearing  force     =fo/, 
«'  being  weight  jK-r  foot  oi'  length  and  /  the  moment  of  inertia. 
311.  The  horizontal  girder  ABC  is  fixed  at   .1   and  rests  upon  supports 
at  H  ;iiid  V.     If  AB  is  n  times  BC,  show  that  the  bending  moments  at  .1  and 
H  ire  in  the  ratio  of  n'  +  2«'  — 1  to  n'  +  2. 


ill.  A  continuous  girder  of  two  equal  spans  is  fixal  at  one  of  the  end 

-iipii.  -'-;.     The  girder  carries  a  uniformly  distributed  load  of  intensity   i/. 

Ii  liir  Ictiiiii.  of  each  span  is  /,  find  the  reai-tions  and  moment  of  fixture.     How 

i!:iirli  must  the  intermediate  supiH)rt  l)o  lowered  so  that   it  may  bear  n 

(I  i!ic  Ici.id?     How  much  should  the  free  end  supjwrt  then  be  lowered  to  bring 

lip  11  the  sup|)orts  the  same  loads  as  In-fore? 

11    ,   1(5    ,   13    ,       W     4  i/Z'     4  I//' 
j"'.  ,-4"'.  .1;^"''; 


Ann 


2S"-  14" 


14  '  4S/iV  24  A'/- 


31  V 

^|l.■|ll-,  ci 


A  continuous-girdei  bridge  has  a  centre  span  of  .'{(K)  ft.  and  two  >.i.lc 

'.ch  of  2(K)  ft.     The  dead  load  tipon  ea -h  of  the  main  girders  is  )2."»()  lbs. 

I"!  Iui(>m1  foot.     In  one  ot   the  .side  spans  there  is  also  an  additional  load  of 

.'■'i«i  lbs.  jKT  lineal  foot  upon  each  girder.     Find  (n)  the  reactions  a?i(l  points 

It  liidrciion.     How  much   (/<)  must  the  third  sup|x»rt   from  the  loaded  end 

Ik'  Idwcred  so  that  the  pressure  U|)on  it  may  iK>  just  zero? 

Am.  (a)  W(vic//«m,v.  .m">.l(i()  lbs.;  f).">r),S.-.()  lbs. ;  322,.')l."i  llw.;  01,r,V) 

lbs. 

BM.'».  13.!I0.S,(X)0  ft.-llw.;  fi.TrM.IOO  ft.-llw. 

PoinU  of  iiijlrrtiim.   For  side  .xpans,  102}}  and  H.'iJ}  ft. 

from  end  supports;  for  centre  span,  distance  r  from  end 

...         ,        ,     I3!)00       4t)2.-|()00 
HupiM.rt  IS  given  by  j;'-— ;vir-'  +  — ^~  -0. 


21 


39 


(6) 


5(»12S(K)O000 
El 


588 


THEORY  OF  STRUCTURES, 


:!i4.  A  continuous  girdor  AC  consists  of  two  equal  spans  AB,  BC,  each 
of  Iciiiitli  /.  riiid  curries  a  uniformlj-  distributed  load  of  intensity  w  uiwn  Ali, 
and  of  iiitensily  lur  uimhi  BV.  Determine  the  bending  moments  at  the 
supports,  th(!  maximiun  intermediate  IxMiding  moments,  and  the  reactions 
(a)  when  both  ends  of  the  girder  are  fixed;  (h)  when  one  end  .1  is  fixed  and 
the  other  free.  If  ic,  =  »\  =  ir,  find  (c)  the  points  of  inflection  and  (d)  maxi- 
mum deflection  for  each  span  in  each  case. 

1(7'  «•/'  1(7' 

Am.  (a)B.M.'s,^(n-ry);    -^(n+1);   ;i^(l+5H): 


1(7= 


1(7, 


^(S3 


-2-2n  +3/i') ;  7'I^(243«'-214n  +35). 
1. ■).}() 


Reartionn,  77.  (0  -  >i) ; 
It) 


2S ' 
1(7' 


f(l+");    ^(-l+9n). 
1(7' 


(b)  B.M.'s,  —:^(^-n);  0;    -,j^il +2n); 


ul' 


,.,.^(.S8-30n+%');  ,.,,, 
Reactions,  ^^(U)-3h);  ^(13  +  19m^: 


(-l+12n)'. 


^|(-l+12n). 


(c)  Points  0/  inflection  at  .212/  and  .788/  from  fixed  ends. 

1    ((7* 
Maximum  defleelion  at  middle  point  of  span  and  -  — -  -  . 

tl84  t,l 

(d)  Points  of  nt flection  for  AB  at  .10.")/  and  .734/  from  .4;  for 

RV  at  rj/  from  C 
Maximum  deflection  for  AB  in   (c)  at  .64/  from  .4  .-iiui 

'"473A7' 

7, 


for  Br  in  I 
from  f  and 


(/)  at  ^,/  (approximat;'ly 


1    (-/• 


(approximately' 


315.  A  vertical  row  of  water-tight  sheet  piling  .30  ft.  high  is  supjM)rtcd 
by  a  series  of  uprights  j>l;ired  H  ft  'entre  to  centre  and  securely  fixed  at  the 
base,  while  the  up|K'r  ends  are  kept  in  the  vertical  by  struts  sloping  at  4."i''. 
If  the  water  rises  to  the  to|)  of  the  piling.  !itid  («)  the  thrust  on  a  strtit;  (''> 
the  mavinnnn  iiiteiisity  of  stress  in  an  upright;  (o  tiie  amount  and  positimi 
of  the  tnaxinium  deviation  of  an  u|iright  from  the  vertical.  If  the  piliii!{ 
is  strenirt helled  by  a  second  scries  of  strut-"  sloping  at  4.')°  from  the  jmints 
of  in.ixiniuin  dcvi.-ilion,  find  dh  the  normal  reactions  ui>on  an  upright  .hkI 
the  bcndiiii:  monicnt  at  its  foot.  Wh.at  (< )  will  l)e  the  reactions  and  iMMMJitig 
moment  if  the  seiiqid  row  of  struts  starts  from  the  niidille  of  the  uprights? 

Ans.  (.j)  4."),(KKI\'2  lbs.; 

{!))  -    (."»02.'i  i     72,000N^'j  '■  n>!i./8<i.  in.  at  On  r>  ft.  from  upper  end; 


EXAMPLES. 


589 


316 

<■.  ,11.:. I 

/;.  (  .  M 

1..-    ;rr,' 
.1 /.':/;( 


(r) 


6220S000\/5 
El 


at  fiN'o  ft.  from  upper  end; 


W)  3045,  110,r)S7.r,  lbs.,  3,077,400  iii.-lbs. 

(()  SS.W.3  ll)s.,  11.-),714.3  lbs.,  2,!»S9,285.7  i.-lbs. 
.  .\  coiitiriuous  girder  MiCl)  i.s  fixed  at  .4,  rests  upon  support.s  at  B, 
I),  and  carries  a  uniformly  distributed  loud.  If  the  reactions  at  A, 
iid  I)  are  o(|ual,  show  that  the  ratio  of  the  length  of  CD  to  AB  must 
Iter  than  unity  and  less  than  J.  Also,  if  CD  =  IAB,  show  that 
':r/>::4:7:r,. 


317.  Tl.*>  Osse  iron  viaduct  eonsi.sts  of  seven  spans,  viz.,  two  end  span.s 
111  .'>.S  III.  ind  live  intermediate  spans  of  3S  m.;  each  main  girder  is  continuous 
:iimI  carric!  a  dead  load  of  14.iO  k.  per  litieal  metre.     P'ind  (n)  the  bending 

I milts  at  the  supiKirts  when  a  proof  load  of  22.")0  k.  jM-r  lineal  metre  for 

.:i' li  fiirder  covers  all  the  spans;  and  also  find  (6)  the  absolute  maximum 
l«  niling  moment  at  the  Imirth  support.  Is  (r)  the  following  sectiim  of  suffi- 
11(1.1  strength?— two  eciual  flanges,  each  composed  of  a  (iOO-mm.XS-mm. 
I'liU'  riveted  by  means  of  two  lOO-nim.  X  HH)-mm.X  12-mm.  angles  to  a 
i>o(l-iiini.X  lO-mm.  vertical  web  plate  and  two  SO-mm.  X  SO-mni.  X  11-mm. 
:iML:lc-i  rivet(-d  to  each  horizontal  plate  with  the  ends  of  the  horiztintal  arms 
I"'  mill,  from  the  edges  of  the  plates,  the  whole  depth  of  the  .sj'ction  l)eing 
I  iiKi  m..  .111(1  the  distance  Ix'tweei.  tne  web  plates,  which  is  open,  being  2.8  m. 
li  insutficienf ,  how  would  you  strengthen  it? 

Alls.  (,()  B..l/.'.s,  410,.'J18   k.m.,  4.-.2,7!)0  k.m.,  443,722    k.m.;    (b) 

.'>42,im»  k.m. 

(r)  /  =.14074 440407  and  max.  (lange  stres-s/sq.  mm.  =7.73  k. 

This  is  much  too  large.     The  section  may  Ik-  strengthened  by  adding 

two   (iOO-mni.XS-nim.   plates  to    ea<h   llange.      /  is   thus   increased  by 

.ii7s:U'J.'>.">30,  and  the  flange  unit  stress  iM'comes  ■')  k./sd.  mm. 

318.  The  Ivstressol  viaduct  consists  of  four  spans  of  2.")  'ii.;  the  main  girders 
:i!v  rnniiiiiioiis  and  their  ends  rest  upon  abutments;  the  dead  load  upon  each 
(.'ii'lrr  is  17tK)  k.  |)er  lineal  metre.  Deterniine  the  po-ition  of  jMiints  of  in- 
1!'  I  loll  ill  each  span,  th(<  reactions  and  iMMiding  moments  at  the  siip[Hirts 
»ii'  II  .III  additioiiMl  load  of  20(K)  k.  jier  lini'al  metre  erosm-s  {a)  the  first  s]ian; 

I  lie  lirst  and  second  spans;  (c)  all  the  spans.  .\lso,  find  {(/)  the  (iliynlule 
ii.ivnhiiin  iM'iiding  moments  at  the  <nterm(>diate  siip|iorts,  and  (r)  determine 
till   iii;i\imum  (lange  stress  at  the  piers,  /  in-ing  .0".1;!(I2!I232I04. 

Ans,  lA't  X,,  X;,  i-i,  X,  U-  the  distances  of  the  |)oint.s  of  inflection  in 

the   first,  second,  third,  and  fourth  spans  from  the  first, 

wcoiul,  fourth,  and  lifth  spans,  respectively. 

(n)j,-l».4    m.,    x,'-28j-,+  140-0,  i','-24.16x,  +  120.S-0, 

x«-20.10.Sm. 

Bearli.ms,  .3.S,.348,',   k.,  81,l«0f  k.,  34 1  (Hi  J   k.,  4'Htllf   k., 

10  473,',  k. 
B.M.'x,  197,.'>44,\  k.m.,  r)3,.-)71 ,'  k.m.,  119,419,»4  k.m. 
(6)  X, -20.72  m.;   x,' - 31. 7Kr,  + 232.4-0,  x,'-28.1x,  + 140.5 
»0,x«- 19.38  m. 


! 


590 


THEORY  OF  STRUCTURES. 


licartions,  30,1 78J  k..  107,821  J  k.,  62,004!  k.,  45,892|  k., 

17.14'.'5. 
B.M.'s,  2r)S,92Sl  k.m.,  120,53")^  k.,  102,0781  k.m. 

(c)  J-,  =  1!).C.4  m.  =j-.:    J-,'  -2().Sx,  (- 134  =0  ami  x,  -  x,. 
I{<(icli<m^,  :i(,.33'.»r  k.,  U)."),714if  k.,  S.'),S!)2?  k. 
fi.-U.'.v,  247,707;  k.m.,  lti.").17SJ  k.m. 

(d)  204,.")<)S]J  k.m.  at  second  support  when  first,  second,  and 

fourth  spans  arc  loadt'<l.     209,S2IJ  k.m.  at  third  sup- 
{M)rt  when  s«'cond  and  third  spans  are  loaded. 
(f)  4..')  k./mni.'. 
319.  If  the  three  sup|X)rts  of  any  two  equal  eon.serutive  spans  of  a  con- 
tinuous girder  of  any  numhcr  of  si)ans  are  depre.ss«'d  l)elow  the  horizontal 
show  that  the  relation  between  the  three  Ix'ndinjf  moments  at  the  supjnirts 
will  Ih>  unaffected  if  the  depre.ssion  of  the  centre  support  is  a  mean  between 
the  depressions  of  the  other  two  sui)ports. 

3J0.  Four  weiphts,  each  of  0  tons,  follow  each  other  at  fixed  distances  of 
5  ft.  over  a  continuous  girder  of  two  spans,  each  equal  to  !yO  ft.  If  the  second 
and  third  sup[K)rts  are  1  in.  and  IJ  ins.,  res(K»ctively,  vertically  below  the 
first  support,  find  the  maximum  B.M.  at  the  intermediate  supjwrt. 


(PI   \ 
■'^■-^•'^ -40000)  ^^-*^"«- 


321.  .\  continuous  girder  of  two  equal  .TO-ft.  spans  is  fixed  at  one  of  the 
end  sup[)orts.  The  girtier  carries  a  uniformly  distributed  load  of  1000  ll>s. 
n<'r  lineal  foot.  Find  the  reactions  and  l)ending  moments  at  the  points  of 
support.  How  much  must  the  intermediate  su|)|K)rt  Ik'  lowered  so  that  it 
may  Ix'ar  none  of  th.'  load?  How  much  should  the  free  end  Ik;  then  lowered 
to  bring  ujK)n  the  supiM)rts  the  .same  loads  as  at  the  first? 

.l«.s.  Heaciions=23,21?4.  •'■>7,142^  19,042f  lbs.; 

Heiuling  moments  =  17S,.">71  J,  207,S,">7|  ft. -lbs. 
3*2.  Each  of  the  main  girders  of  a  railway  bridge  resting  uix)n  two  end 
8upi«trts  and  five  intermediate  supiH)rts  is  fixed  .-it  the  centre  su|)i>ort,  is  3  ft. 
deep  throughout,  and  is  designed  to  carry  a  uniformly  distributed  diiul  loud 
of  \  ton  aiicl  a  livi'  load  of  J  ton  |ier  lineal  foot.  The  end  spans  are  each 
.")!  ft.  S  in-i.  and  the  intermedi.'tte  spans  e.'ich  ."><)  ft.  in  the  dear.  Find  (a)  the 
reactions  at  the  sup|Hirts.  The  ginlers  are  single-weblK-d  and  double-flanged; 
the  flanges  are  12  ins.  wide  and  ecpi.'il  in  sectional  area,  tin-  areas  for  the  inter- 
mediate spiiiii  iM'int;  13  sq.  ins.  and  17  s<|.  ins.  at  the  centre  and  piers  res|iec- 
tively.  I'iml  (^i  ilie  corres|H)nding  moments  of  resistance  and  flange  strcMes, 
the  web  bei.  g  i  in.  thick. 

Aii.i.  (11)  |.-).3.-.0,  13.40."..  3."..7t(),  3H.37S;oiis. 

(b)Mnnln,     1717,   714,  1042  in.-tons;   3.2,    1.3,  and    I.80 

tiiiis/»q.  in. 

"  pnr.s,  24HS,  10!tM.  I«>(i3  in-tms;  .3..5<t,  2.4.5,  and  2>3 

tons  sq.  in. 

3J3.  In  n  certain  Howe-truss  i>rid!ie  of  eijiht  panels  the  liinlxT  cros^-i n- 

arc  directly  siip|H.rieil  by  the  lower  chords,  and  are  placed  sutficiently  close 


EXAMPLES. 


591 


to  distribute  the  load  in  an  approximately  uniform  manner  over  the  whole 
Ic  ii^rth  of  these  chords,  thus  produ.  ;ag  an  additional  stress  due  to  flexure. 
AssiimiT  ■•■  that  the  chords  may  Ix-  regarded  as  girders  supiwrted  at  the  ends 
■i"'l  "'US  over  seven  intermediate  supjxirts  coincident  with  the  panel- 

|.niii!  Ill  that  thes«'  panel-points  are  in  a  truly  horizontal  line,  determine 
i./i  ihr  iH-ndiug  moments  and  reactions  at  the  panel-points;  (h)  the  maximum 
i)iliTn)cdi:itc  bending  moments;  and  (r)  the  pf>ints  of  inflection  corresponding 

to  a  load  of  w  per  unit  of  length,  /  being  the  length  of  a  panel.     (lake  -  =388.) 

Ans.  (a)  B..l/.',s,  0,   --ilrivP,   -SOruP,   -S^irwl,   -32rwP. 
Rcactionn,  lUSrwl,  440rul,  374rwl,  392rivl,  3H6rwl. 
(6)  11,704..-,  wiriy;  5104..5ic(r;)';   6600.5if(r0';   6208..5u-(r?)» 
(c)  The  points  are  defined  by  the  values  of  x  in  x=306r/- 

ff,a+x)+/?,r_!£.(/+a.)'=0;  li,(2l  +  x)+R,il+x),  and 


Ra--^{2l+x)'=0; 


III 

of 

sil| 

:lll 


R,m  +  x)+R,{2l  +  x)+R,{l  +  x)+Ra-^(3l  +  xy=0. 

3J4-  If  d„  d„  d,,  d,  are  respectively  the  deflections  of  the  first,  second, 
tliinl,  and  fourth  panel-jwints  in  question  323,  show  that  the  bending  moment 
at  the  middle  panel-point  (Mt)  is  given  by 

-d7M,--'^/{7ld,-90d,  +  24d,  -(Jd,)+8u;/'. 

325.  Four  loads,  each  of  12  tons  and  spaced  .'">,  4,  and  5  ft.  apart,  travel 
order  over  a  continuous  girder  of  two  spans,  the  one  of  30  and  the  other 
-'0  ft.  Place  the  wheels  so  as  to  throw  a  maximum  H.M.  u|K)n  the  centre 
i|«irt,  and  find  the  corresponding  reactions.     Draw  a  diagram  of  B.M. 

I  find  the  maximum  deflection  of  each  span. 
Alls.  -.'A  wheel   concent  rated  on  intermediate  support;    2.6133   42  2067 

3t>42       .'M)3.4  . 
ss.nns;      --,-    ^,^      ,n. 

m6.  The  loads  upon  the  wheels  of  a  truck,  locomotive,  and  tender,  eount- 

^  in  order  from  the  front,  are  7,  7,  10,  10.  10,  10,  8,  8,  8,  8  tons,  the  intervals 

iii;  .">,  .■),  .■>,  .-),  .'),  0,  .■),  4,  .■)  ft.    The  loads  travel  over  a  contimious  girder 

UM)  .Vj-ft.  spans  Afi,  HC.     I'lacr  the  locomotive,  etc.,  («)  on  the  span    Ifl, 

-  to  give  a  maximum  H.M.  at  B.  Am.  1st  wheel  .8  ft.  from  .1. 

M7-  An  eight-wheel  locomotive  travels  o%er  a  continuous  girder  of  two 
I-'  .  >|Mi:-<:  the  truck-wheels  are  (\  ft.  centre  to  centre,  the  h.ud  upon  each 
r  Uiiiig  S(MM)  lbs.;    the  driving-wheels  are  8J  ft.  centre  to  centre,  the  load 

II  c^icli  pair  l)eing  1«),(X)0  lbs.;  the  distance  centre  to  centre  iH-tween  the 
Ml  drivers  and  the  tiearest  truck-whe«'ls  is  also  KJ  ft.  I'lace  the  locomotive 
I-  ii«  throw  a  maximum  H.M.  upon  the  centre  supiH)rt,  and  find  the  corre- 
-  ling  rcjK  tions.  .i(i.,.   l.st  wheel  43.(i2  ft.  from  end  ;  I()ltMi.:<,  .Wim  7. 

I7ti  lbs. 

U«.  A  single  weight  travels  over  the  span  AB  of  a  girder  of  two  equal 
lis,  Mi.  nc,  continuous  over  a  (rntre  pier  B.     Show  that   the  reaction 


liw 

tl|.. 

tr.. 


592 


TIIEOHY  OF  STRiXTUHi:S. 


AB  . 
fit  ('  is  a  maximum  when  the  distance  of  the  weight  from  A  is  — ;=  if  the  endrf 

.1  and  C  rest  upon  their  sui)]K)rts,  and  when  the  distanee  is  5. IB  if  the  two 

ends  are  fixed.     Find  tlie  eorresiwnding  tending  moments  at  the  central  pier. 

PI        2 
Ai>s.        — ;  -PL 
()\  3     -'' 

329.  The  weiglits  7,  7,  10,  10,  10,  10,  S,  8,  S,  8  tons,  taken  in  order,  pass 
over  a  continous  girder  of  two  spans,  each  of  50  ft.  and  fixed  at  both  ends, 
the  successive  intervals  being  .">,  5,  5,  5,  5,  9,  0,  4,  5  ft.  Place  the  wheels  so 
as  to  give  the  maximum  bending  moment  at  the  centre  sui)iK)rt,  and  find  its 
value.  Ans.  First  wheel  24.3  ft.  from  nearest  abutment. 

Max.  B.M.=  3.32.3  ft.-tons. 

330.  A  swing-bridge  of  200  ft.  length,  with  arms  each  of  90  ft.,  revolves 
a  turntable  of  20  ft.  diameter.  When  the  bridge  is  closed  find  the  reactions 
at  the  four  supports  due  to  u  load  of  3  tons  at  30  ft.  from  an  end  support. 

Alls,  hi  Ions  is,  4i,  -15,  A. 

331.  Three  horizontal  In-ams,  each  ft  m.  long,  36  cm.  deep,  and  having 
an  I  =  l!),7t)(>  cm.',  rest  ui)on  su>ports  and  are  rigidly  tied  together  by  a  trans- 
verse l)e:iin  at  the  centre,  of  similar  section,  2  m.  long  X 20  em.  deep,  and 
having  an  I  =21()2  cm.^  The  main  beams  are  1  m.  c.  to  e.  Disregarding 
the  weight  of  the  l)eams,  find  the  greatest  weight  which  can  be  placed  at  the 
point  of  intersection  of  the  middle  and  the  transverse  beams  so  that  the  stress 
due  to  flexure  mav  ri  wLere  exceed  1000  k./cm.'.  Ans.  7948  k. 


CHAITER  VIII. 


PILLARS. 


I.  Classification.— The  manner  in  which  a  material  fails  under 
imssuro  depends  not  merely  upon  its  nature  but  also  upon  its  dimen- 
yions  and  form.  The  actual  compressive  strength  of  a  material  nmst 
hv  determined  by  the  use  of  very  short  specimens,  in  which  there  is 
hn  tendency  to  buckle  or  l)end.  An  accurate  determination  of  the 
rclaiion  between  the  stress  and  the  strain  cannot,  however,  well  be 
iii;id<'  unless  the  length  of  the  specimen  exceeds  four  or  five  times  its 
liast  transverse  dimension.  The  breaking  load  does  not  change 
iim.li  until  the  length  is  about  ten  times  this  dimension,  and,  if 
liiiiding  is  prevented  by  means  of  slight  lateral  restraints,  will  remain 
prMctieally  the  same  whatever  the  length  may  be.  If  lateral  re- 
.«tr;iints  are  not  employetl,  the  tendency  to  buckle  or  bend  increases 
wit'  the  ratio  of  the  length  to  the  least  transverse  dimension.  De- 
nntiiifi  this  ratio  by  .v,  all  pillars  for  columns)  tinth  ends  truly  flat 
inul  jirwhf  secured  may  in  general  l)e  divided  into  three  classes,  viz.: 

A.  Short  jrillars,  in  which  .s  does  not  exceed  4  or  5,  and  which 
fiiil  under  a  direct  pressure. 

H.  Medium  pillars,  in  whi'-h  s  lies  between  4  and  about  30  for 
tiiiilxr.  cast  iron,  and  hard  steel,  and  between  4  and  about  60  for 
win,,;:!,,  ir,„i  and  mild  steel.  The  failure  of  pillars  of  this  class  is 
'liH  partly  to  a  chreet  pressure  and  partly  to  bending. 

< '.  ion,/  pillars,  in  which  «  exceeds  about  30  for  timber,  cast  i-on, 
aiil  li.ird  steel,  and  alrout  60  for  wrought  iron  and  mild  steel.  The 
t:nliiiv  of  the.se  pillars  is  practically  wholly  due  to  flexure. 

2.  Manner  of  Failure.— The  manner  in  which  a  short  pillar  fails 
'1'  p'nds  uj)on  the  character  of  the  material. 

A  gramdar  or  crystalline  mass  may  give  way  suddenly  under 
(uuipression  and  Ik*  reduced  to  {wwder. 

503 


594 


THEORY  OF  STHlCTi'RES. 


More  frequently  hard  and  brittle  substances  like  glass,  bricks, 
stones,  dry  timber,  cast  iron,  etc.,  which  possess  little  (luctility, 
split  into  fragments,  and  sometimes  a  liaril  vitreous  material  breaks 
up  into  prisms  (Fig.  552). 


Fit;.  i)o2 


Fig.  .WS. 


Fig.  .554. 


Fig  555. 


The  failure  of  fibrous  and  granular  materials  like  timber,  bricks^ 
cement  and  concrete  cubes,  artificial  and  certain  natural  stones,  cast 
iron,  etc.,  is  usually  by  shearing  along  planes  oblique  to  the  line  of 
thrust,  and  the  more  homogeneous  the  material  is  the  more  nearly 
do  these  planes  coincide  with  the  planes  of  least  resistance  to  shear, 
which  theoretically  makes  an  angle  of  45°  with  the  axis  of  the  pil- 
lar (Chapter  V).  Thus  the  portions  of  the  fractured  pillar  assume 
the  forms  of  wedges  or  pyramids  (Figs.  553-558),  and  in  the  testing 
of  cement  and  concrete  cuIm-s  it  not  iufn^juently  happens  that  a 
very  nearly  perfect  double  pyramid  ( Fig.  556)  is  protluced. 

None  of  the  materials  of  construction,  however,  are  truly  homo- 


Fig.  5.56.     FiR.  557. 


FiR.  558. 


mj 


Fig.  559.         Fig.  560. 


geneous.  and  in  the  ca.se  of  ca.st  iron  the  irregularity  of  the  texture 
and  the  hardness  of  the  skin  cau.se  the  angle  between  the  plane  of 
shear  and  the  direction  of  the  thrust  to  vary  from  32°  to  42°.  Brick 
chimneys  again  sometimes  fail  by  the  shearing  of  the  mortar,  the 
upper  i)ortion  sliding  over  an  oblique  plane. 

liuliiimi,  i«'-,  a  lateral  spreading  out  (Figs.  559,  560),  is  character- 
istic of  blocks  of  fibrous  materials,  e.g.,  wrought  iron,  copper,  lead, 
and  timber,  and  fracture  occurs  in  the  form  of  longitudinal  crack."*. 

.\il  .-ill ..stance  s,  pv.'ii  the  iiio^t  ery.-^t.ilHne,  will  l?ulge  .^ilightly  be- 
iure  they  fail,  if  they  possc'ss  .some  degree  of  ductility. 


V  } 


ItODGKINSOXS  FORMULA. 


595 


Buckling  is  characteristic  of  fibrous  materials,  and  the  resistance 
(if  a  piUar  to  buckling  is  always  less  than  its  resistance  to  direct 
crushing,  and  is  independent  of  length. 

Thin  malleable  plates  usually  fail  by  the  bending,  puckering, 
wrinkling,  or  crumpling  up  of  the  fibres,  and  the  same  phenomena 
may  be  observed  in  tl  «  case  of  timber  and  of  long  bars. 

In  the  transverse  testing  of  timber  b?ams  the  initial  failure  ifs 
usually  made  evident  by  the  crippling  of  the  fibrous  layers  on  the 
compression  side. 

Timber  offers  about  twice  the  resistance  to  crushing  when  dry 
tiiat  it  does  when  wet,  as  the  presence  of  moisture  diminishes  the 
lateral  adhesion  of  the  fibres. 

Long  plate  tubes,  when  compressed  longitudinallv,  first  bend  and 
(Mutually  fail  by  the  buckling  of  a  short  length  on  the  concave  side. 

The  ultimate  resistance  to  buckling  of  a  well-made  and  well- 
sliaped  tube  is  about  27,000  lbs.  per  square  inch  section  of  metal, 
uliich  may  be  increased  to  33,000  or  36,000  lbs.  per  square  inch  by 
ili\  itling  the  tube  into  two  or  more  compartments. 

.V  rectangular  wrought-iron  or  steel  tube  offers  the  greatest  resist- 
aiKv  to  buckling  when  the  mass  of  the  material  is  concentrated  at 
III-'  angles,  while  the  sides  consist  of  thin  plates  or  lattice-worlr  suffi- 
nciitly  strong  to  i)revent  the  bending  of  the  angles. 

3.  Hodgkinson's  Fonnulae  for  the  Ultimate  Strength  of  Long  and 
Medium  Pillars.-When  a  lom,  pillar  is  subjected  to  a  crushing  force 
I'  lirst  yields  sideways,  and  eventually  breaks  in  a  manner  appar- 
•nlly  snnHar  to  the  fracture  of  a  beam  under  a  transverse  load 
I  Ins  similarity,  however,  is  modified  by  the  fact  that  an  initial  lon- 
ptu.linal  compression  is  imluced  in  the  pillar  by  tlu    ^uper-imposed 


loail 


Ilodgkinson  deduced  e.xperimentallv  that  the  strength  of  lonff 
-/"/  round  iron  and  square  timber  pillars  with  flat  ends  is  given  by 
an  expression  of  the  form 

d" 

\y  being  the  breaking  weight  in  tons  of  2240  lbs.; 
(I      "       "    diameter  or  siiie  of  the  pillar  in  inches; 
J        ■•        "    longth  of  th- pillar  in /ee/; 
n  and  m  being  numerical  indices; 


596  THEORY  OF  STRUCTURES. 

A  being  a  constant  varying  with  the  material  and  with  the  sectional 
form  of  the  pillar. 
Such  pillars,  which  are  more  than  thirty  times  the  least  trans- 
verse dimension  in  length,  fail  by  bending  and  not  by  simple  crush- 
ing. The  load  is  distributed  more  or  "ess  uniformly  over  the  ends, 
which  are  finished  in  planes  at  right  angles  to  the  axis  of  the  pillar. 

For  iron  pillars n  =  3.55  and  m  =     1.7 

"  timber  pillars n=4       andw=     2 

*  ■  ca.><t  iron A=  44.16 

' '  wrought  iron A  =  133.75 

' '  dry  Dantzic  oak /I  =  10.95 

"  dry  red  deal A=     7.81 

"  dry  French  oak A=     6.9 

The  strength  of  long  hollou-  round  cast-iron  pillars  was  found  to 
be  given  by 

W=  44-34 jTt .  • 

d  being  the  external  and  rfi  the  internal  diameter,  both  in  inches. 

Thus  the  strength  of  a  hollow  cast-iron  pillar  is  approximately 
vi\\ii\\  to  the  difference  between  the  strengths  of  two  solid  cast-iron 
pillars  whose  diameters  are  equal  to  the  external  and  internal  diam- 
eters of  the  hollow  i)illar. 

The  .'Strength  of  medhim  pillars,  i.e.,  of  pillars  which  are  leas  than 
thirty  times  the  least  transverse  dimension  in  length,  may  be  ob- 
obtained  by  the  formula 

W  +  ffS' 

W  being  the  breaking  weight  in  tons  of  2240  lbs.; 

]Y      "      '  •  "  "       <<.(<((<      ( <  j^g  derived  from  the 

formula  for  long  pillars; 
/  being  the  ultimate  cru.shing  strength  in  tons  per  square  inch; 
,S      "      "     sectional  area  of  the  pillar  in  square  inches. 

Acain,  if  the  ends  of  a  casti-iron  pillar  are  rounded,  the  above 
formula;  may  be  still  employed  to  determine  its  strength,  .4  being 
14.9  for  a  nolid  and  13  for  a  hollow  pillar. 


noDCKixsos's  ForaiUL^E. 


597 


According  to  Hotlgkinson,  the  relative  strengths  of  long  cast-iroa 
pillars  of  ecjual  weight  and  length  may  be  tabulated  as  follows: 

(a)  Pillars  with  flat  ends: 

The  strength  of  a  solid  round  pillar  being  100, 
"     "     square    "      is  93, 

"     "     triangular  pillar  is  110. 

(b)  Pillars  with  round  ends,  i.e.,  ends  for  hinging  or  pin  connec- 
tions: 

The  strength  of  a  hollow  cylindrical  pillar  being  100, 
"  "an  H-shaped  ' '      is        74.6, 

"a  +-shaped  "      "         44.2. 

The  strengths  of  a  long  solid  round  pillar  with  flat  ends  anu  a 
long  hollow  cylindrical  pillar  with  round  ends  are  approximately  in 
the  ratio  of  2.3  to  1. 

The  stiffext  kind  of  wrought-iron  strut  is  a  built  tube,  the  section 
(•(insisting  of  a  cell  or  of  cells,  which  may  be  circular,  rectangular 
triangular,  or  of  any  convenient  form. 

In  experimenting  upon  hollow  tubes,  Hodgkinson  found  that, 
other  conditions  remaining  the  same,  the  circular  was  the  strongest, 
and  was  followed  in  ordei  of  strength  by  the  i^quare  in  four  com- 
partments g  ;  the  rectangular  in  two  compartments,  Hi  ;  the 
rectangular,  □;  and  the  square.  ~ 

The  addition  of  a  diaphragm  acro.ss  the  middle  of  the  rectangle 
doubled  its  resistance  to  crippling. 

.\s  the  results  of  his  experiments  Hodgkinson  also  made  the  fol- 
lowing inferences: 

\  pillar  wifh  both  ends  rough  from  the  foundry  so  that  a  load 
•  an  he  applied  only  at  a  few  i.solated  points,  and  a  pillar  with  a 
rounded  end  so  that  the  load  can  be  applied  only  along  the  axis,  are 
each  one  third  of  the  strength  of  a  pillar  of  class  B,  and  from  one 
third  to  two  thirds  of  the  strength  of  a  pillar  of  class  C,  the  pdlars 
liciiig  of  the  same  dimensions. 

The  strength  of  a  pillar  with  one  end  flat  and  the  other  round  is 
an  arithmetical  mean  between  the  strengths  of  two  pillars  of  the 
i^anie  dimensions,  the  one  having  both  ends  fiat  and  the  other  both 
ends  round. 


598 


THEORY  OF  STRUCTURES. 


Disks  at  the  ends  of  pillars  only  slightly  increase  their  strength, 
but  facilitate  the  formation  of  connections. 

An  enlargement  of  the  middle  section  of  a  pillar  sometimes  in- 
creases its  strength  in  a  small  degree,  as  in  the  case  of  solid  cast- 
iron  pillars  with  roimiled  ends,  which  are  made  stronger  by  about 
oneseretith:  hollow  cast-iron  pillars  are  not  affected.  The  strength 
of  a  disk-ended  pillar  is  increaseil  by  about  one  eighth  or  one  ninth 
when  the  middle  diameter  is  lengthened  by  50  j^er  cent,  but  for 
slight  enla.'gements  the  increiise  is  imperceptible. 

The  strength  of  hollow  cast-iron  pillars  is  not  affected  by  a  slight 
v.iriation  in  the  thickness  of  the  metal,  as  a  thin  shell  is  much  harder 
tlrui  a  thick  one.  The  excess  above  or  ileficiency  below  the  average 
thickness  should  not  e.xceed  25  per  cent. 

In  metric  measurement,  if  d  is  the  diameter  and  /  the  height  of  a 
solid  iron  colunui  in  millimeters,  the  safe  load  in  kilogrammes 

J"" 


and 


=  1900-T^  for  cast  iron, 


d* 
=^3800  .T,  for  wrought  iron. 


Again,  if  a  is  a  side  of  a  sfpiare  timl)er  jwst  in  centimeters  and  I  its 
height  in  decimeters,  tiie  smIc  loud  in  kilogrammes 

a* 
=  256.5 X-w  for  first -(juality  well-sea.soned  oak 

a* 

=  180-^         "   oak  of  ordinary  quality 

a* 

=214.2  X-Tj  "   fir.st-(|uality  well-seasoned  pine 


=  160 


P 


pine  of  ordinary  quality. 


Ex.  1.  A  solid  cast-iron  pillar  9  ft.  in  height  and  4  ins.  in  diameter  supports 
a  loud  of  55,000  lbs.  Find  the  normal  and  shearing  intensity  of  stress  per  square^ 
inch  in  a  plane  ncrtion  inclined  at  30°  to  the  axis. 


GORDON'S  FORMULA. 


599 


//  the  ends  of  a  pillar  are  flat  and  firmly  bedded,  determine  its  breaking  weight 
bij  Hoihjkinmm'n  jnrmula. 

55000 
Axial  intensity  of  stress/sq.  in. -..  .,  '-, ^rr^s -2187i lbs. 

Therefore 

normal  intensity  of  stress  =2187J  sin  ."JO"  =1093i  Ibs./sq.  in. 
ami      tangential  intensity  of  stress  =:il87i  eos  30°  =  1894J  Ibs./sq.  in. 
iJy  Hodgkinson's  formula  for  short  pillars, 


the  breaking  weight  in  tons  =44.16 


(4)3.. 
(9) 


1.7 


■154.97. 


■141  J, 


llciice,  since  pillar  is  of  medium  length, 

..       ,     ,u      .•  •  u.  •    .  154.97X-VWW- V4 

Its  actual  breakmg  weight  in  tons  = -^^y^^^^^^-^^ 

takiiifr  80,000  lbs.  per  sq.  in.  as  the  ultimate  crushing  strength  of  the  cast  iron. 
i;x.  2.  Find  the  side  oj  a  square  post  of  ordinary  oak  which  is  to  be  5  metres 
Ii'kjIi  and  to  carry  a  load  of  10,000  kilogrammes. 

10000 -18o£,. 
50 

Therefore  o-19cm. 

l',x.  3.  Determine  the  diameter  of  a  solid  cast-iron  column  which  is  to  be 
5  metres  high  and  to  carry  a  load  of  50,000  kilogrammes. 

Therefore  d  =  160mm. 

4.  Gordon's  Formula  for  the  Strength  of   a  Pillar  of  Medium 

Length.— Consider  a  pillar  with  flat  ends,  of  length  I,  and  sectional 
ana  S,  h  being  the  least  transverse  dimension  in  the  plane 
of  llcxuro. 

The  effect  of  P  is  twofold. 

In  the  first  place,  P  produces  along  the  pillar  a  direct 
p 
tliru.st  of  intensity /i  =^. 

I))  the  second  place,  it  tends  to  bend  the  pillar  and 
tlrvcloi)S  a  skin  compressive  stress  /2  which  is  necessarily 
jrnatcst  at  the  section  in  which  the  neutral  axis  has 
il(>viatod  farthest  from  its  initial  vertical  position. 

Denoting  this  deviation  by  y, 

Py  =  B.M.=/2j, 


,'i't 


J...  J 


^IIIEIil 


U 


hm. 


600 


THEORY  OF  STRivrURES. 


c  being  the  distance  of  the  most  comprcsyeci  fibre  from  the  neutral 
axis  and  /  the  moment  of  inertia  of  the  section  in  the  plane  of  flexure. 

Therefore  '^^    l^S  'r^^'^r^' 

r  being  the  least  radius  of  gyration. 

P 
But  coch,  rah,  and  by  Chapter  MI,  t/oc^. 


Therefore 


a  being  a  coefficient  which  is  to  be  determined  by  experiment. 

Hence,  denoting  by  /  the  total  maximum  stress  in  the  most  strained 
fibre 

/  =  /l+/2  =  /l(l+«p)=^(l+«^2). 


and 


I-.. 


i  +  a- 


J2. 


which  is  known  as  Gordon's  formula. 

The  coeflficient  a  is  by  no  means  constant,  and  not  only  varies 
with  the  nature  of  the  material,  with  the  length  of  the  pillar,  with 
the  condition  of  its  ends,  etc.,  but  also  with  the  sectional  form  of 
the  pillar.  The  variation  due  to  this  latter  cause  may  Im-  eliminated, 
and  the  fornmla  rendered  somewhat  more  exact,  by  substituting  the 
least  radius  of  gyration  for  the  least  transverse  dimension. 

It  has  been  shown  that 


h-lS. 


and  since  c<xh  and  y^Tj 
therefore 


fictfi-^^fxax-^. 


VALUES  OF  COEFFICIENTS. 
Hence  Gordon's  formula  now  becomes 


eoi 


=fi 


12' 


in  which  Oi  is  independent  of  the  sectional  form,  all  variations  of  the 
latter  being  included  in  r^.  This  modified  form  of  Gordon's  formula 
was  first  suggested  by  Rankine. 

A  pillar  (or  column)  is  square-bearing  or  flat-ended  when  it  has 
scjuare  ends  which  butt  against  or  are  securely  connected  with  aa 
immovable  surface;  it  is  pin-  and  square-hearing  when  one  end  is 
square-bearing  while  the  other  presses  against  a  close-fitting  pin; 
it  is  pin-bearing  when  both  ends  are  pin-jointed,  the  axes  of  the  [)ins 
iK'ing  parallel.  In  pm-  and  square-bearing  pillars  \a  and  |oi  should 
Ik?  substituted  in  the  preceding  fornmla?  for  a  and  a\,  while  the  co- 
efficients should  be  4a  and  Aa\  if  the  pillar  is  pin-bearing. 

5.  Values  of  a,  ai  and  f.— The  following  tabl.^,  giving  the  valuea 
of  the  constants  a,  Oi,  and  /,  has  been  prepared  by  taking  an  average 
of  the  best-known  results,  and  is  applicable  to  square-bearing  round 
and  square  pillars: 


Uaterikl  and  Form  of  SMtion. 

/  in  Lb*. 
IwrSq.  In. 

1 

a 

at 

Fi)r  cast-iron  solid  rectaii(?ular  pillars 

80,000 

80,000 

80,000 

80,000 

36,000 

36,000 

36,000 

67,200 

67,2(M> 

67,'200 

114,000 

114,000 

IHiOOO 

5,000 

5,000 

7,'200 

450 

400 

500 

600 

3,000 

2,2.'>0 

5,500 

2,000 

1,400 

2,500 

1,400 

000 

1,500 

250 

2.50 

250 

"          "          "     round  pillars 

"          "        hollow  rectangular  pillars 

"          "            "       round   pillars 

,     6,400 
(36,000 

For  wroxight-iron  solid  rectangular  pillars 

"     round  pillars 

"           thick  hollow  round  pillars 

lor  mild-steel  solid  rectangular  pillars 

"          "           "     round  pillars 

"         hollow  round  pillars 

36,000 

I'lir  Htrong-nteel  solid  rectangular  pillars 

"          solid  round  pillars 

"           "          hollow  round  pillars 

For  pine-timber  solid  rectangular  pillars 

"            "     round  pillars 

3,000 

For  dry  oak  timber 

Uraphical  comparv>on  of  the  crushing  unit  strengUi  of  solid  round 
cast-iron,  wrougfU-iron,  and  mild-med  pillars. 


602 


THEORY  OF  STRUCTURES. 


The  crushing  unit  stress  is  given  by  p  = ^  . 


0,010  lb*. 


"40  50         60 

Fio.  562 

Take  the  different  values  of  -r  as  abscisssD,  and  the  corresponding 

values  of  p  as  orilinates;   the  resulting  curves  for  the  values  of  / 
given  in  the  table  are  shown  in  Fig.  562. 

Hence  the  strength  of  a  mild-steel  pillar  exceeds  that  of  a  wrought- 

iron  pillar,  but  is  less  than  that  of  a  cast-iron  pillar  when  t<  10.7; 

a  wrought-iron  pillar  is  stronger  or  weaker  than  a  cjist-iron  pillar 

according  as  y  >  or  <28.5. 

Kx.  4.  Compare  thr  hrrnkinq  u-riqhls  of  round  cast-iron,  ur ought-iron,  and 
mild-ilrrl  pillars  with  flat  end*,  rarh  fxiiig  9  It.  in  length  and  6  in*,  in  diamtler. 
How  wiU  tht  results  fe*  nuhlilinl  if  I'oth  ewls  are  roundf 


I      12X9 


FitM,  h^fi  mrli  ftfii: 


6 

80000 


-18. 


For  cMt-iron,      B.W.  - ^^,  ^^-7^:^ -  l,2riO,197  lb>. 


i+»iai8)- 


EXAMPLES.  Q03 

Forwrought-in,n,B.W.-?^^Jip.'-   890,1091b..  " 

Second,  both  ends  round: 

For  cast-iron,        B.W.-?00p0><Vi(6)'_    533.^,.. 

"  wrought-iron.B.W.-36000xVi(62'_    646  120" 

i+nWi8)'        o^e.i-io 

1+tAt»(18)'  »»7,0W 

Ex.  5.  A  hollow  mild-steel  pillar  10  //.  hif,h  and  '*  ins.  outside  diameter  has 
I"  carry  a  load  of  33  000  lbs.  Taking  G  as  a  factor  0}  safety  and  assuming  that 
h,>th  ends  are  to  be  pin.c»nnecled,  find  the  proper  thickness  of  the  metal. 

J     4  /[nxi2y  ' 

2500V      4      / 
t  Ix'ing  the  required  thickness.    This  reduces  to 

l«-4< +2.2875-0, 
niid  therefore  /,  the  thickness,  -.69  in. 

i:x.  6.  A  pillar  of  diameter  D  suppoHs  a  given  load;  if  N  piUars,  each  of 
<lmmftrr  d,  are  substil'Ucd  for  this  single  piUar,  show  that  d  must  lie  between 

V» »'"'  Yf    f*^  »"<"*  economical  to  employ  few  or  many  pillars  of  given  height 
t'l  Kiipport  a  given  loadt 

—     /  rvi-the  given  load X factor  of  safety-  ^^')^*.^, 


«hi(li  reduces  to 
Tliorcfore 

ilcnce 


al\D*-Nd*)  -DV(Nd*~D% 
D*>Nd*    and    Nd*>  D*. 


D 


D 


d<iii     and     >T7! 


m 


NV 


604 


THEORY  OF  STRUCTURES. 


If  w  is  the  specific  weight  of  the  material, 

the  weight  of  the  A^  pillars      ^Nn^dHw  -=P„ 
"        "      "    "    single  pillar  =riOTu) -P., 
and  P,>  P2,  since  Xd'>  D'. 

It  is  therefore  more  economical  to  use  a  single  pillar. 

Ex.  7.  Find  the  side  oj  a  xquare  pout  of  ordinary  quality  oak  which  is  to 
he  IL'J  ft.  high  and  is  to  carry  a  load  of  'J0,000  /6s.,  10  being  a  [actor  of  safety. 
The  ends  are  to  he  flat. 

If  X  ins.  is  a  side  of  the  square  .section, 


10X20000- 


7200x» 


7200x« 


12xr2i\'    x'+90* 


Therefore 


•nd 


_L/ilxi2i\ 

'  ^250V       X      I 


2.50 
X* — ij-x' -2.500-0 


X  =8.12  ins. 


6.  Values  of  a  and  ai  for  Shapes  and  Built-up  Members. — Further 
cxiH'riinents  arc  still  ripflfd  for  the  proptT  determination  of  the 
r(X'ttieients  a  and  «i  in  the  eji.se  of  built-up  ineniljers  and  of  .such 
shajM's  a.s  are  shown  by  Figs.  563  to  5()8.     At  pn-sent  every  engineer 


.\4 


fl 


[liJ 


^ 


Flo.  563      Ki(i.  5(i4.      Fkj.  .505.       Fio.  :m\. 


Fi<i.  5f.7. 


Fill.  568. 


ases  the  values  for  which  his  own  ex|)erience  gives  liim  a  preference, 
but  it  is  usually  foun(i  that  tliere  is  no  j)raetieal  difTerence  in  the 
sectional  area«  of  pillars  (or  struts)  designed  in  accordance  with  any 
good  standard  sj)eeification. 

The  lejist  transverse  dimension  (i.e.,  h)  is  to  Ih'  meimured  in  the 
•l-ne  (»f  greatest  flexure. 

Thus  it  may  lie  taken  as  the  leant  side  of  the  rectangle  circum- 
scribing tee  (Fig.  5t);i),  II  channel  (Fig.  .50.')),  and  rrunform  (Fig.  .5«i7) 
wctioris,  and  as  the  |H'r|H'ndicular  from  the  angle  to  the  opposite 
sijle  of  a  triangle  circumscribing  amjle  (Fig.  •'iGK)  sections. 

The  following  table  gives  the  values  of  /  anil  a  for  the  shapes  in 
<|uestion,  the  memlxTs  b'ing  8<|uare-lK'aring.     These  valura  have 


VALUES  OF  COEFFICIENTS  FOR  SHAPES.  605 

bo<>n  deduced  experimentally,  but  must  be  regarded  as  approximate 
only : 

Material.  /  in  Ibs./sq.  in.        - 

Cast  iron 80,000  134 

Wrought  iron 42,500         900 

Mild  steel 45,000        1300 

The  effect  of  a  change  of  form  is  best  provided  for  by  the  use  of 
till"  formula  which  involves  the  least  radius  (r)  of  gyration.  Thus 
til.'  ultimate  strength  in  pounds  per  square  inch  of  medium-steel  col- 
iiiiint!!  is 

50,000 

'-(^ 

-  Ixing  36,000,  24,000,  or  IS.OOO  according  as  the  column  is  square- 
Ixariiig,  pin-  and  square-bearing,  or  p.n-bearing. 

The  factor  of  safety  may  be  4  for  quiescent  loads,  as  in  buildings, 
uiul  a  for  moving  loads,  as  in  bridges. 

Kx.  H  Find  the  breaking  weight  of  a  square-bearing  cw^t-iron  column  10  ft. 
h.,h  and  of  cruaform  section,  the  mHnl  Imng  2  in...  thick  throughout  and  the 
oiifxidr  length  of  each  arm  being  8  in.s.     Then 

„,,      ,  i4-288q.  ins.    and    A'-3'  +  3'-18. 

I  luTofore 

the  breaking  weight  -  -^^^^— -,  -mm  lbs. 

'"^■•tooiTy 

V.x.  9.  IMermine  the  aUowalUe  .xtrc.s  in  pounds  per  square  inch  of  a  10-ft 
T',..l„nn„g  slnU  composed  of  two  5"X3"XA"  angles  placed  back  to  back,  5 
l-riij  a  factor  of  safety. 

1'ir.t.  let  the  long  legs  be  back  to  bark  an.i  spread  i  in.  to  admit  of  a 
MMu'l.'  line  of  i-in.  lattice  bars  along  the  central  plane.     Then 

/-T.fim-.;    .4-4.H»q.in«:     r' -^*^^^-i..5876, 

4.3 

Hihi  the  allowable  stresa  in  pounds  i)rr  square  inch 

10000 

*^18000V   r  / 


606 


THEORY  OF  STRUCTURES 


Second,  let  the  short  legs  be  back  to  back.    Then  /  with  respect  to  neutral 
axis  through  centre  of  gravity  =3.468.    Therefore 

and  the  allowable  stress  in  pounds  per  square  inch 

10000 


1  + 


1     /120\ 
18000V  r  / 


,-4746. 


Again,  the  American  Bridge  Co.'s  specifications  limit  the  length 

of  wind-bracing  jjin-boaring  struts  to  120  times  the  least  radius  of 

gyration,  while  the  lengths  of  other  pin-bearing  members  must  not 

exceed  100  times  this  radius.    They  allow  a  working  stress  in  pounds 

per  square  inch  of 

15()(K) 

for  soft  steel, 


1  + 


l 
13500 


C7 


and  of 


17000 


llOOOVr/ 


for  mild  steel. 


l-h 


Straight  Line  Fortnula. — ^The  coefhcients  in  pillar  formulae,  gener- 
ally speaking,  have  l)een  determined  by  experiments  in  which  the 
pillar  has  been  ultimately  crushed  under  a  steadily  applied  and 
gradually  increased  load.  Recent  exf)erience  has  shown  that  the 
working  stress  of  framed  and  shaped  compression  members  may  be 
conveniently  expres.sed  by  the  "straight-line"  formula 

working  stress  per  square  inch  =  f  — //-, 

where  /  =  12,000  lbs.  and  /i  =  4.'j,  .W.  or  00,  according  as  the  member 
is  .xquare-lM'aring,  pin-  and  .s(iuare-lM>aring,  or  pin-beari!ig. 

No  provision  is  made  in  this  formula  for  the  detrimental  effect  of 
tlie  re[K'ated  stresses  to  which  a  member  is  often  subjected  by  the 
action  of  a  live  load,  as  when  a  train  pas-ses  over  a  bridge.  This 
"impact"  effect  might  1k'  allowed  ff)r  by  reducing  the  live  load  to 
an  equivalent  dead  load  and  making  a  corres|)onding  change  in  the 


I  it 


Pi 


STRAIGHT  LINE  FORMULA. 


607 


coofficients.  Thus,  in  bridge-work,  take  /  =  18,000  lbs.,  and  ft =60, 
70,  and  80,  instead  of  the  values  just  given. 

Again,  for  top  chords  (square  ends  with  or  without  play)  take  /— 
IS.OOOlbs.  and  /t  =  70. 

For  end  posts  or  batter  braces  (with  pin  ends  or  ends  tending  to 
rotate)  take  /  =  18,000  lbs.  and  ;<=  80. 

For  intermediate  posts  (allowance  being  made  for  pin  ends,  dis- 
tortion, and  greater  liability  to  shock)  take  /  =  16,000  lbs.  and  /i=  80. 

Ex.  10.  Design  a  medium-steel  top-chord  section,  the  unsupported  length 
hcing  26  ft.  and  the  total  thrust  along  the  chord  being  800,000  lbs.  Try  a  section 
(Fig.  569)  made  up  of 

One  24"  xi"  cover  plates  =12      sq.  ins. 

Two  3i"  X  3\"  X  f "  top  angles  =  4.96  "  " 
"     18" X}"  webs  -22..50  "    " 

"    6"  X  4"  X  J"  bottom  angles  - 1 3.88  "    " 

so  that  the  total  sectional  area  -53.34  "    '  • 


For  practical  reasons  a  clearance  of  J  in.  may  be  left  between  th"  edge* 


of  the  web  plates  and  the  backs  of  the  top  and  bottom 
angles,  so  that  the  depth  over  ail-i+2(4)  +18-18J 
ins. 

I^t  X  be  the  distance  of  the  centre  of  gravity  of  the 
section  from  00.     Then 

x(53.34) -12X18.5+4.96X17.24  + 

22.5X9.125  +  13.88  X  1.08  -527.81, 


and  therefore 


X  -9.9  ins. 


Tlic  moment  of  inertia  about  the  horizontal  axis  GO  through  the  centra 
I  if  gravity 

-T^X24(i)«  + 12(8.6)' +2X2.87+4.96(7.34)» 

+  2X|Xl8'xA+22.5(.775)»+2X8.68  +  13.88(8.82)' 
-2878.9, . 

and  the  radius  of  gyration  with  respect  to  GG  is  given  by 

l287Jr9 
•■-Vi  53.34- -7-3-*«™- 

Tho  moment  of  inertia  with  respect  to  the  vertical  axis  A  A  will  be  found 
t"  1m'  greater  than  that  with  respect  to  GG,  so  that  7.346  is  the  host  radiua 
of  gyration,  and  therefore 

the  allowable  stress  per  square  inch  -18000-70^"^ 'j^  -16.026  lbs. 

/■<>40 


60S 


THEORY  OF  STRUCTURES. 


sooooo 

Ilorice       the  sectional  area  required  =  -.  t.,.,j.  =53.24  sq.  Ins., 
which  agrees  very  closely  with  the  area  of  the  assumed  section. 

In  Cooper's  bridge  speeificatioiis  for  medium  steel  and  for  live  loads 

/  =  10,000  and  /<  =  45  for  chord  segments; 
/=  8, 500   "    /<  =  45  "    all  posts  of  through-bridges; 
/=  0,000    "    ^  =  40"    all  posts  of  deck  bridges  and  trestles; 
7  =  13,000     '    /<  =  00  "    wind  (i.e.,  live)  loads  on  laterals  and  rigid 
bracing. 
These  results  are  to  lx>  doubled  for  dead  loadx. 

The  ratio  -  should  not   excecil  100  for  main  members  or  120  for 
r 

laterals. 

The  allowable  stresses  for  soft  steel  should  1k"  about  15  per  cent 
less  than  those  for  medium  steel. 

7.  Bending  of  Struts.— The  strut  of  length  /  and  constant  sec- 
tional area  .S  is  a.ssumed  to  be  homogeneous  and  its  weight  is  disre- 
garded (Kuler's  Theory). 

Cask  I.  Strut  Hinged  at  Both  Ends.— l^ct  DBA,  Fig.  570,  represent 
the  neutral  axis  of  the  l)ent  strut  and  assume  that  the  Une  of  action 
of  the  resultant  thrust  P  on  the  strut  coincides  with  the  axis  before 
bending  commences. 

U't  X,  )/  Im'  the  vertical  and  horizontal  distances  respectively  of 
any  |K)int  C  with  respect  to  0.    Then 


^1 


I'y^B.U.&tC^-EI 
which  may  be  written  in  the  form 


d^y 
dx^' 


■H° 


where 


Er 


Fio  .570         ^^^^  complete  solution  of  any  such  equation  is 

y  =  b  sin  ux  +  c  cos  ax, 


EULER'S  FORMULAE. 


609 


b  and  c  being  arbitrary  constants  whose  values  are  governed  by  the 
coiulitions  of  the  particular  problem  under  consideration. 

In  the  present  case  y=(i  when  x=0,  i.e.,  at  0,  so  that  c  is  also  nil, 

and  y =6  sin  ax. 

Again,  y  =  0  when  x = 04  =  OB  A ,  approximately  =  I.    Therefore 

0  =  6  sin  al. 

Now,  h  cannot  be  nil,  as  in  such  case  y  would  be  always  nil,  and 
lateral  bending  would  be  impossible.     Hence 


and 


sin  ai=0 
al=n7:  =  lyl^, 


so  that 


P-n'El'-, 


n  being  a  whole  number. 

The  least  value  of  P  corresponds  to  n  =  l,  and  therefore 

is  the  minimum  thrust  which  will  produce  bending.    It  is  often  called 
the  buckling  load  of  the  strut. 

The  deviation  of  the  axis  from  the  vertical,  i.e.,  the  deflection,  is 
greatest  when 

dy    - 

^  =  0=00  cos  ax, 


or  wlien 
and  then 


ax  =  2' 


y-fesin^ -6, 


so  that  the  coefficient  6  is  the  maximum  deflection,  which,  in  the 
present  case,  is  the  deflection  at  the  middle  of  the  strut. 


610  THEORY  OF  STRUCTURES. 

The  meaning  of  the  values  2,  3,  4, . . . ,  for  n  in  the  equation 

m!=al 

is  that  y  is  assumed  to  be  0  at  one  or  more  points  between  0  and  A,  so  that 
the  strut  has  one  or  more  points  of  inflection. 

If  the  column  is  made  to  pass  through  N  points  dividing  the  vertical  OA 
into  N  +  1  equal  divisions,  then 

I 


j/=Owhenx=-j^r^, 


al 


and  therefore,  by  eq.  (4),  0=6  ^'"jv+T 


or 


iV  +  1 


and  hence 


Fig.  571. 


P  =  «'£/-(Ar  +  l)'. 
As  before,  the  least  value  of  P  corresponds  to  n  -1,  and 

P=EI^(H+i)' 


is  the  least  force  which  will  bend  the  column  laterally. 

Hence  the  strength  of  the  column  is  increased  in  the  ratio  of  4,  9,  16,  etc., 
by  causing  it  to  pass  through  points  which  divide  its  length  into  2,  3,  4,  etc., 
equal  parts,  respectively. 

Effect  of  a  Lateral  Load.— Let  Mx  be  the  B.M.  at  C  due  to  a  lateral  load 
on  the  strut.    Then 


dx 


:,  +  a^y  + 


El 


-0, 


which  may  be  easily  integrated  as  soon  as  Mx  has  been  determined.     For 
example,  if  the  lateral  load  is  of  intensity  w  and  uniformly  distributed 

Af,-|x(/-x), 

a  law  which  may  be  approximately  and  more  conveniently  expressed  in  th» 
form 

Mx'-^smj, 

Q  being  the  total  lateral  load.    Therefore 


rf'.V 


\  Ql    .    nx 


Integrating, 


Ql  .    KX 
(P.-P).y-^8'ny, 


EULER'S  FORMVLJE. 
where  p,  —EI~ 

The  deaection  (y)  is  greatest  at  the  middle,  and  therefore 


611 


Vmaz.  ""; 


Ql 


8(P,  -py 

The  B.M.  ia  also  greatest  at  the  middle,  and  therefore,  since 

Mx  -—  at  the  middle, 


the  max.  B.M.  -Elahj        +^ 

'max.       3 


8  P,-P^8^ 
1    P,Ql 


8  P^-P"'^^'^' 

Z  being  the  least  strength  modulus  of  the  section  and  /,  the  maximum  8tre». 
due  to  bending.  The  upper  sign  denotes  a  compression  and  the  lower  a  ten- 
sion Thus  the  total  maximum  and  minimum  stress  p  to  which  any  part 
of  the  strut  is  subjected  is  given  by  ^  *^ 


Let  g-  -f  -the  true  breaking  load  per  unit  of  area; 


|i-/,-Euler's 


<<       »      <i      «     n 


Then 


p't± 


which  may  be  written  in  the  form 

(-l)i-T)-^l- 

Greenhill  has  shown  (Proc.  Inst.  Mech.  Eng.,  1883)  that  if  the  column 


612 


THEORY  OF  STRUCTURES. 


18  also  subjected  to  a  torsional  couple  T,  the  maximum  thrust  required  to 
bend  the  column  is 

In  practice  the  last  term  is  generally  so  small  that  it  may  be  disregarded. 

Case  II.  Strut  Fixed  at  the  Ends. — Let  fi  be  the  moment  of  fixture 
at  0.  The  Hne  of  action  of  the  resultant  thrust  P  no  longer  passes 
through  the  centre  of  gravity  of  the  end  section.  The  B.M.  equa- 
tion is  now 


or 

where 

Put  b-y=z. 
Then 


a2  =  ^  and  b  =  p. 


rfx2 


+ah=0, 


Kid,  572. 
of  which,  as  before,  the  complete  solution  is 

z=c  sin  ax+d  cosax  =  b-y, 

c  and  d  being  arbitrary  coefficients  whose  values  are  to  be  deter- 
mined. 


dy  , 

Differentiatmg,     --~=ca  cos  ax  -da  sm  ax. 


But 
Therefore 


-^  =  0    when    z=Ooi-  =^  or  =»i. 
dx  2 

0  =  ca,    so  that    c=0, 


and 


Also, 


d  coa  ax  =  b-y. 
0=  +da  sin  ;r-=da  sin  al. 


EVLER'S  FORMUUE.  613 

aitil  tliprefore  al  =  2mt, 

n  being  a  whole  number.    Hence 

The  least  value  of  P  correspond i  to  n  =  l,  and 

is  the  minimum  thruct,  or  the  buckling  load,  which  will  produce  bend- 
iiif:  wiien  both  ends  of  the  strut  are  fixed. 
The  deflection  (y)  is  greatest  when 

dy 

^  =  0=  —da  sin  ax, 

i.e.,  when  ax=0,  and  therefore 

ym»x.  =  fc-d  =  -p-d. 

Case  III.  Strut  Fixed  at  One  End  and  Hinged  at  the  Other.— As 
ill  Case  II, 

J-aHb-y), 

tlic  complete  solution  being  of  the  form 

b—y=c  cos  ax+d  sin  ax. 

At  A,  i.e.,  when  x  =  /,  B.M.=0=6-y.    Therefore 

0  =  c  cos  aZ+rf  sin  al. 

dy 
At  0,  i.e.,  when  isO,  ^=0,  and  therefore 

0-  -ca  sin  al+da  cos  aZ. 


U 


CJICf 


(/=0  and  also  c  cos  aZ-0  -  -ca  sin  aZ,  so  that 
b—y  =  c  cos  ax 


614 
and 


THEORY  OF  STRUCTURES. 


,  2n  +  l 
al= — ?:—?:. 


2 


or 


The  least  value  of  P  corresponds  to  n  =  1,  and 

is  the  minimum  thrust,  or  buckling  load,  which  will  produce  bending 
M-hen  one  end  of  the  strut  is  fixed  and  the  other  hinged. 
For  the  maximum  deflection  (y), 

dy    „ 

^=0— ra  smox, 


.so  I 


that 


ax  =  0, 


and 


Va 


'b—c-a-c. 


Tfie  results  of  Caties  I,  II.  and  III  shoir  that  the  buckling  load  of 
n  long  stmt  hinijed  at  both  ends  is  the  same  as  that  of  a  strut  of  IJ  timen 
the  length  hinijed  at  one  end  and  fixed  at  the  other,  and  is  also  the  same 
us  tluit  of  a  strut  of  twice  the  length  fixed  at  bath  ends. 

If  /  is  tin*  ronipn-ssivc  strciipth  jxt  unit  of  urea  of  a  strut  matprial, 
tlicn  a  short  strut  of  sectional  an'a  N  thccin'tically  fails  by  direct 
••rusliiriK  under  the  load  fS.  It  is  iiiipractical)l(',  lunvovcr,  to  insure, 
rvcii  with  tlic  greatest  care,  that  the  material  is  homogeneous,  that 
the  load  is  |)rojK<rly  distributed  f)ver  the  end,  or  that  the  strut  is 
IM'rfoctiy  straijjlit.  Thus  the  strut  will  actually  fail  und«'r  a  less  load 
than  fS,  ainl  for  the  same  n-asons  the  actual  buckling  load  is  !e!» 

than  r'l''l  ij'  •I'f  «'o<'fHcient  /i  iH'ing  1.  J,  or  4.  aecordiiig  as  the  strut 

irt  »!i!isiM!  !ij  iMiih  ond",  hingfd  nt  fm*'  rnd  and  i'ixinl  at  the  otlvr, 
nr  fix«Ml  at  lH)tli  ends.     Kx|M'riment  shows  that  the  strength  of  struti 


EXAMPLES. 


616 


.ntermediate  in  length  between  these  two  extremes  depends  in  some 
way  upon  the  length,  and  the  empirical  formula 


P  = 


A^ 


fs 


1+ 


fS 


1+ 


IiEk^ 


has  l)een  found  to  give,  with  a  fair  degree  of  accuracy,  the  strength 
1)1'  all  struts. 

It  may  be  noted  that  in  the  case  of  very  long  struts,  the  first 
tiTin  of  the  denominator  may  be  disregarded  as  compared  with  the 
second,  and  the  buckling  load  is  obtained,  while  in  the  case  of  very 
short  struts  the  second  term  of  the  denominator  becomes  small  and 
may  be  disregarded  as  compared  with  unity. 

This  empirical  formula  is  identical  with  Rankine's  modification 

/    . 


of  (Gordon's  formula  if  oi 


f:E7c» 


Ex.  11.  Find  by  Euler'g  mdhod  the  buckling  load  o*  a  160"X3"X2"  ateti 
find  when  (o)  the  turn  ends  arf  hinged  (h)  one  end  is  h;  1  and  the  other  fixed, 
(( )  both  ends  are  fixed,  taking  £-29,400,000  Ihs.  per  sq.  m. 

The  buckling  load  -  /.«:/^  -  ^  X  29,400,000  X  ^'  ^  t^ 

-/«X  22,687 J  lbs. 

For  case  a,  /« -1  and  buckling  load  -22,687 J  lbs. 

"      "    6, /'-I '    -51,04«l  " 

"      "     c,  /I-4 -90,750    " 

I'.x.  12.  Find  the  brmkinii  weight  oj  a  Mrui  of  lh<  name  section  and  of  the 
-i'«r  iiiatlity  .»/  firrl  ,ih  that  in  Ihr  prrcfHiiij  rxnmjilp  but  only  80  int.  in  length, 
II:'  lompressive  strength  of  thv  ntvd  Mng  60,000  /6.t.  per  sq.  in. 


mil  I  hen-fore 


the 


fS  -60,000 X  6  -360,000  Ibi., 
breaking  weight 


360000 


1  + 


_3«nooo 

.'iX  •.'26871 


Thus,  in  raae  a,  /<  -1  and  breaking  weight  -21  ,.342  Iba. 


e,  f 


M   '       *' 


'*      -44.707 
"      -72,479 


616 


THEORY  OF  STRUCTURES. 


Ex.  13.  Find  the  limiting  ratio  of  the  length  (I)  to  the  thickness  (d)  of  a  fltU 
pin-hearing  xteel  bar,  E  living  30,000,000  lbs.  per  sq.  in.  and  the  allowable  uxtrk- 
ing  strrtis  10,000  lbs.  per  nq.  in. 

The  bar  will  not  bend  laterally  so  long  as 


10000< 


30000000  &/»  tt'  1 2100000000 /d 


S 


1-.'  p 


49 


(f). 


i.e.,8olongaH-<       vio<49.69. 

Hence  the  length  of  a  flat  bar  in  compression  seems  to  be  comparatively 
limited.  If,  however,  both  ends  arc  securely  fixed,  the  strength  is  quad- 
rupled and  the  admi.ssiblc  length  of  the  bar  is  doubled,  wliilc  it  may  be  still 
further  increast-d  by  fixing  the  bar  at  intermediate  points  as  indicated  in 
Case  I.  This  shows  the  marked  advantage  to  be  gained  by  riveting  together 
the  diagonals  of  lattice  girders  at  the  points  where  they  cross  each  other. 


Lonq 


im',\ 


Stmts.— Let  OBA  be  the  bent  axis  of  a  long  strut  of  length 
/  under  the  buckling  load  P,  the  line  of  action  of  P  coin- 
ciding with  OA  the  position  of  the  axis  before  bending 
commences.  If  y  is  the  maximum  deflection,  the  maxi- 
mum B.M. 

-Py^pEl~y. 


Fio.  574 


Allowance  may  be  made  for  any  slight  initial  cur\'a- 
ture  in  the  strut,  for  an  uneven  distribution  of  the  load 
over  the  end,  and  for  a  want  of  homogeneousness  in  the 
material  by  assuming  that  the  san»e  maximum  deflection 
is  prcMluccd  by  a  load  \\'{<P)  at  C,  where  AC  is  nece*" 
sarily  a  sma!!  quantity. 


Ijct  AC^x. 
Then 

and 


W(x  +  y)'Py 
Wx 

yp^w 


If  the  line  nf  action  of  the  load  W  coincides  with  the  axis  of  the 

^tnt.  «!•,<•?»  r  \^  «-,'.    ,'<'>  !<>!!»  n^  \hv  load  i.=  h^s  than  P,  d-0,  ainl 
tlu«  failun.'  of  tin-  pillar  would  U-  due  to  direct  crushing.     If  the  load 


LONG  STRUTS. 


617 


re- 


i-x  equal  to  P.  rf  becomes  indeterminate  (=^j  and   the  strut 

mains  in  a  state  of  neutral  equilibrium  at  any  inclination  to  the 
vertical.  It  is  impossible  that  IF  should  exceed  P,  as  d  would 
tl.cii  be  negative;  and  therefore  a  load  greater  than  P  would  cause 
the  strut  to  bend  over  laterally  until  it  broke. 

Thus  P=^fiEI^   must  be  the  theoretical  maximum  buckling 
strength  of  the  strut. 

, ,     ir  p 

Take  ^=/  and  ^=/i.    Also,  let  /z  be  the  maximum  stress  de- 
\<1()Ixh1  in  the  strut  by  bending.    Then 

W{x-^y)^Py^f^Z, 

Z  iK'ing  the  strength  modulus  of  the  section. 

Hence,  if  P  is  the  total  maximum  or  minimum  stress  in  the  most 
(leilected  section, 

p-/±/2-/±f(.o+y)-/±E_^ 


■/± 


W  Ar 


ztx-r 

which  may  be  written  in  the  form 

Ex  14.  Find  the  crmking  load  of  a  solid  mild^el  pillar  3  tn*.  in  diameter 
"' '/  H)  t.  long,  with  two  pin  ends.  Also  find  the  deviation  U)  ol  the  line  of 
•X' !<■'<,, , I  aU>ad  of  20,000  lU  from  the  axU  of  the  piUar,  ,„  that  the  maximum 
lolnmiy  „/  Mreia  ntay  not  exceed  10,000  lb».  per  square  inch. 

Hy  Cionlon'a  formula  and  the  Ublt,  Art.  4, 

the  rru-hing  Io«l  -?^^,-m,2«»2.3  lb.. 
.Again,  the  tbeorftiral  maximum  burklinit  strength  P 
.*/^*-2800C«00xfgf  ,^L_ 
-80,216  Iba. 


618 


Therefore 


THEORY  OF  STRUCTURES. 
P  80216  80216 


P-W     80216-20000    60216 


1.332. 


Hence 


10000 


,f°«-(,.1.332><i 


4(3)' 


and 


Z-.714  in. 


8.  Uniformly  Varying  Stress. — The  load  upon  a  pillar  is  rarely, 
if  ever,  uniformly  distributed,  and  in  practice  it  is  often  considered 

sufficient  to  assume  that  the  pressure  in 
any  transverst^  section  varies  uniformly. 
Any  variable  external  force  j'pplied 
normally  to  a  plane  surface  .4.4  of  area 
S  may  be  graphically  represented  by  a 
cylinder  A  ABB,  the  end  BB  being  the 
locus  of  the  extrenjities  of  ordinatcs 
erected  ujwn  ^4.4,  eai'i  ordinate  being 
projx)rtional  to  the  intensity  of  pi-essure 
at  the  point  on  which  it  is  erected. 

I/'t  P  be  the  total  force  upon  ^4.4, 
and  let  the  line  of  its  resultant  intersect 
.4.4  in  C;  C  is  the  centre  oj  pressure  of 
A  A,  and  the  ordinate  CC  necessstrily 
passes  through  the  centre  of  gravity  of  the  cylinder. 

A.ssume  that  the  ])ressure  upon  .4.4  \aries  uniformly;  the  sUi- 
face  BB  is  then  a  plane  inclined  at  a  certain  angle  to  .4.1. 

Take  0,  the  centre  of  figure  of  .4^1,  as  the  origin,  and  AA  as  the 
plane  of  i.  y. 

l/'t  (»'.  the  axi.'<  of  y,  l)e  parallel  to  that  line  IJE  of  the  plane  BB 
which  is  parallel  to  the  plane  .1.1. 

Through  A'A,'  draw  a  plane  DI)  parallel  to  .4.4  and  form  the 
cylinder  .4.l/>/>, 

The  two  rylinders  AAHB  and  .t.l/>/^  are  evi<lently  equal  in 
vohniie,  ami  OF,  tlv  average  ordinate,  repre.-«'nts  the  mean  pressure 
over  .1.1 :  let  it  Ik-  denoti'd  by  /)o. 

.\f  ;uiy  |niitit  !\  of  t!i«-  phuif-  .4.4  rrrrt  the  ofilinate  RQP,  inter- 
secting the  planes  I)D,  BB,  in  Q  and  /'  respectively. 


Fm.  575. 


UNIFORMLY  VARYING  PRESSURE. 

Let  T,  y  be  the  co-ordinates  of  R. 
The  pressure  at  R 

='P''PR=±PQ+QR^±PQ+OF=±ax+po, 

a  l)eing  a  constant  depending  upon  the  variation. 

I^t  Xo,  yo  be  the  co-ordinates  of  the  centre  of  pressure  C. 
Let  iS  be  an  elementary  area  at  any  point  R. 


«1» 


Q      t- 


m 


■  >  iji « 


R      A      fc 

Fro.  577. 


Fio.  576. 

Then  pJS  is  the  pressure  upon  J.Sf.  and  KpJS)  is  the  total 
prtssure  upon  the  surface  AA,  I  being  the  symbol  of  summation. 
Hence 

io^(pJS)^I(pxJS)    and    yo^ip-^f^-IipyJS). 
Rut  p^po±ax. 

riK'refore  XoSl (p„±ax)JS\  - l-\(p„x±aj»)JS\ 

"'" '  yo^  Kpo  ±  OJ-)  J.S  I  =.  J I  :p„,,  ±  axy)  JS  | . 

\"vv  0  is  the  centre  of  figure  of  .1.1,  ,ind  therefore  I^xJS)  and 
^  !lJS)  are  each  zero. 

Also,  J(J.S)  =.S'.  ^(r>jff)  j„  (}„,  ninmcut  of  inertia  (J)  of  4  1  w'fh 
r-i.rt  to  01',  and  I{xyJS)  is  the  product  of  inertia  (K)  about  the 
axis  OZ. 


620 


Therefore 


aiul 


THEORY'  OF  STRUCTURES. 
XoPo^=aI=XoP    . 


(1) 
(2) 


Fio.  678. 


In  any  symmetrical  section  i/„  is  zero,  and  x,  is  the  deviation  of  the  centre 

of  pressure  C  from  the  centre  of  figure  0. 

Let  X,  be  the  distance  from  0  of  the 
extreme  jx>ints  .4  of  the  section. 

The  greatest  stress  in  ^4.1  is  Po  +  oXt  "Pi, 
suppose. 


-i'^ 


Buta-^=f^,  byeq.  (1). 


Fio.  678. 


Therefore 


ZoT.Sp, 


or 


Pi 


I  + 


(3) 


It  ia  generally  advisable,  especially  in  masonry  structures,  to  limit  z, 
by  the  condition  that  the  stress  shall  be  nowhere  negative,  i.e.,  a  tension. 
Now  the  minimum  stress  is  p,— ax„  so  that  to  fulfil  this  condition 


But 

Hence,  by  eq.  (3), 


Po>or-az,. 
p,  -  ox,  +  po ;    therefore  p,  <  or  -  2p^ 


2p„ 


<  or  - 


i^'-fs 


Md 


—j--<  or  -1;    I.e.,  «o<or  -— . 


The  uniformly  varying  stress  is  equivalent  to  a  single  force  P  along  the 
axis,  and  a  couple  uf  moment 

PXCO-  P^lr^i,'  -  a  v7'»TS«. 
The  line  CO  is  said  to  be  conjugate  to  OY. 

X        I 
If  the  angle  COX  -  8,    then    cot  «  -  —  -  p. 

9.  Weyrauch'i  Theory  of  the  Reaistance  to  Buckling. — In  ordfr 
to  nmke  allowance  for  buckling,  VVcyruuch  pro|)o»e»  llu-  two  follow- 
ing iiu-thncU: 


WEYRAUCH'S  THEORY  OF  BVCKUNO.  621 

Method  I.  Let  F,  be  the  necessary  sectional  area,  and  6,  the 
adiiussible  unit  stress  for  a  strut  subjected  to  loads  varying  from  a 
maximum  compression  B,  to  a  minimum  compression  Bj. 

Let  F'  be  the  necessary  sectional  area,  and  V  the  admissible  unit 
stress  for  a  strut  subjected  to  loads  which  vary  between  a  given 
maximum  tension  and  a  given  maximum  compression,  B'  being  the 
miinerically  absolute  maximum  load  and  B"  the  maximum  load  of 
the  opposite  kind. 

According  to  Chapter  IV,  if  there  is  no  tendency  to  buckling,  and 
...  t-u      ,    u-s 

puttuig  "»-—-,  "»  ""-^i  and  r'  =  u-5-factor  of  safety, 


Fv 


B, 


Bx 


'■  K— t) 


(1) 


and 


6' 


B' 


.(.-w|:)- 


(2) 


If  there  is  a  tendency  to  buckling,  let  /  be  the  length  of  the  strut, 
F  Its  required  sectional  area,  and  T  the  mean  unit  stress  at  the 
moment  of  buckling. 

Then,  according  to  the  theory  of  long  struts, 


(3) 


''i  !)oing  a  coefficient  depending  upon  the  method  atiopted  for  secur- 
ir.R  the  ends,  E  the  coefficient  of  elasticity,  and  /  the  least  moment 
<>l  inertia  of  the  section. 

.VLso,  let  t  be  the  statical  compressive  strength  of  the  material 
of  the  strut,  and  take  <-/ir.    Then 


±    tFP     FP 
'""  T' dEl  ~  kl ' 


whore 


9E 

t  • 


(4) 
(6) 


622 


THEORY  OF  STRUCTURES. 


If  the  strut  under  a  pressure  B  were  not  liable  to  buckling,  it 
would  be  subjected  to  a  direct  thrust  only.    The  required  sectional 

n 

area  of  the  strut  would  then  be  — ,  and  the  unit  stress  for  an  area  F 

would  be  -5. 
F 

If  the  strut  under  the  pressure  B  is  liable  to  buckling,  its  required 
sectional  area  will  be  m,  since  T  is  the  mean  unit  stress  at  the  mo- 
ment of  buckling.  Let  x  be  the  unit  stress,  at  the  moment  of  buck- 
ling for  tlie  area  F. 

Assuming  that  the  unit  stresses  in  the  tv/o  cases  are  in  the  same 
ratio  as  the  required  sectional  areas,  then 


x: 


B    B  B 

f''t  f 


Therefore 


B  t       B 


FT 


=  fl 


(6) 


The  force  which,  when  uniforndy  distributed  over  the  area  F 
will  proilucc  this  stress,  is  Fx  =  /iB. 

Hence  allowance  may  be  made  for  buckling  by  substituting  for 
the  compressive  forces  in  P(iuations  (1)  and  (2)  their  values  multi- 
plied by  /I.     Thus  equation  (1)  becomes 


/iBi 


fiBi 


uBi 


*'  ^(>-»-.t)  •'(•+-.I:) 


-fiFi, 


(7) 


and  equation  (2)  becomes 


F- 


/iB' 


,S' 


'(!-'#«;) 


-j7;jr ,  if  B'  is  a  compression, 


fiB' 


(8) 


and 


F- 


B' 


B' 


,„.  ,  if  B"  is  a  c-'  .pression. 


(9) 


WEYRAUCH'S  THEORY  OF  BVCKUNG. 


62a 


If  ti<\,  equations  (1)  and  (2)  give  larger  sectional  areas  than 
pfiuations  (7),  (8),  and  (9),  so  that  the  latter  are  to  be  applied  only 
;vhen;u>l. 

Mcthod  II.  General  formulae  applicable  to  all  values  of  n  ma\r 
be  obtained  by  following  the  same  line  of  reasoning  as  that  adopted 
in  the  proof  of  Gordon's  formula.    It  is  there  assumed  that  the 

total  unit  stress  in  the  most  strained  fibre  is  pi(l+a^),  p,  being 

n 

the  stress  due  to  direct  compression,  and  pio^  that  due  to  the  bend-^ 
ing  action. 

So,  instead  of  employing  equations  (1)  and  (2)  when  fi<l,  and 
equations  (7),  (8),  and  (9)  when  //>!,  formulae  including  all  cases 
may  be  obtained  by  substituting  for  the  compressive  forces  in  equa- 
tions (1)  and  (2)  their  values  multiplied  by  1  +  ^. 

Thus  equation  (1)  becomes 


(l+;i)gi        ,,  ,    .  _, 


■.'(l+.,|) 


(10> 


and  equation  (2)  becomes 

(1  +  /«)B' 


F  = 


if  B'  is  a  compression,    .    .    (H) 


or 


~7  {\  +  u)B"\  *^  ^" '®  *  compression.  .    .    (12) 

v'[l-m'—f~) 


Equations  (7),  (8),  (9),  respectively,  give  larger  values  of  F  thai 
the  corresponding  equations  (10),  (11),  and  (12). 

For  WTought-iron  bare  it  may  be  assumed,  as  in  Chapter  IV,  that  t>,  -v*  -70  > 
k.  [irr  sq.  cm.,  and  m, -m'-}. 

Tho  value  of  I  is  given  by  formula  (5),  but  is  unreliable,  and  varies  iu 
l)ru(tioe  from  10,000  to  36,000  for  struts  with  fixed  ends. 

When  the  ends  are  fixed,  »-4«',  according  to  theory.    Heaoe 


l~An 


.E 


024 


THEORY  OF  STRUCTURES. 


Therefore  if  £-2,000,000  k.  per  sq.  cm.,  and  1-3300  k.  per  sq.  cm., 
^=23,926,  or  in  round  numbers  23,900;  24,000  is  the  value  usually  adopted 
by  Weyrauch. 

Ex.  15.  The  load  uptm  a  urought-iron  column  360  cm.  long  varies  bHtoeen 
a  cuinprcssion  of  50,000  k:  and  a  compresuwn  of  25,000  k.  Calculate  the  sec- 
tional area  of  the  column,  cmsum:  iq  it  to  be  Ist  solid  and  2d  hoUmv,  alloxvanct 
being  made  for  buckling. 

first.  By  eq.  (1), 

TjOOOO  400 


•■  1 

700(1 

+ix!nn) 

being  the  radius  of  the 

section. 

Also, 

'-t' 

Therefore 

F,      4     11 
/  "r"  "50" 

Hence,  by  eq. 

(4). 

360X360     11  _ 
'^       24000       50 

Thus  />>  1 .  and  by  eq.  (7)  the  required  sectional  area  is 
f, X  1.188 -♦^*X  1.188 -67.9  sq.  cm. 

Second.  F,-*f»-ff(r,'-r,'), 

r,  being  the  external  and  r.  the  internal  radius  of  the  section. 

Let  r,  -fl  cm.  and  r,-  7.92  cm.    Then 

«(»"i'-»"»')-57  43sq.  cm. 


Also, 


/-«^^'-: 


(r,'-r.') 


Therefore 
Hence,  by  eq.  (4), 


F    _4  4 

/  "r,'  +  r,'"l43.72ti4' 


360X.360  4  ,- 

''"    24000       143.7264" 


Thus,  in  the  latter   rase,  since    /<<1,  there  is  ao  tendency  to  buck- 


ling. 


:'f 


FLEXURE  OF  COLUMNS.  625 

If  the  area  is  determined  by  equation  (10),  ita  value  becomes  1.15X*?"- 

ii>  >c|.  cm. 

10.  Flexure  of  Coluiim8.-In  Art.  7  the  moment  equation  has 
been  expressed  m  the  form 

and  this  is  sufficiently  accurate  if  the  deviation  of  the  axis  of  the 
.trut  from  the  vertical  is  so  small  that  (|)'may  be  neglected  with- 
out  sonsible  error.  ' 

The  more  correct  equation  is 

EI 

P  ' 

P  Ixing  the  radius  of  curvature. 

Consider,  e.g.,  the  strut  in  Art.  7,  Case  I.     Then 


— a^w=  — 


/'      1     dd    do 


^°  my-p^Ts^Ty'"''^' 


"  iHing  the  inclination  of  the  tangent  at  M  to  the  axis  of  x.  and  da 
an  ("lenient  of  the  bent  strut  at  M.    Then 


Integrating, 


-a'ydy  =  m\  Odd. 

— „— =  COS^-C0S^n 


(1) 


I"  ing  the  value  of  (?  at  a  strut  end. 


I/'tsin:j=/i    and    sin 


=  /isin^.    Then 


aV 


2-  =  2;<2(l-8in2^), 


y-^cos^. 


(2) 


826  THEORY  OF  STRUCTURES. 

Let  Y  he  the  niaxiinum  dpviation  of  the  axis  of  the  strut  from 
the  vertical,  i.e.,  the  value  of  y  when  0  =  0  or  ^=0.     Then 

1  =  —  = (3) 

a  a 

1  d4> 

Again,  d<f  -=pdO=  -  -  .===. 

Hence,  if  I  is  the  length  of  the  strut, 

1='^  f^—^i-^^J-F,(4>) (4) 

Fj<p)  being  an  elliptic  integral  of  the  firitt  kind. 

Let  P'  be  the  least  thrust  which  will  make  the  strut  bend.    As 
.shown  in  Art.  7, 

and,  by  eq.  (4),  the  corresponding  value  of  the  modulas  ft  is  given  by 

FM)-l (5) 

I>et  the  actual  thrust  on  the  strut  be 

P^ri'P', (6) 

,;2  being  a  coefficient  >  unity. 

The  corre.sjKinding  value  of  the  modulus  is  given  by 

^■.(0)-=r,\;/7  =  r,"«=«2 (7) 

Bv  reference  to  Ix'gendre's  Tables  it  is  found  that  a  large  in- 
crea.se  in  the  value  of  /(,  i.e.,  of  sin  ^j  or  ^^o-  !>*  iieces.sary  m  order  to 
produce  even  a  small  increa.se  in  the  value  of  F^(<^)  and  therefore 


FLEXURE  OF  COLUMNS. 


•627 


of 


"(=J> 


Hence  as  soon  as  the  thrust  P  exceeds  the  least 

thrust  which  will  bend  the  column,  viz.,  P',  Oq  rapidly  increases. 

The  total  maximum  intensity  of  stress  in  the  skin  of  the  strut 
at  the  most  deflected  point 


P     Mz     P    PYz     .    2z   .    d„   ,— 


=  J+T=I  + 


/-=/  +  f-?-^/^. 


(8) 


z  Ijoing  the  distance  of  the  skin  from  the  neutral  axis,  and  /  being 
ciiual  to  -7. 

The  last  term  of  this  equation  includes  the  product  /J?,  which  is 
vcrx-  large,  and  also  the  factor  sin  ^,  which  increases  with  Oo  so 

that  the  ultimate  strength  of  the  material  is  rapidly  approached, 
and,  in  fact,  rupture  usually  takes  place  U]ore  the  column  has  as- 
sumed the  position  of  equilibrium  defined  by  the  slope  ^o  at  the 

I  nds. 

If  there  were  no  limit  to  the  flexure,  the  column  would  take  its 
position  of  equilibrium  only  after  a  n^imber  of  oscillations  about  this 
position,  and  the  maximum  stress  in  tht.  in-.lerial  would  be  neces- 
sarily greater  than  that  given  by  eq.  (8). 


^     V  1  —  n'^  sin^  <f) 


-Again, 

I.«'t  A'  be  the  vertical  distance  between  the  strut  ends.     Then 
\  =  -  n  y-V  sin'  <j> 

--J2E,J4,)-FJi>)\, 


!■■   v)  being  an  elHptic  integral  of  the  second  kind. 


628  THEORY  OF  STRUCTURES 

Hence  the  diminution  in  the  length  of  the  strut 


=  L-.Y=jF,(^)-£.(«^)t. 

If  the  column  has  an  initial  curvature  of  -,  the  moment  equa- 

po 

tion  may  be  expressed  in  the  form 

<-^)="-«(g-t) "> 

Po  and  -.-  Ix-ing  the  values  of  p  and  ~  when  M  =0. 

HinfjeH  /vh^/n.— It  is  assumed  that  the  line  of  action  of  the  thrust 
P  is  at  a  distance  d  from  the  axis  of  the  strut.    Then 


or 


d^}f    drya  ,  p-f  ,n^ 


where  a^  -  .!.   /» =  total  stress  at  the  distance  z  from  the  neutral  axis, 
hi 

and  /  =  streM  du«  to  .lirect  thrust  (- ^.).  x"  thiit  the  stress  due  tr 

it  is  also  assinncil  that  the  form  of  the  axis  of  the  column  l)eforo 
it  is  acti'(|  n\m\  by  the  thrust  /'  is  a  cun-e  of  shies  define*!  by  the 
(Mjuation 

y„  =  Jcosj 


(4) 


thi"  <iriKiii  lx'ii>K  half-way  iM-tween  th.-  t-iuls  of  the  strut,  and  J  bein? 
thr  iiia\iiniim  iniliol  deviation  of  .'..•■  axis  from  the  vertical,  i.e.,  the 
valiK'  nf  v„  when  /  ==(). 


riiercfore 


p   cosy, 


FLEXURE  OP  COLUMNS. 


629 


and  hence,  by  eq.  (3), 


—  =  _a2(y +rf)  _  j_  cos -j- 


A  solut'  jn  of  this  equation  is 


COS-p 

.   ,     .COS  ox      ^  / 


...     (5) 


.    .     (6) 


Now  ^  is  always  small  for  such  values  of  /  as  would  constitute  a 
safe  working  load,  and  therefore 

cos  2"  - 1  -  -g-,  approximately, 
wi  that  eq.  (6)  becomes 

,7  +  f/=rfco8ajn-   —  1      +Jcosy(l--Y)     , 

or 

y  +  d  - d  c(»  ar^l  +  -y  j  +  J  cos  y  (^  1  +    j  j ,  approximately.    (7) 

t-<t  Y  be  the  maximum  value  of  y,  i.e.,  the  value  of  y  when 
/    I).    Then 

^-«'K^+i)'-' ^«> 

ilcnce,  by  oq.  (3),  the  total  inaxiimnn  intensity  of  stress 

^-%^~))    and    r^^t^, 

i:<l.  (9)  is  a  «iua<lrutic  from  which  /  niiiy  Iw  found  in  terms  of  yt. 
A-  .1  first  appn)ximution,  />  may  Ix*  «ul>stitiitf>d  for  /  in  the  last  term 


.  v 


(m 


THEORY  OF  STRUCTURES 


of  tho  jjortion  witliiti  brackets,  the  error  iK'ing  in  the  direction  of 
safety. 

Fixed  £«f/s.— Let  M^  he  tlie  Jiioment  of  fixture. 

Kij.  (3j  now  lieconies 

Assumliip  Mc.iiii  I'iMt  til"'  initial  form  of  the  axis  is  a  curve  of  sines, 
the  sohitioii  ot  llic  l:i>I  r<|ii;ition  is 


cos  -7- 


Mi     (       .Ui\posflX  ■   / 

C«»S  .y  1 J 


(11) 


Initially, 


and 


j/„  =  J  cos  -J, 

'/'/•»    ,  /  / 

,  -  wlien  x=  -,  or  =  — -;. 
(//  2  2 

«  sin  .r        Jt 


d    .  .,,,., 

r  Is  e(iii!il  to  ,  -  wlien  x=  -,  or  =  — -;. 
ax        '  (//  ?  '* 


Hence 


1  f  \  •>  / 


IMT 


COS  -      1  -    r- 

2  r^ 


.      W,  JJ  12-,t2 


y 


.    .     (12) 


Again,  the  value  of  y  at  the  |Niint  /  --=()  is 

,.     /,     M:\a-P       /,     f,-TA 

Also,  if  /i|,  /);.  are  the  int.il  iiiaximuni  intensities  of  stress  at  the 
enil  and  at  the  most  dilli-ctcd  |Miinl,  then 

'-/...,,(,,.';■)„.,. ,„. 

'■-'..  -»(r,,/,'/,i-,.,o.;    ....     05, 


iiiid 


two  ein'jitions  from  wliich  /  mav  In-  found  as  Ix-fore, 


FLEXURE  OF  COLUMNS. 


631 


Tho  following  conclusions  are  drawn  from  the  above  investi- 

pitioii: 

Firxt.  The  actual  strength  of  a  column  depends  partly  upon  known 
facts  as  to  dimensions,  material,  etc.,  and  jmrtly  upon  accidental  cir- 
cuiiislances. 

Second.  Kxperiments  upon  the  crippling  or  destruction  of 
coliiiiins  cannot  be  expected  to  give  coherent  results  when 
ii|>l>li('il  to  the  determination  of  the  constants  in  such  an  equation 

a^  m. 

Third.  It  is  a  question  whether  p  should  be  made  the  elastic 
hunt  of  t'.e  material  and  the  working  stress  a  definite  fraction  of  the 
<ornsiK)nding  value  of  /  derived  from  eq.  (9),  or  whether  p  should 
!>«•  i!u"  allowable  skin  working  stress,  and  the  value  of/  be  found 
\>y  means  of  the  same  equation.  The  former  seems  to  be  the  more 
Injrital  iussumption. 

Fourth.  It  would  api)ear  that  the  strength  of  hinged  colunms  is 
lik.lv  to  Im'  nmch  more  variable  than  the  strength  of  colunms  with 
tix.il  .n(|s,  a.><  it  de|)end.s  u|jon  two  variable  elements,  d  and  J,  while 
1 1 II  tiid  fixture  eliminates  d. 


PiiiBT  Eluptic  Intforal,  F^(<fl). 


1^ 


.-(» 


<i  iim)  I)  mid 

II  IIH7  (I  I1K7 

II  i::.  ti  I7.^ 

11  Mi  it  ■-••« 

I  :i4tt  (■  ,<iw 

II  ».tfl  o  4;w 

II  Mi  O  .W4 

II  -ill  II  III) 

II  tittx  o  •Miy 

il  rVi  I)  Twi 


'■ '        II  >t7.1  1   0  K74 

•^         <>  <NU)  j   (I  M)t 

'►'        I   1117  i    1   (MV 

"        !    114  I    1S7 

I  :ii  1  tn 

t  um  I  I  3\2 

"          I    MM  I    I    4<||| 

««         I   4H4  I   41IT 

1  .'iTi  1  ara 


■•8  i    ,-J 


It  INMI 
CI  (W7 
(I  17,', 
II  Md 


0  4.W  i 

0  ,52,5  i 

0  1112 

(1  7III1 

(»  7H0 

0  H77 

0  Oft.'. 

1  *»a4 

1    14.1 

1   233 

1  »3I 

1   410 

1   4M 

I  mm 

It  nan 

0  (187 
It  17.5 
II  252 
0  XVt 

0  4.W 
0  ,520 
U  1114 
II  7a« 
<l  792 

n  IW3 

0  »T2 
t  Itli2 

1  I. VI 
1.344 

t  n.'w 

I   427 

I  aiB 
I  nut 


0-* 


§8g  gr 

0   17.5  0   17.5 

0  2H2  0  a» 

0  ,'i,*iii  :  It  ;<5i 


0  43» 
0  827 
0  BI7 
0  707 

0  7«N 

n  HM) 
It  UMt 

1  074 
I    IW 


n  4«i 

ct  •.29 
0  112)1 
(I  712 
O  mn 

0  MM 

n  uwi 

1  iHki 

I  IN7 
I   2HA 


I  M7  i  I  :Ma 

I  4,52  t  4N.5 

1  M7  I  .5M 

I  )t4a  I  flMtt 


»i-.«    ji-.r 


n  ono 

O  (»7 

(I  17.5 

II  2aT 

U  3.52 


0  wm 
0  mr 
0  17.-. 
II  2«n 

(I  3.VI 


I  441       (I  443 

(I  aas  I  o.asA 
u  n34  i  (I  nao 

O  Tin  '  It  737 
0  )tl4  :  0  830 


0  »il 

1  010 
I  113 
I  31,5 
I   .120 

I  43il 

1  5.14 

I  A4.1 

I  7A2 


K-J» 


I  4IIH 

I  HON 

I  7,11 

I  M,54 


0  9S8  O  »47 

1  034  1  000 
I  l«2  >  I  17K 
I  2,54  ;  I  mrj 
1  37U  I    (31 


^-.l 


11  (100 

0  0X7 

0  175 

2«V4 


0  (nm 

0  0«7 
0  175 
0  2   1 

0  351      I)  .i5A 

0  44.5  0  44S 

0  .VW  0  M4 

(I  KM  (I  )M4 

'I  7:«i  !  0  74*. 

0  K.W    0  sa« 


0  974 

1  l/tW 
1  iXi 
I  377 
I   .VM 


1  ,5««  j   1  7U3 

I  70,5  i    I  MM 

I  MM      2  1177 

I  QOa      3  373 


;i-l 


0  IMO 

a  o»7 
o.i7a 

It  3«8 

II  ;«6 

0  4AI 

II  549 
It  IVS3 
(I  7B3 
II  NN1 

1  oil 

I  1.54 
I   317 

I  aoA 

1   71S 

3  on 

Z  43A 
3       I 


63J 


THEORY  OF  STRUCTURES 


ShX'ONU    1;LLII'TIC    lNTE<iHAL,    E^(<j)). 


* 

^-0 

f-.l 

II 

-.2 

;.-.3 

/i".4 

d-.S 

,-.6 

I--.7 

C-.H 

»-.» 

ii-1 

u" 

0  (WO 

0  (MX) 

0 

INN) 

0  (NN) 

0  000 

U  000 

0  000 

0  000 

0  000 

0  000 

0  0(N) 

5' 

(1  i(»7 

0  U»7 

(1 

IIS7 

IMI»7 

U  0N7 

U  UH7 

0.OK7 

0  0X7 

0  0«7 

0  087 

0    (IK7 

10° 

(1  1   S 

0  ;7:. 

0 

174 

U   174 

0  174 

0  174 

0   174 

0   174 

0  174 

0  174 

0   174 

IS" 

0  JliJ 

0  jr.:; 

.1 

202 

U  2)i2 

0  2(>1 

0  2(il 

0  2H1 

(1  2(iO 

0  2C.0 

0  2.19 

0  259 

ao° 

0  lil'.l 

:i  M» 

(1 

.!49 

U  34N 

0  34  s 

0  347 

0  347 

(1  34(i 

0  ..■)41 

0  343 

(1  342 

as" 

0  m\ 

0  43(1 

0 

IHti 

0  43.1 

II  434 

0  4.33 

II  431 

II  43(1 

0  42»* 

0  425 

0  42:1 

30- 

(1  .W4 

0  .'.23 

1) 

.■ij:! 

0  .121 

U  .■j2iJ 

0  SI-* 

0  .1 1.1 

0  .112 

11  .109 

0  505 

0  .VW 

35° 

0  (ill 

(1  (>l(l 

II 

iNm 

0  W7 

11  IKI.'i 

11  tKI2 

1)  .MM 

0  1(13 

0  .1l« 

U  .INI 

0  .174 

40" 

(1  i»».s 

(1   CltK 

0 

IHXi 

0  IVIKt 

II  (ItNl 

II  (IS.-. 

11  (.7(1 

11  (172 

II  (.04 

0  ai4 

0  (H3 

45° 

0  7H.) 

0  7k', 

0 

7M 

II  779 

0   773 

0  71.7 

0  7.19 

0  74»* 

0  737 

0  723 

0  707 

so" 

0  S73 

0  M72 

0 

MAO 

0  ««i4 

(1  S.-.7 

0  H4K 

0  H.37 

0  H2.3 

0  HON 

0  780 

0  7fi« 

55' 

IVWiil 

n  o-w 

1) 

U.").'i 

1)  94» 

(1  939 

0  92H 

0  914 

(1  H9.1 

0  K7.1 

0  H50 

0  NI9 

bj" 

1  017 

l.04ti 

(Ml 

1   032 

1   (121 

1    INM 

(1  9K9 

II  Wkl 

0  940 

0  907 

0  NWi 

65° 

1   i:u 

1    1.12 

I2ti 

1    IKi 

1  \(n 

I    0S« 

1  (Hi,3 

1  1133 

1    (Nil 

0  960 

0  906 

70" 

1   L'22 

1  2IU 

212 

1    2<NI 

1    IM 

1    KM 

1    135 

1  099 

1   (NiO 

1  008 

0  940 

k: 

1   MW 

1  .•«)ti 

297 

1   2(« 

I   264 

I   240 

1   207 

I  lai 

1    117 

1  053 

0  9<'.« 

1  :i9i> 

1  »«( 

3h;» 

1   3(17 

1   344 

1   31A 

1  277 

1  227 

1    172 

1  005 

0  l»>5 

85" 

1   4K4 

1   4HU 

4l>H 

1   4.'>0 

1  424 

1  .392 

1  347 

1   2X9 

1  22.1 

1    135 

0  996 

00- 

1  ri7i 

1    5*M) 

.i.'>4 

1    533 

1  .'>U4 

1   4«I7 

1   417 

1  331 

1   27H 

1    173 

1  UUO 

KXA.MI'LKS. 

1.  Tho  sortional  arrn  of  a  pillar  is  144  s(|.  ins.,  and  thr  pillar  parries  a  load 
of  4(XXI  llis.  Kind  the  noriiial  aii'I  taiif^cntial  iiiteii,><iticM  of  stress  on  a  plane 
inclined  at  20'  to  the  axis.  Anx.  3.'.'.")  llw. ;  K.93  llw. 

2.  A  short  rust-iron  coiiinin  of  0  ins.  external  and  4  ins.  internal  diameter 
'■arries  a  lo.'id  of  JO  tons,  the  line  of  artion  of  the  resultant  U'injf  12  ins.  fnmi 
a\is  (if  coluniii.     Find  niaxiinuni  and  niiniiniiin  stn-ss<>s. 

AiiK.  1.").;J7  and  13.82  ton.s/si|.  in. 
J.   A  rvlindriral  j.illar  fi  ins.  in  diameter  sup|Ktrts  a  load  of  400  llw.,  of 
whii  li  the  cciilre  of  ^cavity  is   J   in.  froin  the  axis.     |)eterinine  tiie  greatest 
ami  li':i-<t  intensities  of  slres.s  uikiii  iiiiy  traiisvers«'  .sertion  of  the  pillar. 

Alls.  J.'.jI  llw.;  25!?lbs. 

4.  Cilculate  the  lireakiii);  weight  l.y  iliHliikinsoii's  formula  of  a  solid 
rill. ml  rjist-iron  pillar  20  ft.  in  leiiRth  and  10  ins.  in  diameter,  (1)  both  ei»d« 
U'lnu  .s«>iiin'ly  tixeil:    i2i  Ixith  eiid-^  U-int;  Iniifccd. 

Alls,  il)  0.")|.»  tons;  (2)  414.04  tofw. 

5.  iK'terrnine  liy  IliHlKkinsiiti'^  fnrniiiiti  the  diaiiu-ter  of  a  solid  wniu|{ht- 
iinii  pillar  i'i|iial  in  length  and  sirengtli  to  that  in  the  pn'e«'din(?  (|iie!<tion. 

.ln.'<.  T.S.'i  ins. 
ft,  CmIi  iihili'   the   lireukiiic   wi'ight    l.y    lliMltrkinson's   formula  of  a  .s<|Uarp- 
lv;irinir  linllow  f.i^i-iron  ciiliiinti  "I   I  i:i»,  cxtiTti:!!  ;iiid  3  ins.  Intern '.I  diaiiii-ter 
.1  n.s    42<fi    /'•'  for  long  and  420.5 h  (2124  +  .T-'W'-')  for  medium  <ok 

7.  A  sqn.'in-lHNirinir  linilier  |Kwt  .Is  m  'AH  in.  in  section  is  4  ni.  high. 
\Vh:it  Imid  will  it  ^.ifi  ly  U-ar'  il^e  Hodgkinson\  formula.)     Ann.  10.41)8  k. 

8.  Find  l.y  llodgkiniwin's  furtnula  tli  loud  wiiich  can  Iw  carried  by  a  caxt- 
iron  column,  with  s(|uare  endit,  .'j  m.  high  and  I'JU  mm.  in  diameter. 

Ann.   I.'i,759k 


EXAMPLES. 


633 


0.  Find  by  Hodgkinson's  method  the  breaking  weight  of  a  pin-bonring 
solid  fusUiron  pillar  6  ins.  in  diar.  and  liO  diars.  in  length. 

Ans.  172  tons. 

10.  \  hollow  cast -iron  squurc-lx-ariiifi  rolunni  of  'J  ins.  external  and  7  inc. 
iMi.irial  (liar,  is  24  ft.  long.  U-tiTniiiic  by  Hodgkinson's  method  the  load 
It  will  safely  carry,  10  being  a  factor  of  safety.  .',„,,.  ;J2  tons. 

11.  What  is  the  length  of  a  solid  cast-iron  |)in-l)earing  eolunin  of  .i  in.s. 
liiatncter  which  fails  under  a  IoikI  of  12,.{20  lbs.?  Ang.  lB.7.5ff. 

II.  With  .S  as  a  fuctorof  safety,  the  .-uife  working  load  on  a  solid  »<ii:.ire- 
Uaring  pillar  10  ft.  long  is  .5  tons.     Find  its  diameter.  An»    2.73  !•.«. 

13.  Find  the  breaking  weight  of  a  solid  square-bearing  aist-iron  coiunm 
71  ins.  in  diameter  and  Iti  diameters  in  length,  the  cru.shing  strength  of  tlie 
iron  iH-ing  40  tons.  j^ns    :()78  ton.s. 

14.  .\  hollow  east-iron  column  of  8  ins.  external  and  G  ins.  internal  diarn- 
(KT  is  18  diameters  long.  If  10  is  a  factor  of  safety,  Hnd  the  load  it  will 
sjifcly  carry  when  both  ends  are  (a)  square-bearing,  ib)  pin-bearing. 

Am.  (a)  57.14  tons;  (h)  28.8  tons. 

15.  Determine  the  breaking  weight  of  a  solid  cast-iron  pillar  9  ft.  in  height 
and  4  ins.  in  diameter  when  the  ends  are  (a)  square-bearing,  (6)  hinged 

Am.  (a)  159  tons;  (h)  .")4.16  tons, 
lu.  Determine  the  breaking  weight  of  a  solid  round  pillar  with  both  ends 
tiniily  s».cure<l,  10  ft.  in  length  and  2  ins.  in  diameter,  (1)  if  of  cast  iron;   (2) 
11  of  wrought  iron;    (3)  if  of  steel  (mild). 

Am.  25,142.8  lbs.;  43,516.48  lbs.;  59,130  lbs. 
17    A  solid   pin-bearing  cast-iron  strut   15  ft.  long    is   to   carry  a    load 
of  .'(» tons.     If  (j  is  a  factor  of  safety,  (ind  the  diameter  of  the  strut. 

.4  ns.  6J  ins. 
i3.  A  sf)lid  or  hollow  pillar  of  cast  iron,  wrought  iron,  or  mild  steel  is  to 
I-  a.  siriuhI  to  carry  a  .Mimly  loml  of  .10,000  11)8.  Determine  the  necessary 
■  i  iiii.i.r  in  each  case,  li  lN>ing  a  factor  of  safely.  (The  pillar  is  to  lie  12  ft, 
li  u'li  and  the  metal  of  the  hollow  pillar  is  to  ire  }  ni.  thick)  Determine  the 
!■  1.1  that  will  pnKluce  a  maximum  stress  of  9000  lbs.  ixr  square  ii-ch  in  tho 
-"iiil  n|,^.|  pillHr, 

Ana.  SiAid:     3.fi8  ins.;  .3..39  ins.;  3  ins.:  24.0.55  lbs. 
Hollow:  4.<.N»  iri.'^. ,  S.."»  nis.;  3.9  ins. 
10.  Design  a  square-lwariiig  pine  pillar  .'O  ft.  long  to  carry  a  load  of  (KKtO 
:'        I  l«ing  a  factor  of  safely.  .\us.  U.^|uinHl  area -33.6  «<].  i,is. 

20.  Ihtermine  the  strength  of  a  mild-steel  pillar  7  ins.  in  diameter  and 
:ii  liameters  in  leiiitth.  (<i-  uS.".  /~fl(t.7.')0  IIh.)  Find  the  diamet.  ;•  of  each 
if  'no  iiillars  e(|iiivalent  to  the  single  pillar.  Ann.  4.8  ins. 

.(.  Determine  the  breaking  weight  of  an  oak  pillar  9  ft  high,  1 1  mh.  wide. 
''"'""*•''"'•''■  .In.v    i;i.S.l(i<)||« 

ii.  What  weight  will  Ix-  safely  iK.rne  by  a  |>illar  of  dry  0,1k  subject  to 
\     i;iii..ii,  10  ft.  high  and  (i  ins,  sspiare,  10  being  a  factor  of  sitfety'' 

.!«'.  9im9nMi 

iS.  Calculate  tin-  breaking  weight  of  a  solid  e.-iHt-iron  pilkr  JO  ft   in  length 


t)34 


THEailY  OF  STRUCTl'RES 


arul  10  ins.  in  dianu<tor  when  (a)  both  ends  arc  square-bearing,  (h)  one  end 
is  sduare-bearine  and  the  other  pin-bcaring. 

'  Ans.  (a)  1150.05  tons;  (b)  tj62  tons. 

J4.  Find  thf  diani.tcr  of  .-i  wooden  column  .'O  ft.  long  to  supfxirt  a  load 
of  10()00  lbs.,  10  lniiin  a  fa.lor  of  safety  and  l>oth  ends  of  the  eolumn  being 

absolutely  (ixeil.  .     ,      ,       ,„      ,,,      ,•■  . 

25  Find  the  breakiiip  >tres8  per  square  ineh  of  a  4"X4"  solid  WToujfht- 
iro.i  pillar  for  lengtlLs  of  :..  10,  15,  and  20  ft.,  the  two  ends  being  al)8olutely 
,,^,.,1  Ans.  :W,4SX  ll)s. ;  -JT.tHVi  lbs. ;  21 ,402  lbs. ;  10,36.'$  lbs. 

a6.  Compare  the  bn-aking  weight  of  a  .solid  scpiare  pillar  of  wrought  iron 
•H)  ft  long  and  •>  ins.  s<iuare  with  that  of  a  solid  rectangular  pillar  of  the  same 
nriterial    the  s,-clion  Ix-ing  9  \m.  by  4  ins.     Anx.  S45.217  lbs.;  589,090  Ibe. 

27.  \  solid  round  pillar  i.f  mild  8t<><-l  10  ft.  high  supfjorts  a  Mewly  load 
of  'iO.tKK)  lbs.     If  the  factor  of  safety  is  «,  what  is  its  diameter?     An».  3  ins. 

"28.  Find  the  diameter  of  each  of  four  i)illars  of  the  same  matirial  which 
laav  U-  substitu.e<i  for  the  single  pillar  in  the  preceding  example. 

Anf.  2.04  ins. 

29.  Find  the  breaking  weight  of  a  cylindrical  strut  of  3  ins.  diameter  and 
!0  diameters  in  leimth.  IvMh  ends  lieing  hingi-d.  Anx.  67,4:10  lU. 

30.  \  hollow  cast-iron  pillar  wi'h  an  external  diameter  of  0  ins.  is  to 
|„  Mibstitut.d  f'T  the  solid  pillar  in  the  preceding  example.  IVterniine  the 
thickness  of  the  metal.  •*"«•   i  '"• 

31.  .\  hollow  cast-iron  pillar  12  ft.  in  height  lias  to  support  a  steady  load 
of  33^000  lbs.;  its  internal   diameter  is   5i  ins.     Find  the  thickness  of  the 

melaV.  the  factor  of  sjifetv  iH-insr  6.  ,  ,         u  ''"",',  ""*  '"1 

32.  A  solid  wn.ught-in.n  pillar  is  to  b.^  sulwtitutinl  for  the  pillar  in  the 
preci.linii  example.     Find  its  diameter.  *"•'<•  3i  ins. 

33  The  exiernal  and  internal  diameters  of  a  hollow  cast-iron  -olunin 
I'  ft  in  length  are  /*  and  il>  resiKctively,  the  load  ui«.n  the  column  is 
2:.00()  lbs.  If  th-  fa<tor  of  -afety  is  1,  find  />,  (n)  when  Imth  end.-!  of  the 
eolumn  are  abs<.liit<lv  lixcd;    ('<)  when  Ijolh  ends  are  hinge.1. 

.\nn.  (a)  4,03  ins.;  (h)  .5.4  ins. 

34  \  hollow  r:i-i-iro!i  pillar  of  (i  ins.  external  diameter  and  20  diameters 
in  h-ngth  has  to  carrv  a  loi.l  of  67.200  lb.s.  If  S  is  a  factor  of  sjifety.  lin.l 
the  thickness  of  the  metal?  /«'•  -67  l"- 

35  \  h..ll..\v  circular  mild-steel  -'quare-lK-aring  column  2**  ft.  long  and  ot 
6  ins  ex'crnal  diameter  has  to  carry  u  loa.l  of  50  tons.  Find  the  thickness 
of  l».e  •aetal.  »  Uing  a  factor  of  safety.  Aw,    AjV\  m. 

35.  .\  squan-U-aring  hollow  wrought -iron  cylindrical  strut  10  It.  long  aiM 
of  li  ins   external  diameter  fails  under  a  l«md  of   2.7  tons.     Find  the  thi.  k 

f  .1.    Mw.f.l  Am    .072  ins. 

IM'ss  of  the  met.ii. 

37  \  «.id  rectangular  wrought-ir.m  strut  10  ft.  long  and  .'  in-.  thi'K 
fHiU  under  :i  ;...i.l  of  :U)  long  tons.  If  the  ends  an>  piu-b  aring.  what  sh.^uM 
»*  the  width  of  the  strut'  Ans    5.4  ms 

jg  The  least  breadth  of  an  angle-iron  is  3  inn.,  and  its  length  i«  90  in* 
It  i«  fixed  at  U.tb  cnd».     What  u  the  safe  load,  6  being  a  laetxir  of  safety? 

Ans,  3|  tona. 


EXAMPLES. 


635 


39.  ^^'hat  18  the  breaking  weight  of  a  east-iron  stanchion  of  a  regular 
( ruciforin  section  and  15  ft.  in  height,  the  arms  being  24  ins.  by  1  in.? 

Ans.  2,811,215  lbs. 

40.  Ead.  of  the  pillars  supporting  the  lowest  floor  of  a  refinery  is  16i  ft. 
hieh,  is  of  a  regular  cruciform  section,  and  carries  a  load  of  240,000  lbs.;  the 
i(.i:il  length  of  an  arm  is  14  ins.  Determine  its  thickness,  the  factor  of  safety 
'"■""e  '«•  Ans.  2..598  ins. 

41.  Find  the  load  which  can  be  safely  carried  by  a  colunui   30  ft.  high 
(if  the  .action    shown  by  the  figure,   which  consists  of 
/./..  14"Xi|"sidc  plates,  /our  Z  bars,  each  6"X34"Xj", 
and  otic  web  plate  8"xi".    Use  straight-line  formula 

/.    17,100-57-.  /I n«.  726,000  lbs. 

4J.  P'ind  the  safe  load  on  a  square-bearing  column  20 

ft,  lung  coin|)osed  of  two  Ti-in  X20-lb.  jier  lineal  foot  me- 

(liiuii  st<fl  channels  placed  back  to  back  and  10  ins.  apart, 

4  Uirn!  a  fiictorof  safety.     S<>ctional  area  of  each  chaiuwl 

1 1  .s  sq.  ins.  Am.  100,350  lbs. 


c 


-^ 


^ 


Fio.  580. 


43.  Design  a  medium  steel  strut  L")  ft.  long  and  consisting  of  four  angles 

t(.  rarry  a  load  of  170,000  lbs.;  use  the  formula  p-16,000-60-. 

r 

Wis.  Try  6"X31"Xi"  angles  with  the  short  legs  back  to  back  and  spaced 

]  m.  uiHirl  for  lacing.    Section  recjuired -IS.-W  sq.  ins. 

44.  Wli.it  should  be  the  distance  Ix'tween  the  faces  of  the  long  legs  of 
the  ancles  in  the  preceding  ex.imple  >m  I  hat  the  moments  of  inertia  with  rc!<pect 
(..  till  :i\es  through  the  ecnlri'  of  gravity  of  the  section,  parallel  to  the  short 
I' .1  Idiic  legs,  may  be  equal'  Am.  7.62  ins. 

45-  The  action  of  a  ehortl  segment  22  ft.  long  consists  of  one  18"  Xi" 
...\.r  and  two  12-in.  v44-ll>.  |)er  lineal  f<x)t  channels.     If  the  allowable  stress 

/■  H  that  given  by  pjl  +  .T^fj^Jy)   \  -10,000,  find  the  load  which  may  be 

applied  at  the  centre  of  gravity  of  the  section,  .in.t.  304,325  Ujs. 

4ft.  What  rtduld  ln»  tla-  additional  .stress  if  the  load  were  applied  at  the 
'I  hire  of  t lie  channels'.'  ^na.  .541011)8. 

47  iMermine  th«>  bu-kling  strength  of  a  2-in.  rouml  steel  bar  (iO  diam- 
«■!.  r~  in  length,  with  pin  en<is.  If  the  stress  is  Tiowhcre  to  exceed  12,000 
'h--  \n-T  .-Minan-  inch,  Ki»l  the  load  for  which  the  nmximum  deviation  of  hne 
<  f    .ad  from  axis  will  not  exceed  .125  in.  An*.  15,845  lbs.;  45,792  lbs. 

I 

4*.  <'ale\ilate  the  buckling  load  of  a  piece  t)f  ea.st-imn  pipe,  length  24  ins.; 
'  \  .nial  diameter,  4.4  ins.;  internal  diameter,  3.0  ins.;  rounded  ends. 

.inn.  l.W  t.ous  for  harrl  to  «Jft  tfxm  for  soft  cast  iron. 

4Q.  .\  steel  strut  10  ';.  long  '-oiisists  of  two  tens  hack  to  liack.  each 
!  I"xl".  Taking/  t»,000  Um.,  a,-Tfb«,  and  6  as  a  factor  of  safety 
11  the  working  load  (a;  when  the  strut  has  two  pin  ends;  (6)  wheu  it  has 
< «     fixed  ends.     (£-29,000,000  lbs.) 


636 


THEORY  OF  STRUCTURES. 


Also  find  the  deviation  of  the  axis  of  the  load  from  the  axis  of  the  strut 
so  that  the  niaximuin  stress  in  the  nietul  may  not  exceetl  10,000  lbs.  per  square 
inch.  Ana.  («)  'J^.-Wti  lbs.;  (6)  52,230  lbs. 

Deviation  =«..>4'J  in.  in  (ti)  and  .l.'>8  in.  in  (6). 

50.  .\  .soliti  \vrouf!ht-iron  strut  20  ft.  Iiich  and  4  ins.  in  diameter  has  one 
eiul  fixiHl  and  the  other  1  erfei'tly  trw.  Find  the  deviation  of  the  line  of 
action  of  a  i<»ad  of  lO.OtX)  lbs.  from  the  axis,  so  that  the  stress  may  not  exceed 
10,000  lbs.  |ier  s<|uan'  inch,  K  beinp  27,000,000  lbs.  .in.t.  l.S  ins. 

51.  Find  the  buckling  load  of  a  |iin-lx>aring  steel  strut  of  3  ins.  solid  square 
section  and  lOS  ins.  long,  E  being  :iO,000,000  lbs.  per  square  inch. 

An*,  hy  Euler,  171,485  lbs.;  by  Cordon,  168,370  lbs. 
5J.  Find  the  safe  load  on  a  rolled  tee-iron  strut  «"X4"Xi",  10  ft.  long, 
fixed  at  one  end,  free  at  the  other.  Ant.  '22.65  tons, 

53.  Find  the  strength  of  a  square-bearing  double-tee  section  column 
22  ft.  in  length.  The  equal  2.V'X.y<>"  flanges  are  connected  by  a  12"X1" 
web.  If  the  allowable  stress  is  not  to  exceed  10,000  llw.  per  scjuare  inch,  find 
the  greatest  deviation  of  the  axis  of  the  load  from  the  axis  of  the  column. 
(£,'=31,.')00,0(J0lb8./sti.  in.;   44,000a, -1  and  / -53,770  Ibs./sq.  in.) 


54.  .\  round  steel  strut  1  in.  in  diameter  and  100  diameters  ia  length 
is  pivoteil  at  Inith  ends.  Find  Ruler's  value  for  its  ultimate  strength,  taking 
/•;  -  2<.>,40O,0OO  lbs.  \n'T  s<juare  inch.  Am.  1420  Uw. 

If  an  end  load  acting  by  its<'lf  produces  a  maximum  fibre  stress  of  1000  Ibfl. 
|NT  square  inch,  and  a  lateral  load  acting  by  itself  pn>duci  a  maximum  fibre 
.stress  of  .VXI  lbs.  [K-r  scpiare  inch,  what  will  lie  the  maximum  fibre  stress  when 
the  two  loatls  act  together?  Ans.  2114  (c.)  and  114  (t.)  llia./sq.  inch. 

55.  .\  hollow  cast-iron  column  with  two  pin  ends  is  24  ft.  high  and  hast 
:;  nican  diameter  of  12  ins.;  it  carries  a  load  of  .SO.tKK)  ll>s.  Find  the  |»ro|>er 
tlii<'kncss  i>f  the  metal,  It)  lieing  a  factor  of  sjifely.  If  the  deviation  of  the 
line  of  action  of  the  loa<i  from  the  axis  is  1  in.,  find  the  maximum  stress  fier 
s«|uiire  inch  in  the  metal,  E  Ix'ing  17,(KJ0,(MM)  ll»s. 

.Im«.  1.2s  ins.;  2661  lbs.  jjer  srj.  in. 

56.  Find  the  crushing  load  of  a  solid  wrought -iron  pillar  3  ins.  in  diameter, 
10  ft.  high,  jifiil  fixed  at  Ixilh  ends.  Calculate  the  deviation  which  will  product 
a  inaxiiniiin  sires-*  in  the  metal  of  !H)tK)  lbs.  [ht  s<)uare  inch  under  loads  of 
(<i)    1.">.(KII)  Ills.,  (/,!   :«),(MH»  Ihs.,  /•;  lH-inK2<,t,»KM),00t)ll)s. 

.\ii».  .■{lC.,.-|tK)  lbs.;  (<i)  l.l.-iS  ins.;  (b)  1.101  in. 

57.  S'llve  t!ie  (irecedinm  example  <in  the  assumption  that  the  column  has 
tw..  |.in  I  Ills.  ,1,,,,.  7!»,|2.')  lbs.:  (fi)  .OS.")  in.;  (6)  .7.V)  in. 

58.  \  tiDJliivv  i:isi-iriiii  column  nf  1<1  iiw.  external  and  Iti  ins.  internal 
diamcler  li.i>  nimnle.l  I'lids,  is  Ml  ft.  in  |erii;tli,  ;irid  is  stayed  laterally  at  iii- 
liTx.iis  (if  .'(»  11.     liml  the  le;i«t  fiirci'  which  will  c.iuse  the  column  to  lieiid. 

.\na.  H.TJlt.lHHt  lbs.,  A'  Ixins  Itl.OdO.OtlO  lbs, 
r,o.   iiiiil  ilie  Kit-alving  wiight  nf  ,1  wniuchl-iron  column  3  ins.  in  diameter, 
lOO  ill-,  in  ii'iiirtli.  .'iiiii  liiiiKcd  at  iHitli  en<ls. 


EXAMPLES. 


637 


Also  find  the  deviation  for  which  the  stress  in  the  metal  vnW  not  exceed 
10,000  lbs.  per  square  inch  when  the  load  ui)on  the  column  is  40,000  lbs. 

Ans.   112,434  Ibs.^  .2.5  in. 
to.  \n  8-in.  channel  section  strut   with  4.02" X-.W  flanges  and  a  web 
I-'  in.  thick  has  both  ends  fixc<l.     Taking  £=.30,000.000  lbs.  per  square  inch, 
MM.l  the  ultimate  stress  at  4.5,000  lbs.  ]x-t  square  inch,  find  the  buckling  strength 
for  lengths  of  8,  10,  12,  and  15  ft.    What  should  the  corresponding  lengths  be 
if  lK)th  ends  are  hingi-d?  Aru.  77.9,  49.9,  34.6,  and  22.2  tons ;  one-half. 

6i.  Find  the  buckling  strengths  of  the  following  tees,  with  pin  ends: 


(a)  Area- 1.67  sq.  ins.' 
(6)     "    -1.95  "      " 


(c) 


-2.04 


--  50; 

r 

-  =  100. 

r 


Take  E  - 15,000  tons/sq.  in. 

Ant.  (a)  99  tons;  (6)  51.4  tons;  (c)  30.2  tons 
62.  The  22-ft.  panel  length  of  the  top  chord  for  a  deckn^pan  has  to  carry 
.1  load  of  .'}.'iO,000  llw.  in  direct  compression  and  a  uniformly  distributed  load 
..f  2<>.»00  ll)s.     Design  the  section,  the  allowable  stress  p  in  compression  being 


„v..nl,y,,]l+,^^,.^ 


\  r '    t 


10,000. 


.\ns.  Try  a  section  consisting  of  onr  24"  X  |"  cover,  tu-o  upper  3i"X.3i"X|" 
AU!i\,-<.  t'l-o  lower  (i"  »:  |"x  I"  angles,  two  24"X  J"  webs,  allowing  play  between 
u(l)  edmes  and  faces  of  angles.     Sectif)n  required  "=.')0  sq.  ins. 

63.  IMiijorrril  roncrik  rolumnn  l,xs  than  25  (litimrtrrs  in  length,  uhirh  in- 
rhiilis  all  ii-iunllij  jound  in  prnrtue,  fail  Ixf  dirrrt  crunhing,  and  the  effect  of  bend- 
:iiil  /(.«</  not  therefore  lie  eonxidered.  Find  4u-  safe  load  up,m  a  eonerete  eolumn 
.  /  M.-'ionrt/  area  S  reinforced  by  four  gteel  nnln  eiich  of  a  mfuare  i  ich  seetion.  fe 
I-  iiifi  the  mfe  unit  Kireas  in  the  cmcrete  and  r  the  rwtio  of  elaMie  moduli  of  the 
^to  land  concrete.  Ex.  A'=S"XS";  a-\  s<|.  ii...  f  f,«-  steel  ■':«),000,000,  and 
IT  corierotc  -3,000,000  ll«s.  /s<|.  in.;  /,  =600  lbs. 

.inn.  /riS-k-*.,);  60.000  llw. 

64.  A  l)ri<lge  diagonal  10  ft.  long  consists  of  two  «"  >  .i"  x  J"  tee  Imrs  placed 
link  to  l>ack.     Find  the  maximum  allowable  8tn>!«.  4  IxMiig  a  factor  of  safctv. 

M  -.vlied  the  ends  are  squarc-U-aring,  (b)  when  one  end  is  s<|uare-  aiMl  the 
'ilier  i.in-lK'aring.  Ans.  (a)  »<  ions;  (b)  21.7  Ums. 

65.  The  web  menilvrs  of  n  Warren  girder  are  bars  of  rectangular  sfctioa 
Mil. I   10  ft.  in  length.     (»iie  of  the  Lars  has  to  carry  loads  varying  Iw^ween 

1  -leiuly  mininiiim  tension  of  20.2  tons  an<l  a  maximum  tension  of  40.4  tons, 
iiid  .•mother  to  carry  loads  varying  U-twceti  a  maxinuim  compression  of  ,S.7 
•"IS  ;i!k!  a  maximum  tension  «f  14.4  tons.     Find  the  st'ctional  area  in  each 
<-\si\  allowan<'e  lieing  made  for  buckling  in  the  latter. 

Ana.  7.4  and  29.fi  »q.  ins 

66.  Determine  the  tendency  to  buckle  of  a  double-tee  strut  which  is  to  carry 
loud  varying  iK'tween  a  maximum  tension  of  80,000  lljg.  and  a  maximum 

.|ii|)re.s!<ion  of   (i(),(HNI  llw.       K;i.h  llaiige  consist.-,  of  two  IJ"X6"x|"  angkj 


638 


THEORY  OP  STRUCTl'RES. 


irons  rivotpd  to  n  12"Xj"  wph  plate.     The  length  of  the  strut  is  to  he  (a) 
6  ft.;  (.!>)  I'J  ft.        Am.  (a)  fi=.0.i9(\,  (h)  .9887  and  the  limit  is  approached. 

67.  What  safe  load  may  Ix-  Ijorne  l)y  a  '.V'X'.V'Xi"  aiiftle  wrought-inm  strut 
18  iiLS.  lc>ii){  when  both  ends  are  (")  s(|uare-bearin(j,  ('<)  pin-lx-arinj;,  4  lieing 
a  factor  of  safety?  Anx.  (a)  10.38  tons;  (/.)  8.49  tons. 

68.  The  r_'J-ft.  column  for  an  elevatetl  railway  etiiisists  of  a  10"X|" 
web  riveted  on  each  side  to  two  ecjual  t>"xSJ"  Z  Iwrs.  The  total  vertical 
load  due  to  train  and  sujierxtructun'^  113,2.S()  Ihs. ;  the  weipht  of  the  column 
and  its  attachments  MW  llw.;  the  wind  h)a<l  is  the  e<|uivaUnt  of  two  hori- 
lontnl  forces  of  12.000  and  4800  lbs.  acting  -•-'}  ft.  and  1(>J  ft.  respectively 
aU)vc  the  column's  base.  The  maximum  tleviation  of  the  up|>i'r  end  from 
the  mean  position  due  to  a  change  of  temix'rature -J  in.  The  eolunui  is 
fi.\ed  at  the  base,  and  the  top  is  so  constrainiHl  that  the  a.xis  is  vertical  at  that 
point.  £-27,000,000  lbs.  Find  the  max.  fibre  stresses  due  to  the  several 
forces,  and  also  the  total  max.  fibre  stress  (Roadway  12' c.  to  c.  of  cols.; 
ht.  of  col.  and  truss  - 18'.) 

Ans.  5041  lbs.;  890  lbs.;  71  lbs.;  2t)8rtlb8.;  10,100  lbs. 

69.  Calculate  the  strength  of  a  compression-chord  panel  member  352  ins. 
long  and  romposed  of  two  lS"x|"  web  plates  IJ  ins.  apart,  riveted  to  an 
upper  21"X  A"  cover  by  means  of  two  3i"X3i"X|"  angles,  and  also  riveted 
at  the  bottom  to  two  5"  (horizontal)  X3J"X  A"  angles. 

Ana.  Safe  load  ~.'>33,200  lbs. 

70.  A  brace  hinged  at  both  ends  is  subject  to  stresses  varying  from  a 
maximum  i-ompn^ssion  of  58,100  lbs.  to  a  maximum  tension  of  144,900  lbs. 
The  lenpth  is  Xi7  ins.,  and  the  brace  is  comiKJsetl  of  two  r_"'X3.l73"X.513" 
channels,  back  to  back.     Is  there  a  sufficient  sectional  area  of  metal? 

Aw.  Sectional  area  required  - 13.3  sq.  inches. 

71.  An    intermtHliate    bridge    vertical    33r)    ins.    long    consists    of    two 

12"  X 3.05" X  .39"  channels  kept  lOj"  a|)art  by  2" X  i" 
lacing.  What  force  will  such  a  strut  safely  bear? 
(Neglect  the  effect  of  the  lacing.)    Ans.  141,000  lbs. 


Fid   .'■.81. 
gravity  of  chaniK-l 


7 J.  .\  column  20  ft.  loni;  consists  of  two  8-in. 
channels  spaivd  0  ins.  hark  to  back,  the  flanges  being 
turned  outwards.  The  area  of  one  channel  is  4  sq.  ins.; 
moment  of  inertia  of  one  channel  about  axis  .40  is  3(>.0; 
moment  of  inertia  alxmt  axis  Cl)  through  centre  of 
is  l..V>;  distance  of  CI)  from  buck  of  chaimel  0.6  in. 
Find  the  load  wiiich  the  column  will  safely  carry  if  the  allowable  stress  per 

square  inch  is  given  by  the  formula  p  ll  •  r^„„,  ,)  "10000. 

Ann.  59,010  lbs. 
73.  \  hollow  cylindrical  tower  of  steel  plate  having  an  external  diam- 
eter of  3  ft.,  a  thickne-<s  of  J  in.,  and  a  lieiKlit  .if  (><)  ft.  carries  a  cj-ntrul  load 
of  .'(0  tons  and  is  sulijected  to  a  liorizoiital  wind  pre.ssun*  of  .".r.  ll>s.  \ivt  foot 
of  its  height.  Calculate  the  v»Tiical  stress4's  at  (he  fixed  base  of  the  tower 
on  tiie  windward  ami  on  the  leeward  suit-  (Alii  «  for  the  weipht  of  the  tower 
itself.)  Ans.  2.094  (c.)  and  .118  (t.)  llw./sq.  inch 


EXAMPLES. 


639 


74.  The  oyhnder  of  an  invertetl  engine  is  supported  on  one  wrtical  rast- 
■  ir..ii  .(.lunin  of  variable  rectangular  section,  the  centre  line  of  the  cylinder 

Ihiiik  S  ins.  from  the  vertical  face  of  the  column.  Find  the  maximum  stress 
iii(lu..d  at  a  section  of  the  column  which  i»  10  ins.  d.-ep  and  0  ins.  w-ide  the 
tliirkiicss  of  metal  being  1  in.  all  round  and  the  driving  force  on  the  piston 
Ih'Iiij;  .")  tons. 

75.  Find  the  limiting  ratio  of  length  to  diameter  for  a  steel  pin-bearing 
strut  stramed  to  the  elastic  limit  of  25,000  lbs.  per  square  inch,  E  being 
;«),0(X),000  lbs.  per  square  inch.  ^j„g    27.2 

76.  Kind  the  limiting  ratio  of  length  to  thickness  for  a  steel  square-bearing 
St  I  lit  .stramed  to  the  elastic  hmit  of  2.5,000  lbs.  per  square  inch,  E  being 
.■5(>.(l(K),0()0  lbs.  jier  square  inch.  yln«    63 

77.  The  allowable  stress  in  a  square-bearing  timber  strut  is  1000  lbs.  per 
s.|imre  inch.  The  strut  is  square  in  section  and  £-1,500,000.  Find  the 
liiiiitmg  ratio  of  length  to  side  of  square.  Ana.  70  28 

78.  Find  the  limiting  length  of  an  I  strut,  fixed  at  both  ends,  24  ins.  de^p 
-.1  -'..  sq.  ins.  sectional  area,  and  weighing  85  lbs.  per  lineal  foot,  the  radiua 
cf  »tyration  in  the  plane  of  flexure  being  1.33  ins.    The  metid  may  be  strained 
t"  12.000  lbs.  per  square  inch,  and  £-30,000,000  lbs.  per  square  inch. 

.    ,  •'"»•  418  ins. 

70.  \  72-in.  pin-bearing  steel  T  strut  has  a  3-in.  flange  X2i-in.  stem 
:.M.  weighs  7.2  lbs.  per  hneal  foot.  Taking  £  -29,400.000  lb«.  per  square  inch, 
1  "1  the  greatest  stress  which  can  be  developed  in  the  strut  without  lateral 
'"•'"'"'«•  Ans.  29,040  lbs. 

8o.  .\  7-in.  square-tearing  steel  channel  strut  weighing  9J  lbs.' per  lineal 
f(".t  .ind  58.t}  ins.  in  length  has  2.09-in.  flanges  and  a  web  .21  in.  thick  Find 
tl..'  greatest  stress  which  could  be  developed  in  the  strut  without  lateral  bend- 
""'■       „^  ,  •>««•  331,600  lbs.,  £  being  29,400,000  Uw./sq.  in. 

8i.  The  ends  of  a  long  strut  are  in  the  same  vertical  line,  one  end  is  fixed 
■Mi'\  the  other  i,s  kept  in  jwsition  by  a  horizontal  force.  Show  that  the  buck- 
liiiir  load  is  2.045»t'^. 

8j.  The  strut  OA  is  fixed  at  O  and  carries  a  load  P  at  the  end  B  of  a  hori- 
f'UUil  ...rm  AH  (  -,„  rigitlly  attached  to  the  strut  at  A.  If  the  strut  is  bent 
iti...  the  form  O.J.  th..  horizontal  distance  AC  of  .1  from  the  vertical  through 
"  l"iii)i  /).  iind  if  It  is  as.sumed  that  the  diffen-nce  in  length  betwwn  0.4  (  -/) 
•"Ml  lie  ,s  sufficiently  small  to  be  disregardeil,  show  that  q-(p  +  q)  cos  al 
.\i<"  liiid  the  buckhng  load  when  q  is  small  enough  to  be  neglected. 

Am.  \Elf-. 

83.  Ill  the  preceding  example,  if  in  addition  to  /'  there  is  a  horizontal 
f  r,,  //  u,  «,  si,„„.  ,|„„  j,„.  .Jeflection  y  of  any  |x,int  in  0.1  at  a  vertical  die- 
111"'  X  from  O  is  given  by 


.V  -?  +  (;  +  77(/  -x)  +  b  sin  (ax  +  c), 


MICROCOPY    RESOLUTION    TEST   CHART 

ANSI  ond  ISO  TEST  CHART  No    2l 


1.0 


I.I 


11.25 


1.4 


m 

2.2 
2.0 

1.8 


1.6 


jA 


/PPLIED  JIVMGE 

'»■'''''    r^ist    Mo-    S'-eel 

'*'jchest»r,    N««    Tofh         U609       u<, 


640  THEORY  OF  STRUCTURES. 

the  coefllcieiits  b  and  c  \kw  ■  given  by 

II  jj 

0==p+q  +  -l  +  'osin  c     and     0  =  -     .  +  aft  cos  c,    where    a^El  —  P. 

84.  Show  how  eqs.  (3)  and  (6),  on  pages  628, 629,  will  be  modified  if  the  line 
of  action  of  P  is  distant  «  +  /3  from  one  end  and  a  -  /J  from  the  other  end  of  the 
column's  axis.    Also,  if  the  coefficient  of  elasticity,  E,  is  variable  and  equal 

to  m±H—  at  a  txiint  distant  z  from  the  axis,  «»  being  the  maximum  value  of 

n  r' 
i  and  m  and  n  coefficients,  show  that  uH must  be  substituted  for  y  in 

cq.  (3). 

85.  In   one   of   Christie's   experiments   an   angle-bar   2"X2"XA",    with 

hinged  ends,  for  which  —  had  the  value  154,  deflected  .01   in.  for  an    ncrease 

r 

d     J 
ill  the  load  of  3000  lbs.    Show  that  -  +  —  =  .0048  in. 


86.  A  long  column  with  pin  ends  is  bent  laterally  until  the  angular  devia- 
tion ('!„)  at  the  ends  is  4°.  Find  the  total  maximum  intensity  of  stress,  the 
section  of  the  column  l)eing  (a)  a  circle,  ('<)  a  square.  (A' =  20,000,000  lbs., 
and  the  stress  due  to  direct  thrast  =1")00  li)s.  \tvT  scpiar  inch.) 

Aiix.  (n)  30,61.j  lbs.;  {b)  26,715  lbs. 

87.  With  the  same  mnjitnum  stress  as  in  the  last  example,  find  the  angular 
ilcvi.'ition  at  the  ends  so  that  the  stress  due  to  direct  thrust  may  l)e  10.0(X) 
ll).s.  |KT  wpiarc  inch.  Aiis.  (d)  1°,V;  {b)  1°  33'. 

83.  Show  that  the  load  ri'(iuircd  to  prfnluce  an  angular  deviation  of  14'' 
;it  the  two  pin  ends  of  a  long  column  is  only  one  per  cent  greater  than  that 
which  just  produces  flexure. 


CHAPTER  IX. 


TORSION. 


I.  Tof-ion  is  the  force  with  which  a  thread,  wire,  or  prismatic 
bar  toncis  to  recover  its  original  state  after  having  been  twisted,  and 
is  produced  when  tiie  external  forces  which  act  ujjon  the  bar  are 
icilucible  to  two  equal  and  ojiposite  couples  (the  ends  of  the  bar 
being  free),  or  to  a  single  couple  (one  end  of  the  bar  being  fixed),  in 
jilanes  jx'rpendicular  to  the  axis  of  the  bar.  The  effect  upon  the 
liar  is  to  make  any  transverse  section  turn  through  an  angle  in  its 
own  i)lane,  and  to  cause  originally  straight  fibres,  as  DE,  to  assume 
JK-licoidal  forms,  as  FG  or  DC.     This  induces  longitudinal   stress 


Fio.  683. 

in  the  fibres  and  transverse  sections  become  war|H'(l.  It  is  found 
Midiciently  accurate,  howe\Tr,  in  the  case  of  cylindrical  and  regular 
I">i>gonal  prisms,  to  assume  that  a  transverse  section  which  is  plane 
hcfiirc  twisting  n'mains  i)lane  while  IxMng  twisted.  In  order  that 
till'  bar  may  not  be  Ixmt,  its  axis  mast  coincide  with  the  axis  of  the 
twisting  couple. 

The  angle  turned  through  by  one  transverse  section  relatively  to 
another  is  called  the  anrjle  of  torKwn  (or  lu-ist),  and  Coulomb,  from 
'■N|MTiments  upon  wires,  niatle  the  following  deductions' 

(n)  That  the  angle  of  twist  ia  directly  proportional  to  the  length  of 


Hire  twisted; 


641 


642 


THEORY  OF  STRUCTURES. 


(b)  That  the  angle  of  twist  is  directly  proportional  to  the  tveistinff 
moment; 

(c)  That  the  angle  of  twist  is  inversely  proportional  to  the  fourth 
power  of  the  diameter: 

id)  That  the  angle  of  twist  for  wires  of  the  same  lengths  and  diameters, 
hut  of  different  materials,  is  inversely  proportional  to  a  certain  coeffi- 
cient G  called  the  coejficient  of  rigidity  of  the  material. 

These  results  may  be  expressed  analytically  by  the  formula 


A=C 


LPp 
GD*' 


A  being  the  angle  (in  radians)  through  which  a  wire  of  length  L 
and  diameter  D  is  twis*  d  by  the  application  of  a  force  P  at  the 
end  of  a  lever-arm  of  length  p.  The  value  of  the  coefficient  C  de- 
pends upon  the  form  of  the  section  only. 

Ex.  1.  ,4  brass  wire  30  ins.  long  and  0.1  in.  in  diameter  is  Itvisled  through 
an  angle  of  195°  by  a  ttnsting  moment  of  4  in.4bs.  If  the  coejficient  of  rigidity 
is  3,600,000  lbs.  per  square  inch,  find  C.  Hence  find  the  moment  which  will 
twist  a  bar  of  the  same  material,  300  ins.  in  length  and  1  in.  in  diameter,  through 
an  angle  of  5°. 

195  g  30X4 

180    "    3600000{.1)«* 


Therefore 


C- 


143 

14  " 


■10.; 


32 


,  very  nearly. 


Again, 


5jr    .32      .300  Pp 
180''  s   360b000Xl«' 


Tlitrefore 


Pp- 102.9  in.-Ib«. 


2.  Torsional  Strength  of  Shafts.— -Consider  an  elementary  prism 
Qli  of  tlic  shaft  l)rt\vc'('n  two  cross-WTtions  at  a  unit  distance  (one 
inch)  apart.  Assunung  the  section  on  the  left  to  be  fixed,  the  twist- 
ing moment  moves  the  prism  from  (/R 
tn  ()K\,  the  angle  ffOft,  lieing  the 
y     \    angle  of  twist  per  unit  of  length  (in 

Let  0  l)c  this  angle  .'measured  in  radians. 


vj--^-^ ----__-. 


TORSIONAL  STRENGTH  OF  SHAFTS. 


643 


The  shear  strain  in  the  prism  =  -^'=^=id,x  being  the  length 
OR. 

The  shear  stress  at  ft  is  therefore  GxO,  and  if  o  is  the  sectional 
area  of  the  prism,  the  shear  fo'ce  a,t  R  at  right  angles  to  OR  =  Gxda. 

This  force  is  equivalent  to  a  parallel  force  on  the  axis  and  a 
couple  of  moment  Gx^da. 

There  is  a  similar  force  and  a  similar  moment  for  every  elemen- 
tary area  into  which  the  cross-section  may  be  divided. 

The  forces  must  necessarily  neutralize  each  other  and  the  total 
twisting  moment  (or  torque)  on  the  shaft  ^'IGx^da,  the  symbol  2 
denoting  algebraic  sum. 

Now  G  and  0  are  the  same  for  each  elementary  area,  and  if  T 
is  the  total  torque, 

T^Geijfla^GdJ^Pj,, 

J  being  the  moment  of  inertia  with  respect  to  the  axis,  or  what  is 
called  the  polar  moment  of  inertia. 

For  a  solid  cylindrical  shaft  of  diameter  D, 

TtD* 

*'"32' 

for  a  hollow  cylindrical  shaft  of  external  diameter  D  and  internal 
diameter  D\, 

7t(D*-Di*) 

.\gain,  the  shear  stress  G0x  increases  with  x  and  is  greatest  at 
tho  surface,  i.e.,  when  x-^. 


Denoting  this  maximum  shear  stress  by  /, 


Hence,  for  th«  solid  shaft, 
Pp'T 


OffyrP*    frD'    fiv 


32 


16      5.1' 


644  THEORY  OF  STRUCTURES. 

and  for  the  hollow  shaft 

IP-^-  32  "=16        D      ~5.1         D     • 

If  the  thickness  n ---  — .^ — )  of  the  hollow  shaft  is  small  as  com- 
pared with  the  diameter  D,  then,  approximately, 

GOr.m_jr.DH 


Thus  the  strength  of  the  solid  shaft  is  defined  by  the  value  of  /  and  de- 
pends on  the  culx*  of  the  diameter,  while  its  torsional  rigidity,  which  is 

T 
measured  by  the  ratio  -^,  depends  upon  the  fourth  power  of  the 

diameter. 

In  order  to  diminish  the  amount  of  twist,  or  sprinp,  a  shaft  is 
often  made  nmch  stronger  than  is  actually  necessary,  and  a  common 
rule  is  to  s{x?cify  that  the  twist  is  not  to  exceed  1°  in  20  diameters. 
Then 


'         2  180    20D   2      7200 


Hence 


for  ca.st  iron,         taking  C;=  0,.300,000  Ibs./sq.  in.,  /  =  27.50 Ibs./in.sq. 
■'   wrought  iron,      "      r;=  10..-)0().000    "       "       /  =  4r)83  "       " 
"steel,  "      f/-12.000.000    "       "       /  =  523S  "       " 

In  shafts  in  which  the  spring  is  not  of  so  nmcli  importance,  as, 
e.g..  when  the  .shafting  is  in  short  lengths,  higher  values  of  /  may  bo 
u.<cd.     Thus 

for  ca.st  iron,         if  /  =  4,.500  Ibs./sq.  in.,  the  twist  is  1°  in  12.2  diams. 
"   xvroughtiron,  if/=  7,200    "       "        "      "       "    "  12.7     " 
"   steel,  if/=ll,2(K)    "       "        "      "       "     "    9.4     " 

Another  rule  in  acconlance  with  good  practice  is  that  the  twist 
must  not  exceed  75  degrw  jxjr  lineal  foot  of  length. 


TORSIONAL  STRLNGTH  OF  SHAFTS. 


645 


Again,    let   II.P.   be   the   horse-power  transmitted   by  a    shaft 
making  N  revoUitions  per  minute.    Then,  for  a  solid  shaft, 

T'lr.N =—r^  2s^A'  =  work  done  in  inch-pounds  per  minute 
=  33000H.P.X12, 


and  tiierefore 


Taking  /  =  4500,  7200,  and  11,200  lbs.  per  sq.  in.  for  cast  iron, 
wrought  iron,  and  steel  respectively,  then,  approximately, 


D=  4  \—h-  for  cast  iron, 
/>  =  3.6>^— TT-^  "   wrought  iron, 


D=  3  q~~  "  steel, 

forniula>  agreeing  with  the  best  practice  in  the  case  of  shafts  sub- 
jected to  torsion  only. 

Ex.  2.  A  slerl  .'ihnft  20  ft.  in  length  and  3  ins.  in  diameter  makes  200  revo- 
luliotiK  per  minute  and  IrnnKmit/f  ")0  //./'.     Thrmqh  what  angle  is  the  nhaft  tmntcdf 

.1  irrimght-irnn  shiift  of  the  name  length  is  to  do  the  mmc  u-ork  at  the  same 
■'liiid.  Find  ilH  diamrtrr  so  that  the  .strc.s.s  at  Ihr  rirriimference  may  not  exceed 
thrrc  filths  of  tttat  at  the  eirmmji  rencc  oj  the  .^teel  shaft. 

If  T  is  the  twisting  moment, 


Therefore 


T  X 2;r  X  200  -.W X  33000  X  1 2  in.-lbs. 

22  3« 


and 


1 57r)0  -  r  - 1 20()00(K)  XOk 


49 


7  32' 


2()7000" 


ISO. 


Therefore  the  total  anule  of  twist  in  de(pve8-20Xl2X — <'"-2".26. 


646 


Again, 


THEORY  OF  STRUCTURES. 
'  IG'o'     16' 


if  D  ia  the  diameter  of  the  wTought-iron  shaft. 

Therefore  Z>'=45    and    Z)=  3.556  ins. 

Ex.  3.  Show  that  a  hollow  shajt  is  both  stiffer  and  stronger  than  a  solid  shaft 
of  the  name  material,  weight,  and  length. 

IvPt  (I  l)e  the  diameter  of  the  solid  shaft,  and  d,,  d,  the  external  and  internal 
diameters  respeetively  of  the  hollow  shaft. 

liCt  0„  Oj  be  the  angles  of  torsion  in  radians  of  the  solid  and  hollow  shaft* 
respectively. 

Then,  for  the  same  twisting  couple  T, 

T 

the  rigidity  of  the  solid  shaft      0,     0, 
the  rigidity  of  the  hollow  shaft  °°  T  "  <?, 


Gr.d* 


d* 


Gi:{d,*-d,*)     d,»+d," 
since  ff(d,'-dj') -rd». 

But  dt*+d,''>d',  and  therefore  0,>0„  so  that  the  solid  shaft  twists  through 
a  greater  angle  than  the  hollow  shaft. 

Again,  if  T,  and  T,  arc  the  twisting  moments  of  the  solid  and  hollow 
shafts  respectively, 


r,^,  d^^,JLM*-d,«\         d4        dr(d,2rJ'')} 
Ti^'^lG' 'lQ\      d,      /"d,»+d,»"°     d,'+d,'    ' 

Hence  T,>T„  since  it  is  evident  that  d,(d,'-d2')*<(d,»+dj'). 

Ex.  4.  In  a  spinning-mill  a  ra-H-iron  shaft  8J  ins.  in  diameter  makes  27 
revolittwii.9  per  minute.  Find  the  work  transmitted  if  the  angle  of  torsion  is 
not  to  exceed  i's°  per  lineal  foot.  What  mil  the  work  be  if  the  maximum  shear 
stress  in  a  section  is  not  to  exceed  4.500  lbs.  per  square  inch  t 

First.  If  T  is  the  twisting  couple, 

7'X2^X27=33000XH.P. 
TT^erefore        H.P.  =  JLr  =  ^  ±(,;.  ^) 

-1.37.49. 


NON-CIRCULAR  SECTIONS. 


647 


Second.  H.P.= 


9 


1750    °1750  12 


L//!^U-^-^-4500  ^  ^*-^' -212  72 
[2Vl6y     1750  12^^    7       16      ^^^-^^^ 


Ex.  5.  T'/ie  external  diameter  of  a  hollow  jhaft  is  p  times  the  internal  diam- 
eter. Compare  its  torsional  strength  with  that  of  a  solid  shaft  of  the  same  male- 
rial  and  weight. 

Let  pd  and  d  be  the  external  and  internal  diameters  of  the  hollow  shaft 


Let  Z)=the  diameter  of  the  solid  shaft. 


Then 

Twisting  moment  of  hollow  shaft 
Twisting  moment  of  solid  shaft 


jr_  /p'd*-d*\ 
/16  \     pd    ) 

'lO 


d>  /p*-l\ 
D'\    p    )' 


But 


Therefore  the  moments  are  in  the  ratio  of 


D*~p'd'-d*-{p'-l)d\ 
p'  +  l 


pVp>-i 


3.  Non-circular  Sections. — Tlie  polar  moment  of  inertia,  /,  of 
any  area  with  respect  to  an  axis  through  the  centre  of  inertia  per- 
|x?ndicular  to  its  plane  is  given  by 

J=Ix+h, 

1 1  and  1 2  being  the  moments  of  inertia  with  respect  to  any  two  axes 
tlirough  the  same  point  at  right  angles  to  each  other.  //  it  is  as- 
sumed that  the  angle  of  torsion  (d)  is  the  same  at  all  points  of  a  non 
circular  section,  then 

T 

the  torsional  rigidity  =  ^  =G J =G(I i  +I2) 

-the  sum  of  the  flexural  rigidities  in  any  two 
planes  at  right  angles  to  one  another. 
St.  Venant  has  shown  that  this  is  in  excess  of  the  torsional  rigid- 
ity derived  from  the  true  theory,  and  it  can  be  easily  shown  that 
the  twisting  couple  produces  a  greater  twist  than  that  ba.sed  upon 
Coulomb's  laws  and  also  warps  the  naturally  plane  sections  of  the 
shaft.  Figs.  585,  586,  587,  given  by  St.  Venant,  show  the  appear- 
ances presented  by  elliptic,  square,  and  rectangular  bars  under  ex- 
aoigerated  torsion  as  may  be  obtained  with  such  substances  as  india- 
rubber.    St.  Venant  also  enunciated  the  important  practical  result 


648 


THEORY  OF  STRUCTURES. 


that  tlio  ribs  introduced  to  incrcaso  the  jlcxurtil  rigidity  of  a  bar  or 
boani  has  a  really  detrimental  effect  upon  its  torsronal  rigidity.  , 


Fiu.  585. 


P'lO.  58(). 


Fig.  5S7. 


The  following  table  gives  a  few  of  St.  Venant's  principal  results, 
the  third  column  giving  the  fraction  which  the  torsional  rigidity  is 
of  that  of  a  circular  cvlinder  of  the  same  .sectional  area; 


Section.  Torsional  Rigidity. 

An  elapse (tjz 


An  equilateral  triangle (la^X 


a^  +  lr^ 
\/3 


80 


Fraction. 

2ab 
a^  +  tP 

.72r)81 


A  square (in*  X  .HOoTG      .88362 

A  rectangle ^/o6'M:r  —  .21-) 

Again,  the  true  torsional  rigidities  of  a  shaft  with  a  square  sec- 
tion having  curved  corners  and  Ix^llow  sides  (Fig.  088),  of  a  shaft  with 


Fic.  r,»H.       Fic.   .589.       Fk!    .'",90. 


a  .square  section  ha\ ing  acute  angles  and  hollow  sides  (Fig.  539),  and 
of  a  shaft  with  a  star  .section  having  rounded  points  (Fig.  ')90j  are 


;  i  t 


VARIABLE  RESISTANCE. 


649 


.SOOG.  .8276,  and  .G'io  times,  respectively,  the  corresponding  tor- 
sional rigidities  based  upon  Coulomb's  laws. 

.St.  Venaiit  wius  the  first  to  call  attention  to  the  fact  that  in  non- 
circular  sections  the  stress  is  more  generally  greatest  at  points  in 
fiK>  hounding  surface  which  are  nearest  to  the  a.xis  and  least  at 
th()S(>  points  which  are  farthest  from  the  axis.    Thus  the  surface  shear 

stress  at  any  point  x,  ?/ of  an   elliptic  section   is    —ir^\/¥x-  +  a^y2 


WllK 


2T 


ich  is  greatest  and  equal  to_"^-p,  when  x=0,  i.e.,  at  the  end  of  the 
minor  axis. 

In  the  rectangle  the  shear  is  greatest  at  the  middle  point  of  the 
longest  side,  while  for  squares  and  equilateral  triangles  there  are  lines 
(.f  maximum  strain  through  the  middle  points  of  the  sides.  It  may 
l.<"  remarked  in  general  that  any  elastic  solid  bounded  by  surfaces 
witii  projecting  edges  or  angles,  or  with  re-entrant  edges  or  angles, 
.aimot  exiwrience  any  finite  stress  or  strain  near  a  projecting  point 
unless  acted  upon  by  external  forces  at  the  point;  the  strain  near 
Mil  edge  can  only  be  in  the  direction  of  the  edge,  while  the  stress  and 
strain  are  increased  indefinitely  in  the  neighborhood  of  a  re-entrant 
edge  or  angle.  This  result  is  in  accordance  with  the  important  and 
\\<ll-known  practical  rule  that  every  re-entering  edge  or  angle  ought 
to  1h'  rounded  to  prevent  risk  of  rupture  in  solid  pieces  designed" to 
bear  stress. 

Ex.  6.  .1  f<quare  wooden  shaft  8  //.  in  length  is  acted  upon  by  a  force  of  200 
/''..,  applied  at  the  cirrumferenre  of  an  S-ft.  wheel  on  the  shaft.     Find  the  length 
../  I  he  side  of  the  shaft,  so  that  the  total  torsion  may  not  exceed  2°  {G  =400  000) 
II  h„l  should  be  the  diameter  of  a  round  shaft  of  equal  strength  and  of  the  same 
iiKilirinl? 

/•  <rst.  200  X4  X 12  =  r  iii.-Ibs.  =400000  j^  =g  a*X  .140576,  and  a  =4.655  ins. 
Second.  If  d  is  the  required  diameter, 


400000 


~   1  i^L' 

180  96  32 


=  400000j-^  ^  a*X. 140576. 


Therefore 


7X32     , 
'—^-   a'X. 140576     and    d- 


'5.1  ins. 


4.  Variable   Resistance.— The  fornmla?  deduced  for  the  twisting 
moment  of  a  shaft  is  based  on  the  assumption  that  the  power  is 


650 


THEORY  OF  STRUCTURES. 


transmitted  against  a  constant  resistance.  In  practice  the  resist- 
ance varies  between  a  maximum  and  a  minimum  limit,  which  are 
sometimes  of  w'dely  different  values,  and  the  shaft  must  be  designed 
for  the  maxin'U.n  moment  to  w'aich  it  may  be  subjected. 

Ex.  7.  The  urouyht-iron  screw  shaft  of  a  steamship  is  driven  by  a  pair  of 
cranks  set  at  riqht  a.igles  and  21.7  ins.  in  length;  the  horizontal  pull  upon  each 
crank-pin  is  170,400  lbs.,  and  the  effective  length  of  the  shaft  is  866  ins.  Find 

the  diameter  of  the  shaft  so  that  (a)  the  circumfer- 
ential stress  may  not  exceed  9000  lbs.  per  square 
inch  and  (b)  the  angle  of  torsion  may  not  exceed 

\° 

r^  per  lineal  foot,  G  being  10,000,000  lbs.   The  actual 

diameter  of  the  shaft  is  14.9  ins.  What  (c)  is  the 
actucU  torsion? 

The  ultimate  tensile  strength  of  the  iron  being  60,000  lbs.  per  square  inch, 
find  (rf)  the  actual  ultimate  strength  under  unlimited  repetitions  of  stress. 
What  (p)  is  the  torsion  when  one  of  the  cranks  passes  a  dead  point? 

(a)  Twisting  moment  =  Pr{s\nO  +sin  ixF+e)  =2Pr sin  0  +45°  cos  45°,  which 
is  a  maximum  and  =PrN/ir,  when  d  =45°. 


Fig.  591. 


Then       i76400x21.7xV2"= 


16 


and    d  =  14.53  ins. 


(6)  176400X21.7 XV^  =  10000000  j^jr2jr^''^^ 

and  ri  =  14.9  ins. 

(c)  If  A°  is  the  total  torsion, 

—  t:     A°   (14  91* 

176400X21.7XV2  =  10000000  f^go6    "^2~ 


and 


.l°=.i°..-)46.j. 


(d)  The  twisting  moment  =2/V  sin  (9  +  45°)  cos  45°,  which  is  a  maximum 
when  9=45°  and  a  minimum  when  0=0.    Therefore 

—  ''2  (14  9)'  /— 

max.  mometit  =  Pr v  2  =  /max.  -y    -j  g—  =  1 76400  X  21 .7  X  v^ 

22  (14.9)» 
and  min.  moment  - Pr       =/min.  -^ — 77: —  =  176400X21.7, 


80  that      /max.  =8331.24  lbs. /sq.  in.    and    /min.  =5891.08  lbs. /sq.  in. 


DISTANCE  BETWEEN  BEARINGS.  651 

Using  Unwin's  formula, 

the  fluctuation  d  =/max.  -/min.  -2440.16, 

and  therefore 

''440  16 
the  max.  stress/sq.  in.  -~^  +  v/60000(60000-3  X 2440. 16)  =59361.2  lbs. 

(c)  On  passing  a  dead  point  the  total  torsion  of  4°  is  giveu  by 
176400X21.7  =  10000000,-^^  ^~  ^^^^ 


180  866  7      32 


and 


A"  "3°.92. 


5.  Distance  Between  Bearings,— The  distance  between  the  bear- 
ings of  a  line  of  shafting  is  limited  by  the* consideration  that  the  stiff- 
ness of  the  shaft  must  be  such  as  will  enable  it  to  resist  e.xcessive 
landing  under  its  own  weight  and  under  any  other  loads  (e.g.,  pul- 
leys, wheels,  etc.)  applied  to  it.  For  this  reason  the  ratio  of  the 
maximum  deviation  of  the  axis  of  the  shaft  from  the  straight  to  the 
eorresjwnding  distance  between  bearings  should  not  exceed  a  certain 
fraction  whose  value  has  been  variously  estimated  by  different 
authorities. 

Let  /  be  the  distance  in  feet  between  bearings,  d  the  diameter  of  the  shaft 
Ml  111,  hes,  w  the  weight  of  the  material  of  the  shaft  per  cubic  foot,  and  let  the 
apphcd  load  be  equivalent  to  a  load  per  lineal  unit  of  length  m  times  that 
of  the  s^aft.  Assume  a  stiffness  ,^,5,  and  that  the  axis  of  the  shaft  is  truly 
111  hne  at  tlie  bearings.  The  maximum  deflection  of  the  shaft  is  given  by  the 
formula 

_L  (m+\)  (weight  of  shaft)/'  x  1 728 
"384  El 

384^"'  +  ^>4   T44''';^«~e— • 


D^ 


Therefore 


D 

I  ' 


1 
100° 


(m  +  l)wl* 
'2E      d" 


\io0u;(m  +  l)' 
For  wrought  iron,  E  -30,000,000  lbs.  and  w  -480  lbs.    Therefore 

1-12.73/-^. 

\m  +  l 


652 


THEORY  OF  STRUCTURES. 


If  the  applied  load,  instead  of  l)eing  uniformly  distributed  is  concentrated 
at  the  centre,  the  maximum  deflection 

1    (m  +  })(\voiahtofshaft)/'X1728 
-i)ms.=— -^, , 


EI 


and  hence 


For  wrought  iron 


\llOO«'(m+i)" 


6.  Efficiency  of  Shafting. — Lot  it  require  the  whole  of  the  driv- 
ing moment  to  overcome  the  friction  in  the  case  of  a  shaft  of  diam- 
eter d  and  length  L.    The  elficienry  of  a  shaft  of  the  same  diameter 

and  length  /  =  l-7-. 


But 


fjuP 
16 


=  (Pp)  =  moment  of  friction  =  /<-^/-^  =  t-^~^  ^^ 


w  being  the  specific  weight    if  the  material  of  the  shaft 
coefficient  of  friction.     Hence 


8 


and  fi  the 


1 


=  9 


W 


-m 


-y   and  the  efficiency  =  1  -2/r 


/' 


Fio    h'M. 


The  edicicncv  mav  also  1)0  found  as  follows: 


\x't  fr,   »'r  1)0   the  stress   and   velocitv  at 


any 
the 


point  distant  r  from  the  axis  of  the  shaft  which  is  taken  as 
axis  of  .r,  tlio  other  axes  l)(>ing  as  shown  in  Fig.  o9'J. 

Then,  l)y  Art.  Jd.  Chajitor  V.  the  work   transmitted  across   a 
small  oloincnt  dintz  of  the  section 


'  iiihjdz  =  (jj^v^  +lxy''y  +J„r,)dydz. 


In  the  proM'tit  case 


hi-  -f.miO,        /x,-/,  cos^, 


IV -0. 


•  I',  sin  0, 


r,  -  jv  cos  ff. 


EFFICIENCY  OF  SHAFTING. 
Therefore 

the  total  work  tr&Dsmitted^ J  fudydz 

=//(/riV  sin2  d+frVr  cos2  d)dydz = f  f  f^Vrdydz 
f  and  V  being  the  stress  and  velocity  at  the  surface.    Again, 


653 


Hence 


the  work  lost  in  friction  =  u~r-  Iv. 

4 


the  efficiency  = 


,vdP        imcP, 


,vcP 


Ex.  8.  The  effunency  of  an  axle  is  i;  the  working  stress  in  the  shaft  is  9000 
/''s.  per  square  inch;  the  coefficient  of  friction  is  .10.  How  far  may  work  be 
trausmittedf     The  shaft  is  of  wrought  iron. 


and 


2°         10  17289000' 
/ -81000  ins. -67.50  ft. 


hx.  9.  Determine  (a)  the  profile  of  n  shaft  of  length  I  which  at  every  point  is 
v..  prnportumed  as  to  be  just  able  h>  hear  the  pourr  it  has  to  transmit  plm  the 
P'mr  rn,uircd  to  overcome  the  friction  beyond  the  point  under  cmsideration 
I- 1  ml  (h)  the  efficiency  of  such  a  si  aft,  and  (c)  the  cffiriency  of  a  shaft  made  up 
-'/  II  series  of  n  divisions  each  of  uniform  diameter. 

in)  Let  s  be  the  maximum  allowable  skin  strea-s.    Then 

_,    IS  the  max.  allowable  driving  moment  at  the  driving  end  of  radius  r,  and 

.«Tr' 

at  X  from  the  driving  end,  the  corro- 
.s|X)ndiiig  radius  l)oing  y. 

I!"nce 

fiimy'-dx-y  -moment  uacd  up  in  overcoming  the  friction  for  a  length  dx., 

and 

P'^f'yMx-      H  ..      ..  „  „      ..       ..  the  length,. 


654 


THEORY  OF  STRUCTURES. 


Therefore 
DifTerentiating, 


or 


fwicy'dx 2"J/'rfy, 

dy        ^lj£j  ^ 

y         3    s^'^'SL' 


where  ^(  "JF/  '^  *^^  length  of  a  shaft  of  uniform  diameter  for  which  the 
whole  driving  moment  is  required  to  overcome  the  friction.    Integrating, 


log.y--T7+c, 


c  being  a  constant  of  integration. 
Whenz-0,  y-r;  therefore 


and 

is  the  profile  required. 
(b)  Again, 


3L 


logf  r-c. 


J^-e~si, 


the  useful  moment  "2* » 


tad 


1  1  SftT 

the  total  moment  ~~9"» 
the  efficiencv — ^  —  e   ^  ■ 


(c)  The  efficiency  of  each  length  - — - —  -  (1  -rr) , 

and  therefore 

the  efficiency  of  the  shaft  -  ( 1  -rr)  • 

7.  Combined  Bending  and  Tortion. — Tt  was  shown  in  Chapter  V 
that  in  a  shaft  subjootod  to  a  Ix^nding  moment  A/^  and  to  a  twisting 
moment  iV/i,  acting  simultaneously, 

the  maximum  t«nHi!e  or  compressive  Htre3!?  =  ^  ^"^   \t 


COMBINED  BENDING  AND  TORSION. 


655 


and 
where 


the  maximum  shear  stress 


=#^. 


Z  being  the  strength  modulus  of  the  section. 
For  a  solid  section  of  diameter  d,  Z  =  ~, 

hollow  "       "  external  diameter  di,  and  internal  diam- 
eter d,,  Z  - '^^^^^^ 
Hence 

the  maximum  tensile  or  compressive  stress  =  ^(M,,  -{-VM^  +  Mjt)^ 
and 

the  maximum  shear  stress  -  Ky^M^ + M?. 

Generally  speaking,  it  is  found  that  the  first  of  these  two  equations 
gives  the  largest  diameter,  and  thus  the  maximum  tensile  or  com- 
pressive stress  is  the  same  as  the  shearing  stress  when  the  shaft  is 
subjected  to  a  twisting  moment  Mb+y/M^'  +  M,''. 

Ex.  10.  Potver  w  UUcen  from  a  skaft  by  means  of  a  piUley  24  inches  in  diameter 
uh,ch  IS  keyed  on  to  ths  shaft  at  a  point  dividing  the  distance  between  two  con- 
■^rrutive  supports  into  segmenU  of  20  and  m  ins.;  the  tangential  force  at  the 
circumference  of  the  puUey  is  5500  lbs.  If  the  shaft  is  of  cast  iron,  determine 
Us  diameter,  taking  into  account  the  bending  action  to  which  it  u  subjected. 

J/6-5500  ^1^- -88,000  in.-lb«. 


Therefore 


Ml  -5500  X 12  -66,000  in.-lbs. 

Af»  +  y/AU'^-M,'  -88000  +  n/(88000)'  +  (66000) 
-198,000  in.-lbs. 


Hciire  if  d  is  the  diameter  required, 


and 


22  d' 
SeOOy  jg-198000, 

d- 5.646  ins. 


li^ 


656 


THEORY  OF  STRUCTURES. 


Again,  another  expression  for  the  maximum  stress  in  the  sec- 
tion of  a  shaft  may  be  obtained  as  follows:  The  stresses  at  a  point 
near  the  circuinforenee  are : 

On  a  cross-section,  (a)  a  normal  stress  p(  ="7" ) ! 


(6)  a  shear        "     9  (  =  9^)  J 


and  on  a  section  parallel  to  the  axis,  (a)  a  normal  stress  l>; 

(b)  a  shear  stress  qi  =  '^] , 

Z  lieing  the  strength  tnodulus  of  the  section. 

Hence,  as  in  Chapter  V,  if  ;>i,  p^  are  the  principal  stresses  at  the 
point, 


+  92 


and 


P2 


'h^'h 


'-^g2. 


Let  ei,  62  be  the  strains  in  the  directions  of  the  principal  stresses, 


then 


JlCi=pi =Tj H \T+9 


and 


a  Ix'ing  Poisson's  ratio. 

The  maximum  stress  /  develojx'd  in  the  material  mast  not  exceed 
the  greater  of  the  two  quantities  AVi  and  Ee2,  and  therefore 


,     p>7-\    n  +  l    \j?      „ 


CENTRIFUGAL  WHIRLING  OF  SHAFTS. 
A  common  value  of  a  is  4,  and  then 


657 


1 


or 


/  =  3P  +  Wf  +  4^2  =  ^  (|Mj  +  |V^/j2  +  Jl/^2) 


This  formula  is  given  by  Grashof,  Cauchy,  and  others  for  com- 
binod  bending  and  twisting. 

Ex.  11.  Solving  Ex.  10  by  the  preceding  method, 

22  d'     3  «; 

SeOOy  jg=j(88000)  +'^v/(88000)'  + (66000)' 


and 


or 


■134,750  In.-lbs., 
d«- 122.5, 
d  -4.967  ins. 


8.  Centrifugal  Whirling  of  Shafts.— It  is  known  that  a  shaft, 
liowover  nearly  balanced,  wlien  driven  at  a  sufficiently  high  sjjeed 
Ih-hiL'^,  or  "whirls,"  as  it  is  termed. 

The  particular  or  "critical"  si)eed  depends  on  the  manner  in 
wliirh  the  shaft  is  su[)ported,  on  its  dimensions,  its  modulus  of 
elasticity,  and  on  the  manner  of  loading. 

In  an  unloaded  shaft  the  jM-riod  of  whirl  coincides  with  the 
natural  period  of  vibration,  as  might  be  exiK>cted;  but  generally 
ill  :i  loaded  shaft  the  fM-riod  of  whirl  is  less  than  the  natural  period 
<il  vibration.  As  in  the  lateral  vibration,  .so  in  whirling,  there  is  a 
M'tics  of  jK'riods  at  which  the  shaft  whirls. 

This  torsional  vibration  often  occurs  in  very  small  .shafting  and 
M.riu'limes  in  long  shafting  of  much  larger  diameter.  If  the  im- 
IHilscs  producing  the  vibration  are  rej)eated  at  the  projjcr  intervals, 
111.'  \ibration  may  continually  increase  until  the  torsion  becomes  of 
siitlicient  magnitude  to  cau.se  rupture. 

Consider  the  cjuse  of  a  uniform  shaft  weighing  w  lbs.  per  unit  of 
I'MRth,  subjected  to  an  endlong  thrust  F  and  revolving  with  an 
angular  v«'locity  of  w  radians  per  second. 


658 


THEORY  OF  STRUCTURES 


Take  the  middle  point  of  the  shaft  as  the  origin  and  let  y  be  the 
deviation  from  straightness  of  a  point  distant  x  from  the  origin. 
Then  if  m  is  the  B.M.  at  this  point  due  to  the  centrifugal  effect, 


d-m     w    „ 


Also, 


,(Ptj 


-£/'—,  =  total  B.M.  at  the  point 
(tx- 


Differentiating  twice, 


d*y       iPy    d^m 
dx*       dx^     dx^ 


or 


d*y      F  d^y    w  cfi       w 

d?^Yidi'~~^Yiy~Yi 


The  general  solution  of  tliis  equation  is 

y  =  A  cos  ax  +  B  sin  ax + Cc"  +  De'^'  —^, 


where 


and 


«'=2k(^+Nl^'+^^^'H 


^  =  2^(-^^  +  n'^'  +  -*<^)- 


Since  y  is  the  same  for  equal  positive  and  negative  values  of 
X,  fl  =  0  and  C  =  D.    Therefore 

Ji 


y  =  A  C0.S  «j  +  C(e"4-c-")  — 


Again,  a.ssuining  that  the  lx>arings  do  not  constrain  the  direction 
of  the  axis  of  the  shaft, 

-pr,  =  0  when  X  =  ;r. 


CENTRIFUGAL  WHIRLING  OF  SHAFTS. 


659 


Also, 

Therefore 

and 
Hence 


y  =  0  when  x=— 


A  = 


al  /  ^         9'\ 

0=-Aa2coa:^+C^le2+e    2J 

al        /  <"  -"X      n 


and 


C= 


a>2(a2+^2)cos^^ 


gaz ga^ 


The  B.M.  is  greatest  at  the  middle  point  and  is  the  value  of 

,<Py 


M„ 


-  EI~2  when  x = 0.    Therefore 

-Aa^-'yr^-    ^"^^    /    1  1 


I  COS-;j- 


cosh  y , 


The  greatest  stress  at  the  middle  is 


Z    "*"<$' 

^  boing  the  strength  modulus  of  the  section  and  S  its  area. 

Kxl2.  A  propelUr  shaft  13.4  t„s.  in  diameter,  98  /<.  ?o„j,  and  making 
'■O  ra'rt?u<ions  ppr  minute  is  sub}ex:ted  to  an  end  thrust  of  .50,000  lbs  Show 
l->t  the  centrifugal  force  effect  is  307  times  greater  than  the  effect  due  to  the  end 


(Vittrifugal  effect      w  m'     /  F  \* 
Kiid-thrust  effect  "'gWl^  \2El} 


_4r£a^       4X.28(2«)'x30000000x»'f1.T4^« 

F'g     '        (50000) '  X  32.2  x  12  X  64  X4 ^'^ 


660 


THEORY  OF  STRUCTURES. 


Suppose  that  there  is  no  end  thrust  (F=0)  and  that  the  effect 
of  the  shaft's  weight  may  be  disregariled  as  compared  with  the  cen- 
trifugal effect.    The  general  equation  and  its  solution  now  become 

d*y    w  u? 

— ^  — 77=0 

dx*    g  El-' 
and  y  =  Acosax, 

•t  being  the  deflection  at  the  middle  and  a^  =  \j—  -Trr. 


9  EI 


Also, 
Therefore 
Hence 


j/  =  0  when  x=^ 


0=^<os—     and    al=7t. 
a       \ww'/ 


Ex.  13.  ^  steel  shaft  of  diameter  d  ins.  and  weighing  .28  lb.  per  cubic  inch 
makes  n  revolutions  per  second.  Take  £=28,000,000  lbs.  per  square  inch. 
Then 


u'-.28' 


)rrf' 


=  .22Xd', 


and 


l~t 


^32Xl2X28000000X>rrf«\t 
^'x.28X(2;rn)»X64  J 


■-(^)'- 


g.  Helical  Springs. — Let  the  figure  represent  a  cylindrical  spiral 
spring  of  length  L  supporting  a  weight  11'.     Consider  a  section  of 
the  spring  at  any  point  B. 

At  this  point  there  is  a  shear  11'  and  a  torque  Wy, 
y  being  the  distance  of  B  from  the  axis  of  the  spring, 
i.e.,  the  radius  of  the  coil. 

The  effect  of  \V  may  generally  lie  neglected  as  com- 
pared with  the  effect  of  the  moment  Wy,  and  it  may 
therefore  be  assumed  that  the  spring  is  under  torsion 
at  every  point.     Let  there  be  n  coils.     Then 
„,       hr^    aOnr* 

r  Ix'ing  the  radius  of  the  spring  and  0  the  twist  in  radians  per  unit 
oT  length. 


HELICAL  SPRINGS.  661 

Also,  L  =  2~yn,  approximately. 

The  change  of  length  in  t':e  spring  (i.e.,  the  deflection) 

^^        Gi:r*  ~  Gr  ~    Gr    ' 

The  energy  stored  up  in  the  spring,  i.e.,  the  work  done  in  stretching 
or  compressing  the  spring, 

2  Gnr*"    4G    ' 

A  weight  at  the  end  of  the  spring  tends  to  turn  as  well  as  to 
change  the  length  of  the  spring,  and  this  is  due  to  a  slight  bending 
action. 

According  to  Hartnell,  /  =  70,000  lbs.  per  square  inch  for  i-in.  steel, 
/  =  G0,000  lbs.  per  square  inch  for  f-in.  steel,  /  =  50,000  lbs.  per  square 
inch  for  i-in.  steel,  and  G  varies  from  13,000,000  lbs.  for  i-in.  steel 
to  11,000,000  lbs.  for  f-in.  steel. 

Also,  for  wire  less  than  J  in.  in  diameter  he  takes 

the  safe  load  =     y-,  and  the  deflection  =  ^g^^. 

Assuming  that  the  laws  upon  which  the  torsion  of  rectangular  and 
f:quare  sections  is  based  are  the  same  as  for  circular  sections,  then  for  a 
rectangular  section  of  breadth  b  and  depth  h, 


Wy  =  YfObh{b^+h^)  =  kfbhVl^+h'' 


anil  the  deflection 


3-      ^^^,     _4^jr^^_12HyL_ 
GVb^  +  h'^'^GVlP+h^     Gbhib'  +  h")' 

For  a  square  section  b  =  h. 

These  results  must  be  modified  in  accordance  with  the  deductions 
of  Art.  3. 


662 


THEORY  OF  STRUCTURES. 


Helical  Spring  in  Torsion. 

Let  a  moinont  of  Mt  in.-lbs.  twist  the  spring  through  an  angle  t) 
(measured  in  radians). 


(I 

(I 


rii,  n2  be  the  number  of  free  coils  before  and  after  twisting. 


yuVi  "    "    r^'dii  of  the  spring 


It        ((        (( 


Then 
and 


2;rj/ini=L=2;:j/2«2 
fl  =  2;r(ni  —nz). 


Therefore         AU^Ell^  -^)  J^{n,  -n^)  =  ^. 
\yi    yzl      L   '  '      '^     L 


The  angle  of  twist  in  degrees  = 


ISJOO 


Tzd* 
For  a  round  wire  of  diameter  d,  7  =  -^. 

d4 


square 


side  d, 


7  =  ^ 
12 


Open  Coiled  Helical  Spring. — In  the  preceding  discussion  the 
coils  are  assumed  to  be  so  flat  that  the  strain  is  taken  to  be  one 


V 


of  torsion  only.  If  the  obliquity  of  the 
helix  is  large,  the  bending  effect  can  no 
longer  be  disregarded.  In  addition  to  the 
load  W,  applied  axially,  and  tending  to 
elongate  the  spring,  let  a  couple  T,  about 
the  axis,  tend  to  increase  the  number  of 
coils. 

Consider    the    portion    of    the    spring 

below   a    section    at    0.    For   equilibrium 

the  molecular  forces  developed  in  this  section  must  balance  W  and  T. 

Drawing  OY  parallel  to  the  axis  and  OX  at  right   angles,  the 

load  W  produces  a  moment  Wy  about  OX. 

Draw  OY'  at  right  angles  to  OX',  the  axis  of  the  wire.    Then 
Wy  and  T  m.ay  o.arh  be  resolved  into  two  components,  viz., 

Wy  cos  a  and  T  sin  a,  producing  bending  about  OY'. 
Wy  sin  a  and  T  cos  a,        ' '        twisting  about  OX'. 


Fio.  594. 


1    4. 


HELICAL  SPRINGS. 


663 


Thus,  taking  as  positive  the  directions  of  closer  winding  and 
axial  elongation, 

the  total  torque  about  OY' =  Wy  sin  a  -  7  cos  a 

and  the  total  torque  about  OX'  =  Wy  cos  a -k-T  sin  a 

=  EiI=Ai, 

J  being  the  polar  moment  of  inertia  and  A(^  =  EI)  and  B^^GJ)  the 
flexural  and  torsional  rigidities  of  the  wire.    Then 

^= angle  of  twist  per  unit  of  length 

Wy  sin  g  —  r  cos  a 
B 

and  t  =  angle  of  bending  per  unit  of  length 

Wy  cos  g  +  r  sin  g 
A 

Hence  if  4>  and  i  are         otal  angular  rotation  and  axial  elonga- 
tion, 

4>    .  . 

•q=i  sm  g  —d  cos  a 

riT     •  I  ^      ^\  ,  m  h^^^ «     sin2  g\ 

=  TTy  sm  g  cos  «  ( J -;gj  +  r  (^-^  + -g- j, 


and  ■cr'=i  cos  a-\-d  sin  a 


_,  /cos  ^a    sin2  a\     „  .  /  1      1\ 

-  Try(^-^  + -g- j  +  r  sm  g  cos  «  ( J --g) . 

A  positive  value  of  4>  indicates  that  as  the  spring  elongates  the 
winding  increases.  If  the  coils  are  very  flat  and  a  therefore  so 
small  that  it  may  be  taken =0,  then 


<t>    T 


Wy 


S~B     *"^    Sy~  A  ' 


004 


THEOhY  OF  STRUCTURES. 


indicating  that  an  axial  force  does  not  tend  to  produce  rotation,  and 
tiiat  no  axial  elongation  will  be  produced  when  the  spring  is  sub- 
jected to  a  coujile  only. 

Ex.  14.  -I  2-1/1.  In  Ileal  spring  with  30  coils  is  made  of  \-in.  steel  irirc.     Find 
the  dcfleetion  uiukr  a  load  «/  1  lb.,  the  ewfjicient  of  disturlion being  12,000,000  lbs. 


Deflertion  = 


Therefore 


2UVL^4nV« 
G-r'    ~    iir*   • 

4x1X1X30 


delieetion  =[.j;,-;^^>'^^  =.04096  in. 


EX.AMPLES. 

1.  A  steel  shaft  4  ins.  in  diameter  is  subjected  to  a  twisting  couple,  which 
produces  a  circumferential  stress  of  15,000  lbs.  What  is  the  stress  (shear)  at 
a  point  1  in.  from  the  centre  of  the  shaft?    Determine  the  twisting  couple. 

Ans.  7.500  lbs.;  188,.';71f  lbs. 

2.  .\  weight  of  2|  tons  at  the  end  of  a  1-ft.  lever  twists  asunder  a  steel 
shaft  1?  ins.  in  diameter.  Find  the  breaking  weight  at  the  end  of  a  2-ft.  lever, 
and  also  the  modulus  of  rupture.  Ans.   li  tons;  53.4  tons/sq.  in. 

3.  A  couple  of  A'  ft  .-tons  twists  asunder  a  shaft  of  diameter  d.  Find  tlie 
couple  which  will  twist  asunder  a  shaft  of  the  same  material  and  diameter 
2d.  Ans.  HX. 

4.  Compare  the  couples  required  to  twist  two  shafts  of  the  same  material 
through  the  same  angle,  the  one  shaft  being  /  ft.  long  and  d  ins.  in  diameter, 
the  other  2/  ft.  long  and  2^/  in.  in  diameter.  Compare  the  couples,  the  diam- 
eter of  the  latter  shaft  being-.  Ans.  I  to  8;  32  to  1. 

5.  .\  shaft  1.5  ft.  long  and  4J  ins.  in  diameter  is  twisted  through  an  angle 
of  2°  under  a  couiile  of  2000  ft  .-lbs.  Find  the  couple  which  will  twist  a  shaft 
of  the  same  material  20  ft.  long  and  7s  ins.  in  diameter  through  an  angle  of 
2i°.  Ans.  12,288  ft.-lbs. 

6.  A  wrought-iron  shaft  20  ft.  long  and  .5  ins.  >n  diameter  is  twisted  through 
an  angle  of  2°.  Find  the  maximum  stress  in  the  material,  G  being  10,500,000 
ft.-lbs.  Ans.  .1S19.2  lbs.  i)er  sq.  ins. 

7.  A  shaft  1  in.  in  dianwter  can  safely  transmit  a  torque  of  2400  Ib.-ins. 
What  diameter  of  shaft  would  l)e  required  for  transmitting  15  H.P.  at  20O 
revolutions  fier  minute?  Ans.  Ijins. 

8.  The  amount  of  twist  in  a  solid  shaft  is  to  be  limited  to  1°  for  each  10  ft. 
of  length.  Find  the  diameter  for  a  twisting  moment  of  50  in.-tons,  the  modu- 
lus of  torsional  rigidity  Ix'ing  10,000,000  lbs.  ix?r  sq.  in.  Ana.  5.143  ins. 

9.  A  crane  chain  exerts  a  pull  of  (iOOO  lbs.  tangentially  to  the  drum  uixm 
which  it  is  wrapped.  F'ind  the  diameter  of  a  wrought-iron  axle  which  will 
transmit  the  resulting  couple,  the  effective  radius  of  the  drum  being  7^  ins., 
and  the  safe  working  stress  per  square  inch  7200  lbs.  Ans.  3.17  ins. 


•  if^l 


EXAMPLES. 


665 


10.  Find  the  diameter  and  the  total  angle  of  torsion  of  a  12-ft.  wrought- 
imii  shaft  driven  by  a  water-wheel  of  20  H.P.,  making  25  revolutions  per 
iiiimite,  G  being  10,000,000  lbs.  and  the  working  stress  7200  lbs.  per  square 
'"'•*>•  Am.  3.29  ins.;  3°.0. 

11.  A  brass  wire  20  ins.  long,  0.1  in.  diameter,  twists  through  a  total  angle 
of  130°  when  a  twisting  moment  of  4  in.-lbs.  is  applied.  Find  G  for  the  ma- 
terial. What  would  be  the  twist  of  a  shaft  of  the  same  material  with  a  twist- 
ing moment  of  600  in.-lbs.,  20  ft.  long,  1.2  ins.  diameter? 

Ans.  3,600,000  lbs. /sq.  in.:   110  28' 

12.  A  round  iron  shaft  15  ft.  long  is  acted  ujjon  by  a  weight  of  2000  lbs. 
ai)piied  at  the  circumference  of  a  24-in.  wheel  on  the  shaft.  Taking 
0' -  6,000,000  lbs.,  find  the  diameter  of  the  shaft  so  that  the  total  angle  of 
torsion  may  not  exceed  2°.  Ana.  3.76  ins. 

13.  A  force  of  200  lbs.  at  the  circumference  of  an  S-ft.  wheel  twists  a  round 
wooden  shaft  8  ft.  long.  The  total  angle  of  torsion  is  not  to  exceed  2°.  Find 
the  diameter  of  the  shaft.  Ans.  5.174  ins.,  G  being  750,000  Ibs./sq.  in. 

14.  Calculate  the  diameter  of  a  steel  shaft  to  transmit  4000  H.P.  at  200 
revolutions  per  minute,  when  the  allowable  stress  on  the  metal  is  12,000  lbs. 
lK>r  square  inch.  Am.  8.12  ins. 

15.  Deduce  the  diameter  of  a  shaft  to  transmit  300  H.P.  at  200  revolutions 
]ier  minute  when  the  allowable  stress  is  10,000  lbs.  per  square  inch. 

Ans.  3.64  ins. 

16.  Fmd  the  diameter  of  a  solid  steel  shaft  which  is  to  transmit  a  moment 
of  392,700  ft.-lbs.,  the  .-naximum  allowable  shear  stress  bein?  11,200  Ibs./sq. 
'"'"h-      ^  .Ins.  12.9  ins. 

17.  Calculate  ihe  diameter  of  a  hollow  shaft  required  to  transmit  1000 
H.P.  at  50  revolutions  jwr  minute,  the  skin  stress  being  6,000  lbs.  |3er  square 
imh,  and  the  internal  diameter  f  of  the  external  diameter. 

.4n«.  Ext.  diar.  =  10.71  ins. 

18.  In  a  4-in.  shaft  10  ft.  long  the  maximum  shear  stress  is  10,000  Ibs./sq. 
in.  and  6"  =  10,000,000  Ibs./sc).  in.     Find  the  twisting  couple  and  the  torsion. 

Ans.  125,714f-in.-n)s. ;  1°.374. 

19.  Determine  the  diameter  of  a  wrought-iron  shaft  for  a  screw  steamer, 
and  the  torsion  per  lineal  foot;  the  indicated  H.P.  =1000,  the  number  of 
revolutions  jier  minute -150,  the  length  of  the  shaft  from  thrust-bearing  to 
.srew  =  75  ft.,  and  the  safe  working  stress  =7200  lbs.  ]ycT  square  inch. 

.!//.«.  6.67  ins.;  10°.5. 

20.  In  a  spmning-mill  a  ca.st-iron  shaft  84  ft.  long  makes  'tO  revolutiona 
per  minute  and  transmits  270  H.P.  Find  its  diameter  (1)  if  the  stress  in  the 
metal  is  not  to  exceed  5000  lbs.  per  square  inch;    (2)  if  the  angle  of  torsion 

JO 

per  lineal  foot  is  not  to  exceed  jg .     Also  (3)  in  the  first  case  find  the  total 

'"••sion.  ,4ns.  (1)  7.02  ins.;  (2)  10.23  ins.;  (3)  13°.048. 

21.  A  shaft  2  ins.  in  diameter  and  140  ft.  long  is  used  to  transmit  30  H.P. 
ai  300  revolutions  per  minute.  Find  the  angle  through  which  the  shaft 
i^prings  nd  the  skin  stress  in  the  material.  Modulus  of  rigidity  -5000  tons 
per  squ^/e  inch.  Ans.  38°.57:  4010  Ibs./sq.  inch. 


!    1 1 


666 


THEORY  OF  STRUCTURES. 


22.  A  wrought-iron  shaft  in  a  rolling-mill  is  2^0  feet  in  length,  makes  95 
revolutions  per  minute,  and  transmits  120  H.P.  to  the  rolls;  the  main  body 
of  the  shaft  is  4  ins.  in  diameter,  and  it  revolves  in  gudgeons  SJ  ins.  in  diam- 
eter. Find  the  greatest  shear  stresses  in  the  actual  shaft  and  also  the  necessary 
diameter  of  the  shaft. 

Ans.  6330   and  76S2   lbs.;  3.50   ins.    if/ =9000  Ibs./sq.   in.;  5.12   ins.   if 

^=13X12. 

23.  A  torce  of  5000  lbs.  at  the  end  of  a  6-in.  lever-arm  twists  a  fiO-in.  shait 
of  2  ins.  diameter  through  an  angle  of  7°.     Find  the  modulus  of  rigidity. 

Ans.  9,372,000  lbs. 

24.  Determine  the  twisting  moment  and  the  torsion  for  a  4-in.  shaft  10 
ft.  long  and  subjected  to  a  maximum  shear  stress  of  10,000  Ibs./sq.  in.,  G 
being  i2,.WO,000  Ibs./sq.  in.  Ans   125,714f  in.-lbs.;  2''.7. 

25.  \  3-in.  shaft  40  ft.  long  springs  6°  when  transmitting  power  at  150 
revolutions  per  minute.  Find  the  H.P.  transmitted,  taking  G  =  12,000,000 
Ibs./sq.  in.  Ans.  49.6. 

26.  A  shaft  transmits  a  given  H.P.  at  .V  revolutions  per  minute  without 
bending.     Find  the  weight  of  the  shaft  in  jwunds  jwr  lineal  foot. 

An...  32.0  f-'^)'. 

27.  The  shafting  of  the  turbines  at  Niagara  Falls  consi.sts  of  a  steel  tube 
3S  ins.  in  diameter  and  J  in.  thick.  Find  what  horse-power  can  be  transmitted 
at  '2oO  revolutions  per  minute  when  the  working  stress  is  limited  to  0000  lbs. 
per  square  inch.  .\lso  find  the  diameter  of  a  solid  shaft  which  will  be  equiv.i- 
lent  to  the  above.  Ans.  57030;  20.11  ins. 

28.  A  vertical  cast-iron  axle  in  the  Saltaire  works  makes  92  revolutions 
per  minute  and  transmits  300  H.P.;  its  diameter  is  10  ins.  Find  the  angle  of 
torsion.  .ln.«.  .022S2'' per  lineal  foot. 

29.  .\  .'■olid  shaft  is  subjected  to  a  twisting  moment  of  .50  in. -tons,  the 
modulus  of  torsional  rigidity  b<:ng  .5000  tons/.square  inch.  Find  its  diam- 
eter, a.ssuming  (a)  that  the  amount  of  twist  is  limited  to  l°for  each  10  ft. 
of  length,  (b)  that  the  working  stress  is  nowhere  to  exceed  3i  tons/square 
inch.  Ans.  (a)  5.144  ins.;  (6)  4.175  ins. 

30.  Find  the  diameter  of  a  .shaft  which  is  to  transmit  25  H.P.  at  50  revo- 
lutions per  minute,  and  in  which  the  working  stress  is  to  be  7000  lbs.  per  square 
inch.  Ans.  2.84  ins. 

31.  .\  line  of  steel  shafting  is  80  ft.  long;  if  a  twisting  moment  of  4000 
lb. -ins.  is  applied  at  one  end,  what  will  l)e  the  total  angle  of  twist,  the  diam- 
eter of  the  shaft  Iwing  2i  ins.?  What  horse-jwwer  will  the  shaft  transmit  at 
220  revolutions  per  minute?  .1m,v.  5". 2;  14  H.P. 

32.  .A  turbine  makes  114  revolutions  per  minute,  and  transmits  !)2  H.P. 
through  the  medium  of  a  shaft  8  ft.  0  ins.  in  length.      What  mu.st  l)e  the 

.)0 


diameter  of  the  shaft  so  that  the  total  angle  of  torsion  may  not  exceed  * 


3* 


a  being  10,500,000  lbs.? 


An.i.  4.7  ins. 


EXAMPLES. 


667 


Determine  the  side  of  a  square  pine  shaft  that  might  be  substituted  for 
the  iron  shaft      (0=525,000  lbs.)  .4ns.  8.45  ins. 

33.  Iiiul  the  diameter  of  a  shaft  for  a  winding-drum  which  works  under 
the  following  conditions:  The  load  lifted  is  IJ  tons;  diameter  of  drum,  5  ft.- 
width  of  face  of  drum,  2G  in.s.;  distance  from  inner  face  of  drum  to  the  middle 
of  the  l-aring  of  shaft,  1.3  ias.;  maximum  stress,  7000  lbs.  per  square  inch 
<"'"=-;'»•' '^''*-)  ^ns.  4.82  ins. 

34.  Tne  diameter  of  one  shaft  is  double  that  of  another  of  the  same 
mutcrul;  the  smaller  gave  way  when  subjected  to  a  twisting  moment  of  2 
ft.-tons.    What  twsting  moment  will  be  required  to  wrench  the  other? 

.     .         .,,,.,,  -^"•^-  lO  ft.-tons. 

35-  An  iron  shaft  of  which  the  working  stress  must  not  exceed  548  k  /cm 
18  acted  ui)on  by  a  couple  equivalent  to  a  foice  of  .50  k.  at  the  end  of  a  lever 
0.1  m.  in  length.     Find  the  diameter  of  t'c  .shaft.  1,,,,    o  05  (.,„ 

36.  A  malleable  iron  shaft  20  ft.  long  and  fi  ins.  diameter  is  subiectedto 
a  moment  which  twists  the  ends  through  an  angle  of  2°;  taking  r/,  the  coefficient 
of  transverse  elasticity,  as  9,000,000,  find  /,  the  stress  at  the  skin. 

,-    IT    ,1  «u    J-  ,  ^'**-  •''^2Si  Ibs./aq.  inch. 

37.  Und  the  diameter  of  an  iron  shaft  which  is  to  transmit  l'>0  H  P  at 
<i0  revolutions  per  minute,  the  safe  working  strength  being  548  k.,/cm.'! 

38.  A  water-wheel  of  20  H.P.  makes  5  revolutions  per  minute-  find"the 
duiineter  suitable  for  the  maleable  iron  shaft  which  transn-i-s  this 'force  If 
the  shaft  IS  12  ft.  long,  what  is  the  angle  of  torsion?  (/  =  0000  lbs.,  C  -9,000  000 

''•''^*'-  '"-^  Aru,.  5.224  ins.;  3'.  156 

39.  How  many  H.P.  may  be  tran.smitted  by  a  shaft  of  1.50  mm.  diameter 

40.  A  shaft  12  ft.  long  and  6  ins.  diameter  is  subjected  to  a  twrstinK  mo- 
ment of  K,  ft  -tons  and  the  two  ends  are  thus  twiste.1  through  a  certain  angle- 
a  second  shaft  of  the  siime  material,  16  ft.  long  and  9  ins.  diameter,  is  twisL 
•so  that  Its  angle  of  torsion  is  exactly  the  same  as  that  of  the  firs  •   fi„,l  ,.,0 
.w.s„ng  moment  required  to  ao  this.  ,,„,.  ooj  i'^.t,.,,: 

41.  Find  the  maximum  stress  developed  in  a  shaft  of  120  mm  diai.w.ter 
wh.cb  transnnt.s  200  H.P.  at  .50  revolutions  per  minute.     Ans.  Zk    Z' 

y.  If  /,  ,8  thp  .s,if,.  torsional  working  stress  of  «  shaft,  and  /,  is'the  safe 

of  the  shaft  IS  to  Its  l)en(ling  resistance  in  the  ratio  of  2/,  to  /,. 

43.  A  circular  shaft  is  twisted  Ix-yond  the  limit  of  elasticity.  If  the  enu.li. 
ration  of  stress  is  rn-rfcct , show  that  for  u  given  maximum  st».ss  the  tZvL 
.oupK.  IS  greater  in  the  ratio  of  4  to  3  than  it  would  be  if  the  elasnVitn.  rf 

44.  Show  that  the  resilience  of  a  twisted  shaft  is  pro,K,rtional  to  its  weight. 

An».  Kesilience-i-' ""''""'•• 
G       4 


668 


THEORY  OF  STRUCTURES. 


45-  If  a  round  bar  of  any  material  is  subjected  to  a  twristing  rouple,  show- 
that  its  maximum  re.<='=ence  is  two  thirds  the  maximum  resilience  of  the 
material. 

46.  A  shaft  nioving  with  a  surface  veiocity  of  10  ft.  ikt  second  transmits 
100()  H.r.  Find  the  diameter  if  the  shear  stress  is  not  to  ex<ced  10,()00  lbs. 
I)er  s(iuare  inch.  .(„,,    :{.74  i„s. 

47.  A  windiiiK-drum  '20  ft.  in  diameter  is  used  to  raise  a  load  of  ,')  tons.  If 
the  driving-shaft  were  in  |)ure  torsion,  find  the  diameter  for  a  stress  of  .3  tons 
I)tr  square  inch.  .i,,.,    ,u  j  ;„„ 

48.  Find  the  thicknes.s  of  a  hollow  shaft  when  (a)  its  rigidity,  (6)  its 
strength  is  L'.j' ,  greater  than  that  of  a  solid  shaft  of  diameter  t/ of  the  same 
length  and  weight.  ^„,v.  („)  ,;  x  •  ^535,  (ft)  d  X  •  3316. 

49.  Find  the  percentage  of  weight  saved  by  using  a  hollow  instead  of  r 
solid  shaft. 

•)()Q 

Arts.  If  of  equal  .v/iff>ir.s.s  =  -%  -v. 

in-  + 1 

Ifofequal..v.«,M=,0()jl-^^5^[, 

m  iM'ing  the  ratio  of  the  external  to  the  internal  diameter 
of  hollow  shaft. 

50.  A  solid  and  a  hollow  cylindrical  shaft  of  equal  length  contain  the 
same  amount  of  the  same  kind  of  metal,  the  solid  one  fitting  the  hollow  of  <he 
other.     Compare  their  torsional  stn-ngths  when  used  separately. 

Alls.  Strength  of  solid  shaft  =.471  xthat  of  hollow  shaft. 

5".  Find  the  diameter  of  a  hollow  shaft  required  to  transmit  .")000  H.P.  at 
70  revolutions  [ht  minute;  sli.ss,  7.">00  lbs.  jxt  square  inch;  the  external 
diameter  In'ing  twice  the  inner;  maximum  twistingmoment  =  li  times  the  mean. 

5^-  A  stcl  b.ir  having  a  diatneter  of  .410  in.  and  a  length  under  test  of 
4  ins.  gave  the  following  results  in  the  testing-machine: 

Torcjuo  in  In.  lb.-.  Angle  of  Twici. 

l.W  1°  1' 

.300  -2°  3' 

Calculate  tli"  nuMJulus  of  rigidity. 


Turgue  in  In. -lbs. 

450 
600 


Ancla  of  Twist. 
3°.V 
4°  6' 


53.  The  halves  of  a  flange  couplii.g  for  a  shaft  transmitting  00  H.P.  at 
100  revolutions  |ht  minute  arc  bolted  together  with  six  lK)ltsat  Oins.  from  the 
centre.  I'ind  the  diameter  of  the  bolts,  the  safe  shear  stress  l)eing  8000  lbs. 
pers(|uarein.h.  4,,^     4,  ;„ 

54.  A  steel  shaft  4  ins.  in  diameter  and  weighing  4!H)  lbs.  jkt  cubic-  foot 
makes  100  revolutions  [kt  minute.  If  the  working  stn-ss  in  the  metal  is  1 1  ,L'00 
lbs.  (H-r  s.|uan>  inch,  find  the  twisting  couple  jind  the  distance  to  which  the 
work  can  U-  trai  sunlle.l,  <he  c<H'flicient  of  friction  lieing  .0',  and  the  efTi- 
rien.y  of  the  shaft    J.  ,!„.,.   140,SOO  in.-lbs.;  S-.>-.'Sj  ft. 

$%.  If  the  shaft  i;  of  sfccl,  and  if  the  I..ss  due  to  friction  is  20  |ier  cent, 
find  the  distance  to  which  work  may  Iw  transmitted,  ft  heiuft,  .().'). 

Am.  658'2|  ft. 


EXAMPLES. 


669 


56.  A  wrought-iron  shaft  220  ft.  between  bearings  and  4  ins.  in  diameter 
can  safely  transmit  120  H.P  at  the  rate  of  95  revolutions  per  minute.  What 
is  the  efficiency  of  the  shaft?     (/i  =A.)  Ans.  .976. 

57-  The  efficiency  of  a  wrought-iron  shaft  is  i;  the  working  stress  in  the 
metal  is  7200  lbs.  per  .square  inch;  the  coefficient  of  friction  is  .125.  How 
far  can  the  work  be  transmitted?  Ans.  4.320  ft 

58.  The  workirjg  shear  stress  of  a  shaft  is  15,000  lbs.;  how  far  can  work 
be  transmitted  with  an  efficiccy  of  .5,  the  coefficient  of  friction  being  .5? 

Ans.  24,.">45.4o  ft. 

59.  Take  a  round  shaft  3  ins.  in  diameter  and  find  the  sizes  of  equivalent 
shafts  of  square,  elliptic,  and  rectangular  sections  if  the  breadth  and  thickness 
of  each  of  these  latter  are  as  1  to  2.  If  these  shafts  are  20  ft.  long,  and  they 
are  transmitting  20  H.P.  at  100  revolutions  per  minute,  what  is  the  total  twist 
of  each  of  them?     {G  =  10,.")00,000.) 

.1«.^.  2.7.3  ins.,  1.78;  2.38  ins.,  1°.05;  2.15  ins.X4.3  ins.,  .93". 

60.  A  WTought-iron  shaft  200  ft.  in  length  and  weighing  440  lbs.  per  cubic 
foot  IS  supported  on  l)oarings,  the  cwfficient  of  friction  being  .05.  The  shaft  is 
subjected  to  a  uniform  twisting  coui)l(.  which  developes  a  stress  of  10,000  lbs.  per 
square  inch.     Find  the  efficiency  of  the  shaft.  .4n,s.  .994. 

61.  An  iron  shaft  of  1  in.  diameter  is  .subjected  to  a  turning  effort  of  rm 
ft -lbs.  The  shaft  is  1000  ft.  long;  find  its  efficiencv  in  so  far  as  it  is  affected 
by  its  weight.  .         _„ 

Ats.  .99. 

62.  \  hollow  cast-iron  shaft  of  12  ins.  external  diameter  is  twisted  bv 
u  couple  of  27,000  ft.-lbs.  Find  the  proper  thickness  of  the  metal,  so  that 
the  stros.s  may  not  exceed  5000  lbs.  |)er  square  inch.  .4,,,,.  ..30S  in. 

63.  What  twisting  moment  can  \>c  transmitted  by  a  hollow  steel  shaft 
of  Sins,  internal  and  10  ins.  external  diameter,  the  working  stress  being  5 
tons  ,K.r  square  inch?  ,,„,.  57;,,  -^^^^J 

64.  1  he  inner  and  outer  diameters  of  a  hollow  steel  shaft  are  10  and  12  ins 
and  /,  =.r>  tons  |)er  square  inch,  is  the  working  value  of  the  resistance  to  s'liear- 
ing.     Uhat  IS  the  twisting  moment  this  shaft  is  capable  of  transmitting? 

-,    \\i    .  ,1  •  1  r  ,  .  ■^''•'-  10.51?  in. -tons. 

05.  >\liat  thickness  of  metal  is  r.'quired  for  a  cast-iron  hollow  shaft  of  10 
ins.  outer  diameter  so  jis  to  resist  a  twisting  moment  of  10  ft.-tons? 

66.  A  hollow  shaft,  the  external  and  internal  diameters  of  which  are" '^0 
ins.  an.l  8  ins.  resin-ctively.  r-ns  at  70  revolutions  |K>r  minute  with  a  surface 
stress  (.f  (KXM)  lbs.  per  square  inch.  Find  the  twisting  moment,  and  horse- 
|)ower  tr.insinitted. 


khli 


67.  .\  solid  shaft  IS  10  ins.  in  diameter,  and  the  internal  diameter  of  a 
hollow  shaft  IS  5  ins.,  find  the  ext.  riial  diameter  and  compare  the  torsional 
-irciigths,  the  shafts  l)eing  of  the  same  weight  and  material. 

-a     I  I    11  ,  -l"".  •'•N  5  ins.;  v'5to3. 

08^  A  hollow  steel  shaft  has  an  external  diameter  d  an<l  an  internal  diam- 


eter 


Compare  its  torsional  dlreiigth  with  that  of  (a)  a  solid  steel  shaft 


670 


THEORY  OF  STRUCTURES. 


of  f'liampter  d;  (b)  a  solid  wroupht-iron  shaft  of  diameter  d;  the  safe  working 
stresses  of  steel  and  iron  Ix'iii!:  .")  tons  and  3}  tons  respectively. 

Ans.  (a)U;  (h)  H 
09.  A  sohd  wroupht-iron  shaft  is  to  be  replaced  by  a  hollow  steel  shaft 
of  the   same  diameter.     If  the  material  of  the  latter  is  30  per  rent  stronger 
than  that  of  the  former,  what  must  be  the  ratio  of  internal  to  external  diame- 
ter?    A\hat  is  the  jx'rcentage  saving  in  weight?  Ann.  1.44;  4'^'[. 

70.  A  square  stet>l  shaft  is  required  for  transmitting  power  to  a  30-ton 
overhead  travelling-crane.  The  load  is  lifted  at  the  rate  of  4  ft.  |)er  minute. 
Taking  the  mechanical  efficiency  of  the  crane  gearing  as  35  fx-r  cent,  calcu- 
late the  necessary  size  of  shaft  to  run  at  100  revolutions  per  minute.  The 
twisf  must  not  exceed  1°  in  a  length  equal  to  30  times  the  side  of  the  square. 
('•  =  13,000,(K)0.)  .4 «s:  2  ins.  square. 

71.  A  round  cast-iron  shaft  15  ft.  in  length  is  acted  uixin  bv  a  weight 
of  1.>(K)0  lbs.  ap()lie(l  at  the  circumference  of  awheel  on  the  shaft;  the  diameter 
of  the  wheel  is  2  ft.  Finil  the  diameter  of  the  shaft  so  that  the  total  angle  of 
torsion  may  not  exceed  2°.  ,i„,,.  3.7,;  i,„_ 

72.  A  wrought -iron  shaft  is  subjected  to  a  twisting  couple  of  12,000  ft.-lbs.; 
the  length  of  the  shaft  In-tween  the  .sections  at  which  the  power  is  received' 
and  given  olT  is  30  ft.;  the  total  ailmissible  twist  is  4°.  Find  the  diameter 
of  the  shaft,  /<  (Art.  (1)  IxMUg  J,  and        0,000,000  lbs.  Ans.  5.S  ins. 

73.  Find  the  horse-power  which  -nay  Ik-  transmitted  bv  a  shaft  4  ins. 
in  diameter  when  ruiming  at  l.V)  re\oiutions  jkt  minute,  if  the  stress  due  to 
twisting  l)e  limited  to  0(XK)  lbs.  [x-r  square  inch.  .4,,.,.  270. 

74-  The  working  stress  in  a  steel  shaft  subjected  to  a  twisting  couple  of 
KKK)  m.-tons  is  hmited  to  11,200  lbs.  p<.r  square  inch.  Find  its  diameter- 
also  find  the  diameter  of  the  steel  shaft  whi.h  will  transmit  5000  H.P.  at  (i(i 
revolutions  jx-r  minute,  /<  Ix-ing  J.  .-l,,,,.  10  i„s.;  (i.48  ins. 

75.  A  wrought-iron  shaft  is  twisted  by  a  couple  of  10  ft.-tons.  Find 
its  diamet(  r  (a)  if  the  torsion  is  not  to  exceed  1°  [ht  lineal  foot,  (b)  if  the  safe 
working  stress  is  7200  lbs.  ixt  s(|uan'  inch.     (G  =  10,000,0(K)  lbs.) 

Am.  (a)  3.7  ins.;  (b)  .5.7  ins. 

76.  A  steel  shaft  2  ins.  in  diameter  makes  100  revolutions  |x<r  minute 
and  transmits  25  H.P.  Find  the  maximum  working  stress  and  the  torsion 
|^«r  hncal  foot,  (;  Mne  10.(KH»,(HK)  lbs.  Also  l,.>d  the  diameter  of  a  shaft 
of  the  same  material  which  will  transmit  100  U.V.  with  the 
working  stress.  .t,,,,.  i(),oo_.^,^  n,^ 

77.  A  steel  shaft  .3(K)  ft.  in  length  makes  2(K)  revolut 
transmits   10   |!.l'.       Determine  its  diameter  so  that   the  greatest   st 


the  material  inav  In-  tl 


shaft  1  in.  in  diameter  and  tri 


II'  same  as  the  stress  at  the  circumfei 


same  maximum 
0574°;  3.17  ins. 
[XT  minute  and 

ress  in 


ivnce  ol  an  iron 


msniitting  rm  ft.-lbs.     If  |()  is  a  factor  of  safct^ 


find  the  coefli.ient  of  torsional  ruj.tiire.  .l„.s'.  .,S07  in.  (  =<  i,,.);  (K),()()0. 

78.  A  iouikI  l)ar  of  steel  is  I  in.  in  diameter  and  S  ft.  in  length  (or  /-IS 
ms.).  Take  /•■  1.500  lbs.  Show  that  an  endlong  |„ad  oiilv  sufficient  of  its.>lf 
to  |,ro<lu-e  a  stress  of  MMO  lbs.  iht  square  inch.an.i  a  U-nding  m.m.ent  which 


by  itself  woijld  only  produce  a  stress  of  ,S1(>  11 


together,  produce  a  stress  of  23,1  !K)  lbs.  ix>r  K(|uare  inch 


>s.  |KT  square  inch,  if  lioth 


EXAMPLES. 


671 


79.  A  wooden  shaft  of  d  ins.  diameter  and  I  ins.  length  makes  n  revolutions 
jHT  second  and  "whirls."  If  the  weight  of  the  wood  is  36  lbs.  per  cubic  foot, 
the  moclulus  of   elasticity  being  2,000,000  lbs.   per  square  inch,  show   that 


4- 


8o.  A  pulley  is  keyed  truly  to  a  shaft  which  rotates  with  an  angular 
velocity  <o.     If,  when  rotation  takes  place,  the  shaft  bends  slightly,  show  that 

the  couple  on  the  shaft  is  equal  to  (o'(.4-B)^,  in  which  A  and  B  are  the 

inoiiicnts  of  inertia  of  the  pulley  about  axes  through  its  centre  of  gravity 
luTix-iidicular  to  its  plane  and  perpendicular  to  the  axis  of  the  shaft  respec- 
tively, antl^T^is  the  inclination  of  the  plane  of  the  pulley  to  a  plane  perpen- 
dicular to  the  original  alignment  of  the  shaft. 

8i.  If  a  thin  disk  weighing  10  lbs.  and  of  10  ins.  diameter  rotates  at  1000 
revolutions  per  minute  about  an  axis  through  its  centre,  and  if,  instead  of 
Ixirig  jxTjiendicular  to  the  shaft,  it  is  out  of  truth  by  ^U  of  its  radius,  find 
tlic  couple  on  the  shaft  in  inch-pounds. 

Si.  A  wrought-iron  shaft  is  subjected  to  a  twisting  moment  of  3(i(HH) 
ll..-ins.  and  a  bending  moment  of  18,000  Ib.-ins.;  find  the  diameter  when'the 
tiiaximum  shear  stress  is  80(K)  lbs.  per  square  inch.  Find  also  the  twisting 
inDincnt  which  alone  would  produce  a  shear  stress  of  the  same  numerical 
^ '''"*"•  Atis.  3.322  ins.;  58,250  in.-lbs. 

83.  A  screw  propeller-shaft  10  ins.  in  diameter  is  subjected  to  a  twisting 
moment  of  35  ft.-tons,  and  to  a  l)ending  moment  of  10  ft.-tons,  duo  to  the 
weight  of  the  shaft  and  the  pitching  of  the  ship.  What  is  the  maximum 
compressive  stress  if  the  thrust  of  the  screw  is  10  tons?  .4n.s.  2.92  tons. 

84.  Find  the  diameter  of  a  wrougUt-iron  shaft  to  transmit  90  H.P  at  130 
r.v..luti<.ns  |)cr  minute.  If  tlwre  is  a  bending  moment  equal  to  the  twisting 
iiioiiuMit,  what  ought  to  Ijc  the  diameter?  Ann.  .3..-)4  ins;  4.75  irw 

85.  A  round  bar  I  in.  in  diameter  and  3f)  ins.  between  the  supports  de- 
lists ^1013  in.  under  a  load  of  <K)  lbs.  in  the  middle  and  twi.sts  through  an  angle 
-I    ,..°  when  subjected  to  a  twisting  moment  of  1000  in  -lbs.  on  a  10-iii.  length 

""'''•"""'•  A,i^.  3X10';  133X10». 

86.  A  wrought -iron  shaft  3  ins.  in  diameter  and  making  140  revoliitioM- 
IHT  nimute  is  sup|H)rtcd  on  wall-brackets  U\  ft.  apart.  There  is  a  pulley  on  1  'u- 
.shaft  midway  Ix'tween  the  In-arings.  If  the  resultant  sid.«  pull  due  t<.  the 
".Mirht  of  the  (Milley  and  the  pull  r.f  the  I  elt  is  210  lbs.,  what  is  the  grciKst 
linrM..,M)wcr  the  shaft  will  tran.smit  with  safety?  Safe  shear  stress  7,S0()  lbs 
IHT  square  inch.  _,^,^     ,„.  ' ' 

8r.  A  shaft  12  ins.  m  diameter  tran.^mitting  a  twisting  moment  of  100 
...=.,»..,  m  alw,  subject  toa  brnding  moment  of  20  fl-t..us.  Find  the  maximum 
stress  induced.  Aus.  4.3  torn, '^.  ia. 


M 


ir' 


1  *  1  i ' 


I  i 


THEORY  OF  STRUCTURES. 


88.  If  n  round  bar  1  in.  in  diameter  and  40  ins.  between  supports  de- 
fieits  .0!)3()  in  under  a  load  of  100  lbs.  in  the  middle,  and  twists  through 
an  atigle  of  .037  radian  when  subjected  to  a  twisting  moment  of  1000  in.-lbs. 
throughout   its  length  of  40  ins.    find  K,  (i,  and  A'. 

Ana.   14,510;  5,510,  and  13,191  tons/sq.  in. 

8g.  .\  steel  shaft  earries  a  5-ft.  i>ulley  midway  l)etwecn  the  supports  and 
makes  six  revolutions  (ler  minute,  the  tangential  force  on  the  pulley  being 
500  lbs.  Taking  the  coefficient  of  working  strength  at  11,200  lbs.  per  square 
inch,  find  the  diameter  of  the  shaft  and  the  projjer  distance  between  the  bear- 
ings   the  stiffness  of  the  shaft  being  t/(T(i. 

go.  A  counter-shaft  10  ft.  Ixjtween  bearings  carries  two  24-in.  pulleys,  the 
one  1  ft.  and  the  oth»>r  5  ft.  from  a  bearing.  Assuming  that  the  tight  is  twice 
the  slack  tension,  determine  (n)  the  e(|uivalent  twisting  moment  on  the  shaft, 
(/))  the  diameter  of  the  shnft,  (f)  the  angle  of  torsion  when  one  i)ulley  receives 
and  the  other  transmits  .")0  H.l'.  at  SO  revolutions  per  minute,  the  belts  being 
horizontal  and  on  opposite  sides  of  the  shaft. 

Ans.  ia)  3'.'S1.25ft.-lbs.; 

(b)  7.13  ins.,  /  lieir.g  10,000  lbs.; 
(f)  2.00  minutes,  (i  Ijeing  12,000,000  lbs. 
9'.  Find  the  greatest  torque  for  a  bar  3  ins.  in  diameter,  if  the  longitudinal 
extension  of  the  material  is  to  l)e  limited  to  jAo,  the    modulus  of  rigidity 
being  5000  lbs., sq.  in. 

92.  Power  is  taken  from  a  shaft  by  means  of  a  pulley  24  ins.  in  diameter 
wiiich  is  keyed  on  to  the  shaft  at  a  point  dividing  the  distance  between  two 
consecutive  supports  into  segments  of  40  and  tW  ins.;  the  tangential  force  at 
the  circumference  of  the  jjulley  is  (KiOO  lbs.  If  the  shaft  is  of  steel,  determine 
its  diameter,  taking  into  account  the  bending  action  to  which  it  is  subjected, 
the  working  stress  iM'ing  11,2(K)  lbs.  [xt  scj.  in.  Auk.  ,")..5S  ins. 

03.  A  wrought-iron  shaft  is  sul)je(t<'d  simultaneously  to  a  Ijetiding  moment 
of  S(KX)  in.-lbs.,  and  to  a  twisting  moment  of  15,000  in.-lbs.  Find  the  twist- 
ing-moment equivalent  to  these  two  and  the  least  sat'e  diameter  of  the  shaft, 
the  safe  shear  stress  Ix-ing  taken  at  SOOO  lbs.  [rt  stjuare  inch. 

Aus.  25,(XH)  in.-lbs.;  2.52  ins. 

04.  A  shaft  "t  ins.  in  diameter  and  12  ft.  long  is  sup|)orted  at  its  two  ends 
and  loaded  at  the  two  points  whi "h  divide  its  length  into  three  equal  i)arts 
with  4  tons  at  eacii  point ;  a  twisting  moment  of  20  ft. -tons  is  applied  to  one 
end  of  the  siiaft,  while  the  other  is  hcln  fixed.  Find  the  greatest  intensity 
of  the  tluii>t.  tension,  ami  siicaring  stress,  and  the  angle  that  the  line  of 
greatest  iirincipal  stress  makes  with  tlie  axis  of  the  shaft. 

-b/.v.  3.4!)  and  2.15  tons/sq.  in.;  25°  40'. 
95.  .\  steel  tulM'  of  10  ins.  exicrnal  diameter  and  12  ft.  long  is  sujiiwrted 
hori/onially  at  liie  ends.  \\  a  ponit  4  ft.  from  one  end  a  bracket  is  lixed  at 
right  angles  to  the  axis  of  the  tube  and  supiwirts  at  its  end  a  weight  of  32  tons, 
the  distance  iM'tween  the  centre  of  the  tube  and  the  weight  iK'ing  24  ins.  Find 
the  thickiR'ss  of  the  tube  so  that  the  stress  may  nowhere  exceed  5  tons  \m 
s<iuare  uich.  Ans.  .39  in. 


EXAMPLES. 


873 


96.  An  oncino-rrank  is  12  ins.  long  and  9000  lbs,  is  the  greatest  force 
transmitted  along  the  connecting-rod.  If  the  wought  iron  of  the  shaft  will 
safely  bear  a  shear  stress  of  9000  lbs.,  find  the  diameter  of  the  shaft,  the  hori- 
zontal distance  of  the  centre  of  the  crank-pin  from  the  centre  of  the  nearest 
bearing  being  10  ins.  ,1^    5.072  ins. 

97.  A  shaft  8  ins.  in  diameter  is  subjected  to  a  thrust  of  100  tons  uniformly 
distributed  over  its  two  ends,  and  a  twisting  moment  of  30  ft.-tons.  Find  the 
preatest  intensity  of  thrust  and  shearing  stress  and  the  angle  made  by  the  line 
of  principal  stress  with  the  axis  of  the  shaft. 

Ans.  4.71  and  3.715  tons/sq.  in.;  37°  14'. 

98.  Find  the  diameter  of  a  malleable-iron  shaft  capable  of  bearing  a  tension 
of  50  tons,  and  a  twisting  couple  whose  moment  is  25  ft.-tons,  the  resistance 
of  the  material  to  tension  and  shearing  being  5  and  4  tons  per  square  inch  respec- 
tively. Ann.  7.28  ins.;  the  diameter  for  maximum  shear  =7.08  ins. 

99.  A  steel  shaft  transmits  a  maximum  twisting  moment  of  70,000  in.-lbs., 
and  is  at  the  same  time  subject  to  a  bending  moment  of  25,000  in.-lbs.  Und 
the  necessary  diameter  of  the  shaft  if  the  safe  stress  in  the  material  is  4  tons 
ix^r  square  inch.  .         .  . 

Ann.  4  ms. 

100.  A  shaft  upon  bearings  40  ft.  apart  carries  a  pulley  at  a  point  30  ft 
from  one  bearing.  The  bending  effect  on  the  shaft  at  this  point  is  equivalent 
to  that  produced  by  a  weight  of  75  lbs.  at  the  same  point  when  the  shaft  is 
.subjected  to  a  twisting  moment  of  1000  ft.-lbs.  Find  the  proper  diameter  of 
the  shaft,  so  that  the  stress  may  nowhere  exceed  12,000  lbs.  per  square  inch. 

.  ,.,/..       .  -^n*-  2.056  ins. 

101.  An  cngine-crank  is  16  ins.  long  and  the  distance  between  the  centres 
of  the  pin  and  the  Iwaring  is  20  ins.  If  the  force  on  the  crank-pin  centre  is 
5(MK)  lbs.,  find  the  ma.ximum  intensities  of  thrust  (or  tension)  and  shear  and 
also  the  angle  between  the  line  of  greatest  principal  stress  and  the  axis  of  the 
''"'f'-  -Ins.  7127and4073  1bs./sq.  in.;  27°. 

102.  A  3i-in.  steel  shaft  is  subjected  to  a  twisting  couple  of  T  ft.-lbs  and 
to  a  H.M.  of  i  T  ft.-lbs.  Find  the  value  of  T  so  that  the  maximum  shear  stress 
may  not  exceed  15,000  lbs.  per  inch.     .\l8o  find  the  torsion  of  the  shaft  per 

lineal  foot  of  length,  G  being  12,000,000  lbs.  Ans.  31,582  in.-lbs.;  — . 

103.  Find  the  axial  load  and  the  energy  stored  up  in  a  200-mm.  helical 
;^teol  spring  of  10  coils  for  a  max.  stress  of  2000  k./cm.=,  the  steel  being  20  mm. 
m  diar.  and  taking  A'  =9 x  10»  k./cm.'.  Ans   314  k.;  2200  k. /cm. 

104.  The  fly-whet>l  of  a  direct-driven  generator  has  a  radius  of  gjTation 
of  9  ft.,  weiglis  100,000  lbs.,  and  runs  at  94  revolutions  per  minute  The 
.rank-shaft  is  22  ins.  diameter  at  the  fly-wheel  .seat.  Suppose  an  uecideiit 
Inr.ibly  brings  the  shaft  to  rest  from  full  sfXM?d  in  five  revolutions,  the  rctarda- 
tH.n  Ixjing  uniform,  find  the  stress  on  the  .shaft  due  to  the  inertia  of  the  lly- 
wheol,  acting  along  with  a  torque  of  1,600,000  in.-lbs.  due  to  the  engine  and 
a  Ix'ndmg  moment  of  3,000,000  in.-lbs.  due  to  the  weight  of  the  fly-w  eel  and 
armature  vlna.  29,118  lbs. 

105.  The  amm  shaft  of  a  steamship  transmits  lOOo  H.P.  at  80  revolutions 
l>er  minute,  and  ia  of  mild  steel  9i  ins.  in  diameter.   It  is  subjected  to  a  'ongi- 


SI 


r\ 


674 


THEORY  OF  STRUCTURES. 


tudinal  thrust  of  25,000  lbs.  Find  the  resuhant  simple  tensile  or  compressive 
stress  due  to  the  combined  torsion  and  compression.  Ans.  4856  lbs. 

io6.  A  street-railway  generator  gives  3000  B.H.P.  at  75  revolutions  per 
minute.  The  maximum  bending  moment  on  the  crank-shaft  is  13,500,000 
in.-lbs.  If  the  maximum  torque  is  1.5  times  the  mean  torque,  find  the  diam- 
eter of  the  shaft  (solid).     (/.  =5000.)  An-i.  W.27  ins. 

107.  The  steam  stop-valve  of  a  vertical  engine  is  worked  by  means  of  a 
hand-wheel  14  ins.  in  diameter  at  the  bottom  end  of  a  vertical  rod  or  shaft  (of 
circular  section)  16  ft.  long.  If  a  couple  of  moment  1400  in.-lbs.  is  applied 
to  the  hand-wheel  find  the  diameter  of  the  shaft  (a)  if  the  relative  twist  of 
the  ends  of  the  shaft  is  not  to  exceed  2  degrees;  (6)  if  the  maximum  shear 
stress  in  the  shaft  is  not  to  exceed  9000  lbs.  per  square  inch.  (Take  G" 
11,000,000.)  .4ns.  (a)  IJ  ins.;  (b)  U  in. 

108.  The  hollow  vertical  sleeve  of  a  Weston  centrifugal  of  3^  ins.  external 
and  2  ins.  internal  diameter  makes  1200  revolutions  per  minrte  and  carries 
at  its  lower  end  a  weight  of  1000  lbs.  Show  that  the  centrifugal  effect  is 
58,047  times  that  of  the  weight.      (Take  £  =30,000,000  lbs.  per  square  inch.) 

log.  If  the  critical  length  of  a  shaft  13.4  ins.  in  diameter,  subjected  to 
endlong  thrust  alone,  is  equal  to  the  critical  length  when  subjected  to  cen- 
trifugal force  alone,  show  that  P=  68,500u/;  also  show  that  if  the  length  is  98 
ft.,  the  critical  P  is  327,600,  and  the  critical  a»  is  nearly  4.8  radians  per  second, 
or  46  revolutions  per  minute. 

no.  A  WTOught-iron  propeller  shaft  has  a  diameter  of  22J  ins.,  and  the 
pitch  of  the  screw  is  3oi  ft.  The  indicated  horse-power  at  53  revolutions  per 
minute  is  lO.S.'iO,  and  assuming  thcvrctically  that  the  whole  of  this  is  utilized, 
show  that  the  end  thrust  is  ISl  ,,j30  lbs.  and  that  the  twisting  couple  is  12.;J00,- 

000  in.-lbs.  If  f:  is  29,000,000  lbs.  per  square  inch,  compare  the  twisting 
and  end-thrust  effects.  AIho  find  the  length  of  the  shaft.  If  a  shaft  of  this 
length  and  diameter  is  subject  to  no  end  thrust,  to  no  twisting  moment,  nor 
to  its  own  weight,  show  that  it  will  break  by  centrifugal  force  if  it  revolves  at 
a  greater  sjieed  than  5|  revolutions  per  m-nute.  Ann.  1751;  4454  ins. 

III.  A  4-in.  steel  shaft  (£'=3X10'  Ibe.  per  square  inch)  15  ft.  long, 
with  its  ends  supported  but  not  constrained  as  to  direction,  is  subjected  to 
equal  and  opposite  endlong  forces  (like  a  strut)  each  of  200  lbs.;  taking  into 
account  its  inertia  but  neglecting  its  mere  weight,  what  is  its  critical  speed 
of  rotation?  ^^.  592  r.p.m. 

1'  -  Taking  the  proof  stress  to  be  140,000  lbs.  and  (7  to  be  13,000,000  lbs., 
find  the  axial  proof  load,  the  deflection,  and  the  resilience  of  a  4-in.  spring 
of  coiI.-i  48  ins.  in  length  and  1  in.  in  diameter. 

Am.  K',750lb8.;  2.069  ins.;  14,215.4  m.-lbs. 

113.  A  spring  is  formed  of  steel  wire;   the  mean  diameter  of  the  coils  is 

1  in.;  the  working  stress  of  the  wire  is  .W.OOO  lbs.  |)er  square  inch;  the  elonga- 
tion under  a  weight  of  lOi'^  lbs.  is  2  ins. ;  the  coefficient  of  rigidity  is  12,000,000 
lbs.    Find  the  diameter  of  the  wire  and  the  number  of  coils. 

Ans.  .1  in.;  15.28. 


EXAMPLES. 


675 


114.  '^ind  the  weight  of  a  helical  spring  which  is  to  bear  a  safe  load  of 
fi  tons  with  a  deflection  of  1  in.,  G  being  12,000,000  lbs.  and  /  fiO,000  lbs. 

T,r    Vir,A  *u     *•  ,        .,,     .  ^"«-  -'2.69  ibs. 

115.  Find  the  time  of  oscillation  of  a  spring,  the  normal  displacement 
under  a  given  load  being  J. 

Ans. 


'4 


Ans.  (o) 


■number  of  coils.) 

7    ny'W  .,  V 

A  " 'gF^' "  »> -0  and  y, -yj -; 


116.  Find  the  deflection  under  the  weight  TT  of  a  conical  helical  spring 
(«)  of  circular  section;  (b)  of  rectangular  section,  the  radii  of  the  extreme 
«<"ls  being  y.  and  y„  and  the  radial  distance  from  the  axis  to  a  point  of  the 
spring  at  an  angular  disUnce  <t>  from  the  commenceme  .t  of  the  spiral  being 
given  by  the  relation  ^^^—^=-^      (n. 

n(y,+y,){y,'+y,')W/    ny'W 
Gt* 

(b)  l.8nn(y,+y,)(y.^+y,')^-l±^  g'.  (,  ^nd  h  beingthe 

sides  of  the  rectangular  section. 

snrinlV"'^.i!^^.11*^"'"'  **/  "^'^'*^  ^^^  "^"^  '^^  gravitation  unit  for  a  steel 
spring  from  the  following  obser\'ations : 

Lo«l  in  Scale-pan.         Scale  Reading.         No.  of  Vibration,.  Time  i„  Sec.. 

5                      3.28  _  Z 

10                      3.56  _  _ 

1«                       3.84  _  _ 

S                       ''12  100  36 
26                      4.40 
30                      4.68 
Diam.  of  mre-.257  in.;  length  of  coil=122.59  ins.;   mean  diam.  of  coU 

=3  'ns.;   weight  of  pan,  etc.  -2.72  lbs.                Ans.  11,500,000  lbs.;  32.3 

sion"xper;ment r  ""'"'"  "'  "^''^^  ^^^  '''''^  '""^  ^°"°-"«  --»^«  «f  '^  tor- 


I-oail  in  Soale- 
paii  in  Lbs. 

5 
10 
15 


Ri«ht-hantl 
Scale  Reading. 

7  0 

7.26 

7.44 


Left-hand       I.oa.1  in  Scale- 
Scale  Reading,     pan  in  Lbs. 

8.0  20 

8.18  25-^ 

8.28  30 


Right  hand 
Scale  Reading. 

7.5S 
7.72 

7.85 


Left-hand 
Scale  Readinc 

8.34 

8.40 

8.45 


arm^iS'i^:^""  '"•'  *'''  ^''''"'''  ^'""'"^  '"«-  ^°'''^  '«^'«^-25  ins.;   scale- 

119.  A  con.-l  spring  of  round  wire  40  ins.  long  and  .2  in  in  diameter 
has  ceils  varying  from  4  to  2  ins.  in  diameter.  If  the  proof  strels  isT(S 
lbs.  r-  square  mch,  find  the  proof  load  and  the  shortening  "L  to7eToa^' 


Ans.  17.44  ins. 


704  m. 


mi 
w 

I.  f. 


CHAPTER  X. 


BRIDGES. 


1.  Classification. — Bridges  may  be  divided  into  three  general 
classes,  viz.: 

A.  Bridges  in  which  the  platform  is  carrietl  by  trusses  of  different 

types  in  which  the  chords  (flanges)  are  either  horizontal 
or  curved  or  are  composed  of  slojiing  members. 

B.  Bridges  in  which    the    platform  is  suspemiotl  from  cables 

passing  over  high  piers. 

C.  Bridges  in  which  the  platform  rests  upon  arched  ribs. 

In  the  present  chapter  it  is  proposed  to  deal  with  the  bridges 
of  Cla.'is  .\  only. 

Figs.  59o-614  are  skeleton  diagrams  of  the  various  types  of 
truss  which  are  commonly  employeil  in  bridge  construction.  The 
maximum  depth  of  a  truss  is  governed,  to  some  extent,  by  local  con- 
ditions, but  usually  varies  from  one  fifteenth  to  one  seventh  (and 
even  more)  of  the  span.  Girders  and  trasses  may  require  to  be 
designed  to  meet  conditions  of  a  specified  strength,  or  of  a  specified 
stiffness,  or  of  both,  dei)ending  essc-ntially  on  the  ratio  of  span  to 
depth.  If,  for  example,  this  ratio  should  exceed  twelve,  deflection 
becomes  a  serious  consideration,  and  therefore  stiffness  is  then  a 
most  important  consideration.  In  ordinary  practice  it  has  often 
been  the  custom  to  limit  the  maximum  depth  of  a  truss  to  IJ  times 
the  width  of  the  bridge,  .so  that  the  depth  would  then  l)e  not  more 
than  24  ft.  for  a  .single  anil  40  ft.  for  a  double-track  bridge. 

Position  of  Platform. — The  platform  may  be  supported  either 
at  the  top  or  bottom  flanges,  or  in  some  intermediate  position.  In 
favor  of  Mie  last  it  is  claimed  that  the  main  girders  may  be  braced 
together  below  the  platform  (Fig.  615),  while  the  upper  portions 
serve  as  parapets  or  guards,  and  also  that  the  vibration  communi- 

676 


BRIDGES. 


677 


FINK 


Fio.  S95. 


/^<b^^<!xM\ 


Fio.  sgg. 


POST 


WHIPPL 


Kl».  686. 


md 


% 


w*nnEN 

/V-A/v\A/\A 


Flo.  698. 


^ 


WARREN 


Fio.  eoa 


Fio.  601. 


LENTICULAR 


r 


PRATT 


FIO,  (ny2. 


<^M^  i^^m^ 


Kio.  6113. 
PARABOLIC 


PEGRAM 


Fio.  605. 
TRIANGULAR  LATTICE 


Fio.  607. 


SCHWEDLER 


rmNt<iwuii/i/i>v 


Fio.  608. 


KELLOOO 


xlxIXlXlXl^ 


Fio.  61  \ 


BALTIMORE 


Ai/^IXIXIXIXlXlX 


Flu.  bid. 


FIO.  612. 

EIGHT  PANEL  RIOID 


fiii.   614. 


r.ii  i 


678 


THEORY  OF  STRUCTURES. 


cated  by  a  passing  train  is  diminished.  The  position,  however,  is  not 
conilucive  to  rigidity,  and  a  large  amount  of  metal  is  required  to 
form  the  connections. 


Fio.  615. 


Fio.  616. 


The  method  of  supporting  the  platform  on  the  top  flanges  (Fig. 
616)  renders  the  whole  depth  of  the  girder  available  for  bracing, 
and  is  best  adapted  to  girders  of  shallow  depth.  Heavy  cross- 
girders  may  be  entirely  disj)en.sed  with  in  the  case  of  a  single-track 
bridge,  and  the  load  most  effectively  distributed,  by  laying  the 
rails  directly  upon  the  flanges  and  vertically  above  the  neutral  line- 
Provision  may  be  made  for  side  spaces  by  employing  sufficiently 
long  cross-girders,  or  by  means  of  short  cantilevers  fixe<l  to  the 
flanges,  the  aiUa.  i  ;j;e  of  ihi  former  arrangement  being  that  it 
increases  the  resistance  to  lat(>ral  flexure  and  gives  the  platform  more 
elasticity. 

Figs.  617,  618,  619  show  the  cross-girders  attached  to  the  bot- 
tom flanges,  and  the  desirability  of  tliis  mode  of  suj)port  increases 
with  the  depth  of  the  main  girders,  of  which  the  centres  of  gravity 
should  be  as  low  as  possible.  If  the  cross-girders  are  suspended 
by  hangers  or  bolts  Ix'low  the  flanges  (Fig.  619),  the  depth,  and 
therefore  the  resistance  to  flexure,  is  increa.sed. 

In  order  to  stiff"en  the  main  girders,  braces  and  verticals,  con- 
sisthig  of  angle-  or  tee-iron,  are  introduced  and  connected  with  the 
cross-girders  by  gu.'^set-pieces,  etc.;  also,  for  the  same  purpose,  the 
cro.ss-girders  may  be  prolonged  on  each  side,  and  the  end  joined  to 
the  to|)  liaiigcri  by  suitable  bars. 

When  the  depth  of  tlie  main  girderc  is  more  than  about  5  ft. 


I: -I 


BRIDGES 


679 


tlic  top  flanges  should  be  braced  together.  But  the  minimum 
clear  headway  over  the  rails  is  16  ft.,  so  that  some  other  method 
j^hould  bo  adopted  for  the  support  of  the  platform  when  the  depth 
of  the  main  girders  is  more  than  5  ft.  and  less  than  16  ft. 


Fig.  618. 


Fig.  619. 


Assume  that  the  depth  of  the  platform  below  the  flanges  is 
2  ft.,  and  that  the  depth  of  the  transverse  bracing  at  the  top  is  1  ft.; 
the  total  limiting  depths  are  7  ft.  and  19  ft.,  and  if  1  to  8  is  taken  as 
;i  moan  ratio  of  the  depth  to  the  span,  the  corresponding  limiting 
spans  are  56  and  152  ft. 

Comparative  Ad'>-antages  of  Two,  Three,  and  Four  Main  Girders. 
-  A  bridge  is  generally  constructed  with  two  main  girders,  but  if  it 
i-^  (Topsod  by  a  double  track  a  third  i.s  occasionally  added,  and  some- 
iiiucs  each  track  is  carried  by  two  independent  girders. 

The  omj)loymont  of  four  independent  girders  possesses  the  one 
frivat  advantage  of  facilitating  the  maintenance  of  the  bridge,  as 
'-;i('  lialf  may  bo  closed  for  repairs  without  interrupting  the  traffic. 
<  >n  the  other  hand,  the  rails  at  the  approaches  must  deviate  from 
the  main  linos  in  order  to  enter  the  bridge,  so  that  the  width  of  the 
iiriilfro  is  much  increased,  and  far  more  material  is  required  in  its 
ciinstruction. 

h\'\\,  if  any,  reasons  can  be  urged  in  favor  of  the  introduction 
"i"  a  third  intermediate  girder,  since  it  presents  all  the  objectionable 
features  of  the  last  system  without  any  corresponding  recommontla- 
tion. 

The  two-girder  system  is  to  h  preferred,  as  the  rails,  by  such 
an  arrangement,  may  be  continue  '  over  the  bridge  without  devia- 


i 


1:        '!'< 


'm 


680 


THEORY  OF  STRUCTURES. 


tion  at  the  approaches,  and  a  large  amount  of  material  is  coonomized 
even  taking  into  consideration  the  increased  weight  of  long  cross- 
girders. 

The  up{.er  and  lower  chords  of  a  bridge-truss  are  connected 
together  hy  a  web  which  may  be  close,  i.e.,  may  be  maile  of  pb.U'n 
butting  the  one  against  the  other,  or  may  be  open,  i.e.,  may  be  corn- 
posed  of  a  number  of  .'separate  memlx>rs  in  the  form  of  verticals    nd 
diagonals.     These   verticals  and  diagonals  intersect   the  chords  In 
what  are  calleil  panel-points,  and  the  sjjace  between  two  such  c.i,- 
secutive  points  is  a  panel.  The  whou  of  the  memlxTs  may  be  riveted 
together  or  may  be  connected  together  by  means  of  suitably  designed 
pins.    The  former  method  has  the  advantage  of  .securing  a  much 
stiffer  structure  and  of  making  the  separate  members  interdepen- 
dent and  therefore  also  of  distributing  any  weakness  inherent  in  any 
particular  memlx-r  over  other  memb<'rs  with  which  it  is  rigidly  con- 
nected.    Thus,  in  the  event  of  the  failure  of  a  tie  or  strut,  the'stres.s 
it  was  intended  to  carry  is  taken  up  by  adjoining  members  and 
the  bridge  itself  remains  in  working  order.    The  safety  and  strength 
of  a  pin-connected  bridge,  on  tiie  other  hand,  depends  upon  the 
strength   of  each  individual   memln'r.     Its  construction,   however, 
is  nuich  simpler,  a..  •  the  determination  of  the  stresses  in  the  several 
memb(>rs  is  much  more  definite  and  accurate-.     Although,  for  this 
purpose,  it  is  necessary  to  a.ssunu>  that  the  pin-joints  are  friction- 
less,  this  assumption  is  much  nearer  the  truth  than  the  sissumption 
that  in  a  riveted  structure  the  total  shear  at  any  vertical  section 
is  equally  tlivided  In'tween  all  the  members  intersected  by  this  sec- 
tion, which  is  equi\  dent  to  the  substitution  of  a  iman  stress  for  the 
stres.s(>.s  in  the  several  bars. 

The  tendency  in  recent  American  practice  has  l)een  to  extend 
the  U.S-  of  riveted  bridges  to  si)ans  as  long  lus  2()0  ft.  and  even  beyond. 
Above  this  limit,  however,  the  connections  l)ecome  unduly  large,  if 
a  single  web  .system  is  used.  Some  riveted  spans  for  heavy  railway 
traliic  have  recently  Iwen  built  whose  length  is  about  230  ft.,  l)ut 
in  fiiese  ca.>^es  double  web  sysf(>nis  have  Ix'en  employed. 

I>i(i(l  L«w»/,s.— The  dead  load  on  a  bridge  consists  of  the  entire 
weight  of  the  trus.se8,  floor,  and  track.  less  such  parts  as  the  i)edeBti.|.s 
en.l  lloor-lK'anis,  ami  anchorage  metal,  the  weight  of  which  \b  boriu 
dinctly  hy  th.         >,  wiijioul  causing  any  stresses  in  the  structure 


li; ' 


BRIDGES. 


681 


as  a  whole  In  the  case  of  highway  bridges  in  northern  climates,  a 
Miitable  allowance  must  be  made  for  snow  and  ice.  In  the  case 
of  steel  railway  bridges  the  following  formula'  give  with  sufficient 
afouracy  the  weight  of  structural  steel.  To  thi.s  amount  must  be 
added  from  3M  to  4oO  lbs.  per  lineal  foot  of  span  for  each  track  to 
make  allowance  for  the  ties,  rails,  fastenings,  and  guard-rails 

Let  11  be  the  weight  in  tons  (of  2000  lbs.)  of  one  engine  and 
'.'•nder  ^yhen  the  live  load  consists  of  two  coupled  locomotives 
loUowed  by  a  uniform  train  loail. 

Let  w  =  the  total  weight  of  metal  in  pounds  per  lineal  foot  of  span : 
/  =  length  of  span  in  feet. 

Then  for  deck  plate-girder  spans 

oootr+ion7+23ooi 


w  = 


400-1 

For  half-through  plate-girder  spans 

400ir+1000/ 


w  =  - 


89- 


Wl 
1000 


For  through  riveted  Pratt-truss  spans 

«'  =  8.6(/  +  1.251F-112) +4.31388 -(Pr+20|, 
the  last  term  being  ignored  if  negative. 
,iJ^l  through  pin-connected  Pratt-truss  spans  with  parabolic  top 

«'  =  8.63(/  +  1.3U'-140). 

For  double-track  bridges  as  above  add  85  ptT  cent. 
For  through  pin-connected  Petit-trass  spans 

U'=10.9(/  +  1.1H'-190). 

For  double-track  bri.lges  add  the  following  j)ercentages: 

350  feet  span        add  75  jwr  cent. 


400  " 

450  " 

500  " 

r,r^  " 

000  " 


<  i        r- 


70  " 
65  " 
60  " 
55  " 
50  " 


II 
II 
II 
II 


if 


:  J 


f 


MM 


mil 


TO 


m 


682 


THEORY  OF  STRUCTURES. 


Ill  the  case  of  liighway  bridpos  there  is  such  a  variety  of  types 
of  flooring  tl'.at  general  formula^  for  the  weights  are  of  little  use. 

Live  Loadf.—FoT  railway  bridges  the  live  load  sf)ecifi('  is  the 
lieaviest  locomotive  and  train  load  that  is  likely  to  cross  the  struc- 
ture, or  some  eciuivalent.  As  both  the  weight  and  the  distribution 
of  the  loads  for  actual  locomotives  change  greatly  from  time  to 
time,  a  typical  engine  and  train  load  is  frequently  ailopted.  The 
typical  engine  may  not  corresi)ond  exactly  with  any  engine  in  use, 
but  it  is  so  chosen  as  1o  produce  about  the  same  stresses  in  all  mem- 
bers as  the  actual  engin''s.  Convenience  of  calculation  is  also  borne 
in  mind,  particularly  in  the  wheel  si)acings,   in  which  fractional 


li 
=1 


S'     * 


•  <       ^  :        — 


-'        ?        •! 


KIlHl  |».r  ft. 
I  .  \ 


Fig.  620. 

numbers  are  avoided  as  f'-r  as  i>os.sible.  Among  the  systems  of 
tyjiical  loads  in  general  u.se  are  those  of  Cooper  and  AVaddell. 
Cooimt's  ('la.«s  K  40  load,  as  shown  in  Fig.  020,  consi.sts  of  two  con- 
.s()lidati(tn  locomotives  carrying  40.(KX)  lbs.  on  fach  driving-axle, 
till'  engines  being  followed  by  a  uniform,  train  load  of  -l(KK)  lb.-:,  jxt 
lineal  foot.  Class  V.'M)  would  have  .SO.tKK)  lbs.  on  each  driving-axle 
and  a  train  load  of  ;5(KK)  ll)s.  jwr  lineal  font.  For  all  classes  the 
wheel  spacing  is  tlie  same,  and  the  \n\n\  and  tender  axle  loads  b'ar 
a  constant  ratio  to  tlie  driving-a.xle  loads,  viz.,  ')()  jkt  cent  in  the 
former  case  and  (io  [kt  cent  in  tiie  latter. 
Fig.  (LM  .shows  Waddells  Class  l. 


!l     U  |J  i  |l     f  il  i  i|     s!     t  I!  8'  Ij     il\i\^ 


lio    «21. 


The  wheel  spacing  is  the  same  for  all  (•lass«'s;  and  for  each  con- 
secutive ela.ss  above  or  Jh-Iow,  the  driving-axle  load  varies  by  steps 
of  ;{(MK)  lbs.,  the  pilot  and  tend(>r  axle  l.mds  by  1000  lbs.,  and  the 
unifortn  train  load  by  2(K)  ll)s.  |mt  litieai  foot. 


BRIDGES. 


683 


\^hon  tlio  hvo  load  has  boen  selected,  the  maximum  shears  and 
bending  moments  caused  by  the  actual  wheel  concentrations  n.ay 
he  coniputed    or  any  point  in  a  span,  as  illustrated  in  a  subsequent 
example.     ^^  ,th  a  view  to  reducing  the  labor  of  computation,  several 
<-o..vont;onal  methods  of  treating  the  live  load  have  been  p;oposed. 
Of  these  the  Equivalent  Uniform  Load  method  is,  p^^rhaps,  most 
|:,.'nerally   used.     Evidently   no   single    uniform   load    will   p;oduce 
ho  same  shears  and  bending  moments  at  all  points  in  a  structure  as 
the  actual  wheel  loads.     But  a  close  approximation  may  be  arrived 
^.t  by  computing,  m  the  case  of  pla.e  girders,  the  uniform  load  which 
Will  caase  the  same  bending  moment  at   the  centre  of  the  span  as 
he  actual  wheel  concentrations,  and  by  taking,  in  the  case  of  trusses 
the  uniform  loa.l  wh.  h  will  produce  the  same  B.M.  at  the  quarter^ 
i'omts,  or  by  taking  the  average  of  the  uniform  loads  which  will 
produce  the  .same  lK>nding  moments  at  all  the   panel-points.     As 
'i'.'span  length  increases,  the  equivalent  uniform  load  grows  smaller 
MM.e  the  heavy  engine  loa.ls  th^.,  extend  over  a  smaller  fraction 
"I  the  span  length. 

The  live  l,«ul  for  higliway  briuj,  s  is  taken  from  40  to  120  lbs 
l-r  s,,uare    foot,    acconiing  to    cirevmstances.     For  floor  .systems 
t-  c-oncentrations  caased  by  a  road  roller  are  sometimes  s,>ecified 
I-<'|Hls  arising  from  electric  or  oth,      tram-cars  mav  Ik-  treated  bv" 
"<l.er  of  th<«  methods  indicated  above  for  railwav  structures. 

-«!''■  !;«W.'//"'"1^/'^'  *'"""*"•  •''"'•''  ^'•'■''."'"^''V.  .>23.  i.  the  loading 

_  »    n  Mn„l  ronsoluhhou  vn.jmv  with  jour  drivers  and  am  leading  whJ 

Ihe  weiylu  coneeutrations  being  shown  by  Fig.  622.  ' 


0  e^•o^^•r;;)  .,-|^ 
1     -.^     a     I 


"•ffl 


Fia.  «22. 


: 


span  -239  ft. 

U-nph  of  .ontre  vortirnl.s-40  ft.:  of  rnd  vortiral^-27  ft 

(  -  i;  inl'T'TK'"  '"'  '•";;"'"""""-"""'  ""•  l-'-Kth  of  .nu-h  panel  to  bo  21  ft. 
( '.m  .n..).     The  error  thus  made  i«  .sufficiently  .mall  to  be  dinn^garded 


ilP 


6a4 


THEORY  OF  STRUCTURES. 


Five  distributions  may  be  considered,  viz.: 


re  at  a  panel-point; 


When  the  front  whc 

"        "    first  dri\  ' 

"    second    •  "       "        "         " 

"    third      

"    fourth    ' 

It  maybe  easily  shown  that  the  stresses  in  the  several  members  are  greatest 
when  either  the  siroml  driver  is  at  a  panel-point  or  the  third  driver  is  at  a 
panel-point,  the  c()rresi)onding  panel  loads  in  pounds  for  the  whole  truss 
lieiiig 

11,900,   49..^)00,  3S,700,   45,925,   43,750,   36,225,   36,000,   36,000  36,000, 
and 

4\\rm,  38,700,  45,925,   43,750,   36,225,   36,000.   36,000,   36,000,  36,000. 
The.s<'  results  may  be  obtained  analytically  or  graphically. 

Annlylically.~Vor  example  let  .1,  li,  C,  D,  Fig.  624,  be  four  consecutive 
panel-points,  and  let  the  third  driver  Ix-  at  li.     Then 


Fio.  624. 


Panel  load  at  A  -7500— +12000  ('-~^)  -11823,  say  11,900  lbs. 


Panel  load  at  B  -7r)00.,,^^  + 12000 


180+236  +  288  +  232 


+ 10625 


,    141  +77 


288  /  "^ '"""'''       288 

-49387,  say  49,.500  1b.s.; 
Panel  load  at  (' 

=  .2000p«;f  J)  +,0625C^^^^>l;^^4±^^i)  .7500'i:j 


2.S.S 

-3S445,  say  38,700  lbs. 
etc. 


ctP. 


e'c. 


Grnphirnlbi.  Upon  the  vertical  through  n  (Fig.  624)  take  liM  to  repre- 
sent 7.')()0  lbs.,  and  join  .I.U.  Let  the  vertical  through  d,  meet  .1.1/  in  /»„ 
and  the  horizontal  through  M  in  r,.  Then  a,/;,  repn-sents  the  portion  of 
7.")00  lbs.  borne  at  li,  and  h,r,  the  portion  Ixirne  at  .1. 

Also,  take  US  to  represent  12,(KK)  lbs.;  join  .LV,  CN.  \jp\  the  verticals 
through  <!.,,  ./„  ,1^  meet  .t.V,  C.V  in  t,,,  l,„  l,„  and  the  horizontal  through  A'  in 
Ci,  t\,  c,.  Then  </,'.„  (/,/>„  (i,h,  represent  the  |M>rtions  of  each  12,000  lb<.  l)orric 
at  li,  while  br,.,  I),r,  represent  the  portioiw  iMirne  .-it  .4,  and  ft^f,  the  portion 
borne  at  C. 


1    • 

i;  '■• 


BRIDGES 


685 


Unally  take  BO  to  represent  10,625  lbs    and  ioin  m     T^f  tu         .■     . 
tl.rou.h  a,,  a,  n^eet  CO  in  K  6  .  and  the  horizS  through  Jtn      7"S 
y.,.  .A  are  the  portions  of  each  10.625  lbs.  borne  at  i,  whie  t  '•6/^^ 
tl.e  portions  borne  at  C.     Thus  the  total  weight  at  B  '    '^' 

^^^  =«.<>.  +aA+aA  +  BX  +a,b,  +a,h,  +a,h„ 

.ivJIH'""'^'""''"'  °^''^"  '"''^'  '•""'^^"'-'"^'^  «<  the  panel.point.,  which  will 
grve  the  max.mum  stresses  .n  the  several  members,  may  be  ta'i>ulat;d  i  tw: 


Di«tribu- 
tinns. 


fi*p  1 

2 
"     X'.'.'.  .  . 

••     4 

"     5 

■  •     fi 

■•     7 

'■     H..     . 


End 
Ileac- 

tiori 
nt  .4. 


49500 
11900 


187990 
lt>2920 
124230 

95020 

69410 

47400 

29100 

153801 
121500     27000 


Load 
at  p,. 


Loail 
at  P2- 


Load 
at  Pa. 


I^Kiad 
at  p,. 


Load 
a'  Pi- 


Load 
at /)„, 


I 


Load 
at  p7. 


Load 
at  Ph. 


3C.225 
437.'>() 


■^Si^'      ■«S»25  43750 

49._>00  I  .3S700  4.-,92.->      .,.,,.,11 

MOO  j  47200  ,  402(H)      434(0 

I     6400  47200 

..1  6400 


27000 


2700(i  '   27006 


36000  36000  i  36000 
36225  3(i000  36000 
Ai\o,\r.  I;"?"*'  •''7 '00  36(KK) 
*'.2^  f  i'">  *W0()  37100 
l^  4«-:<>0  43400  4.-,H00 
6400  47200  40200  434(H) 
6400  47200  40200 

27000      27000  27wS  27^5! 


Load 
at  p». 


36000 
36000 
36000 
36000 
37100 
4581 H) 
43400 
40200 
27000 


The  case  giving  the  maximum  stress  in  any  member  is  ir^dicated  in  Fig:  6^. 

2.  Stringers—l-l^oh  length  of  stringor  between  consecutive  floor- 
beams  nmy  x-  regarded  a.s  an  independent  girder  resting  upon 
supports  at  the  end.,  and  should  l,e  designed  to  Ix-ar  with  safety 
tlH'  absolute  maxmum  bending  moment  to  which  it  may  be  sub- 
J--t-l  l.y  the  live  load  If  the  l,eams  are  not  too  far  apart,  the 
.  l.solute  maximum  bendmg  moment  will  1k>  at  the  centre  when  a 
driver  is  at  that  point.  Again,  in  the  ca.se  of  the  Sault  Ste.  Marie 
Hndge  Art.  I,  it  may  be  ea.sily  shown  that  the  maximum  l)ending 
■•"."u-nt  .s  produced  when  the  four  pairs  of  drivers  are  between  the 
ili)()r-l)eams. 

Ix-t  i/=distance  of  first  driver  from  nearest  point  of  support. 
1  he  reaction  at  this  supjiort 

12000  .-j)o 

-  ,,Hs-(«24-4i,)»'|"(20«-v). 

^.n,UrtW:id  S  '  ''^"'''"^'^  -^  '"'-'•""'"  "'  ^''^  --^  -  ^hird  driver. 

500 
-  —(206  -  y)  (.56  +  V )  - 1 2000  X  ,56 

500 
•>t  the  third  driver-~f206-.v)(108+y)-12000(.52  +  108). 

Ill  the  first  case  it  is  an  alyxnlute  maxivium  when  y~75"- 
wcoiul   "       ••  ••  I.  ,,     y_49/,.' 

Its  value  in  each  case  being  2,188,1 66§  in.-lbs 


ifflf 


6cS6 


THEORY  OF  STRUCTURES. 


Hence  th?  IxMiding  moment  is  an  absolute  maximum  and  equal  to  2,188,- 
10(i5  in.-lbs.,  at  two  pojnt.i  distant  75  ins.  from  each  point  of  support. 

Also,  if  /,  is  the  moment  of  inertia  of  the  section  of  the  stringer  at  these 
points,  c,  the  distance  of  the  neutral  axis  from  the  outside  skin,  and  /,  the 
coefficient  of  strength,  then 


§(21881003)  =f~  for  the  inner  stringer, 


and 


i(21SSlG0§)  =/,-^  for  the  outer  stringer. 
The  pontinuity  of  the  stringers  adds  considerably  to  their  strength 

3-  Camber.— Owing  to  tho  play  at  the  joints,  a  girdor  or  truss 
will  ik'Hcct  to  a  nmoli  groattT  extent  than  is  indicated  by  theory, 
and  th(>  material  will  receive  a  permanent  set,  which,  however,  will 
ftot  prove  detrimental  to  the  stability  of  the  structure  luiless  it  is 
increased  by  subsequent  loails.  If  the  chords  wers  initially  made 
straight,  they  wouki  curve  downwards;  and  although  it  does  not 
iiecr>ssarily  follow  that  the  strength  of  the  truss  would  be  sensibly 
impaired,  the  apiK-arance  would  not  be  i)leasing. 

In  practice  it  is  often  specified  that  the  girder  or  truss  is  to  have 
such  a  camlKT  or  upward  convexity  that  under  ordinary  loads  the 
grade  line  will  Im>  true  and  straight;  or,  again,  that  a  camber  shall 
be  given  to  th(>  span  by  making  the  panel  lengths  of  the  top  chord 
greater  than  those  of  the  bottom  chord  by  .125  in.  for  everj-  10  ft. 

The  lengths  of  the  web  inemliers  in  a  camlxTetl  iruss  are  not 
the  .same  lus  if  the  chords  wer(>  horizontal,  and  nmst  l)e  carefully 
calculateil  so  ,is  to  insure  that  the  several  parts  will  fit  together. 

To  find  an  Approximate  Value  for  the  Camlter,  etc. 

lA'i  d  be  the  dejith  of  tiie  truss. 

I^et  si,  .s'a  be  the  lengths  of  the  upjxT  and  lower  chords  respect- 
ively. 

I'ft  /i,  J2  te  the  unit  stresses  in  upixT  and  lower  chords  respect- 
ively. 

Let  f/i,  rfa  be  the  distances  of  the  neutral  axis  from  the  upper  and 
lower  chords  resp(>ctively. 

Let  li  Ix'  the  radius  of  curvature  of  the  neutral  axis. 

Let  I  be  the  span  of  the  truss. 


lihlr 


RIVET^ONNECTION  BETWEEN  FLANGES  AND  WBB.        687 


Then 


W  .1 


di^sij-l  _/i 


B 


I 


E 


and    -~  = 


^2       l—Sj       f. 


II  ~    I     =^)  approximately, 


the  chords  being  assumed  to  be  circular  arcs. 

Hence  the  excess  in  length  of  the  upper  over  the  lower  chord 

=si-s2  =  -^(/,+/2)^z| 


R 


Let  X,,  X2  be  the  cambers  of  the  upper  and  lower  chords  respec- 
tively. «+d,  and  R  -d2  are  the  radii  of  the  upper  and  lower  chorda 
respectively. 

By  similar  triangles, 

the  horizontal  distance  between  1     7?+rf, 
the  ends  of  the  upper  chord     J      ~R~^' 


the  horizontal  distance  between 
the  ends  of  the  lower  chord 


R    '' 


Honce  \2~irV  =^»><2(/i;+rf,),  approximately. 


and 


Therefore 


\2     R     }  °*2X  2(7? -da),  approximately. 


Xi 


rnH)  -  -i(-l-) 


4.  Rivet^onnection  between  Flanges  and  Web.-The  web  is  gener- 
ally rivetetl  to  angle-irons  forming  part  of  the  flanges. 

The  increment  of  the  flange  stress  transmitted  through  the  web 
t^rom  point  to  point  tends  to  make  the  angle-irons  slide  over  the 
tiaiige  surfaces. 

LVnnte  the  increment  by  F,  and  let  h  be  the  effective  depth  of 
the  girdfT  or  truss. 


688 


THEORY  OF  STRUCTURES. 


Then,  if  S  be  the  shearing  force  at  any  point, 

Fh  =  the  increment  of  the  bending  moment  per  unit  of  length 

/dM\     „  .      ,  ,       , 

=  1-1— )  =  .S  in  the  case  of  a  close  web, 

and       Fh  =  the  increment  of  the  be)  .ding  moment 

=  (J.l/)  =Sa  in  the  case  of  an  open  web; 

a  being  the  distance  between  the  two  consecutive  apices  or  panel- 
points  within  which  S  lies. 

Hence,  if  A'  be  the  number  of  rivets  per  unit  of  length  for  the 
close  web,  or  the  number  between  the  two  consecutive  apices  fo. 
the  open  web, 

N—r-ft  =  F  =  Y  for  the  close  web, 


and 


Sa 


=  -jr  for  the  open  web, 


d  being  the  diameter  of  a  rivet,  and  /,  the  safe  coefficient  of  shearing 
strength. 

5.  Eye-bars    and    Pins. — Eye-bars    connected    with    pins   have 
been  commonly  employed  in  the  construction  of  suspension  cables, 


jEq 


FiQ.  626. 


Fio.  627. 


the  tension  chords  of  ordinary  trusses  and  rsntilevcrs,  and  the 
diagonals  of  web  systems.     The  requisite  sectional  area  Is  obtainni 


tl'^ 


DETERMINATION  OF  STRESSES. 


689 

l.y  placing  a  numbor  of  bars  side  by  side  on  the  same  pin,  and,  if 
nrc-ssary,  by  setting  two  or  more  tiers  of  bars  one  above  another 
Ihe  figures  represent  groups  of  eye-bars  as  they  often  occur  in 

jinu'tice. 

If  two  sets  of  2n  bars  puU  upon  the  pin  in  opposite  directions. 
;.<  m  Figs.  626  and  627,  the  bending  moment  on  the  pin  will  be  nPp 
"  iK.mg  the  pull  upon  each  bar,  and  p  the  distance  between  the  centre 
lilies  of  two  consecutive  bars.       Hence 


■.V 


-I'i-J 


nPpJ-I, 

I  iHing  the  stress  in  the  material  of  the  pin  at  a  distance  c  from  the 
iKutral  axis,  and  /  the  moment  of  inertia. 

In  aeneral,  the  ben.ling  action  upon  a  pin  connecting  a  number 
"•  v.Ttical,  horizontal,  and  inclii   d  bars  may  be  determined  a« 

MM  lows: 

Consider  one  half  of  the  pin  only. 

I-«t  V,  Fig.  626,  be  the  resultant  stress  in  the  vertical  bars     It 
IS  ii.r("s.sarily  ecjual  in  magnitude  but  op- 
i'">it('    in    direction    to   the   vertical   com-  ! 

I«"»i't  of  the  resultant  of  the  stresses  in 
111-'  irK-Iin(>d  bars.  Lot  v  be  the  distance 
I"  I  wcon  the  lines  of  action  of  these  two  re- 
sultants. The  corresponding  bending  action 
"I"'"  the  pin  is  that  due  to  a  couple  of 
whiih  the  moment  is  Vr. 

1  .-t  h  h^  the  distance  between  the  lines  of  action  of  the  equal 
-ultants  H  o    the  horizontal  stresses  upon  each  side  of  the  pin 

0. ",',  ;""'"7'^J;":'""^  ^"^•'"g  -^^lon  upon  the  pin  is  that  due  to  a 
c-'iiple  of  which  the  moment  is  Hh. 

H."ee  the  maximum  bending  action  is  that  due  to  a  couple  of 
^^l^.  h  the  moment  is  the  resultant  of  the  two  moments  Vv  and  Hh, 

6.  Determination    of    Stresses-Stresses   are    developed  in  the 
'^'^^  lal  members  of  a  bridge-truss  by 


Fio.  628. 


[ri 


d 


in: 


n-fi 


■  ■  n 


690 


THEORY  OF  STRUCTURES. 


(a)  The  dead  load,  i.e.,  the  weight  of  the  bridge-trusses  and 
platform; 

{h)  The  lire  load,  i.e.,  the  weight  of  a  passing  train  and  also  the 
pressure  of  the  wind; 

(c)  Change.s  of  teiuixTuture. 

For  the  present  it  will  be  assumed  that  the  dead  and  live  loads 
are  uniformly  tlistributed  and  are  efjuivalent  to  d  and  I  respectively 
at  each  panel-point. 

In  any  panel,  if  a  shear  (s)  developes  a  stress  (d)  in  a  sloping 
member  inclined  at  0  to  the  vertical,  then  evidently 

d  cos  0  =  s,    or    d  =  s  sec  6. 

Also,  the  corresponding  streso  induced  in  a  horizontal  chord 

=  dsin  0  =  s  ta,nO. 

Again,  in  the  case  of  a  riveted  bridge-truss  with  horizontal  chords 
it  is  assumed  that  the  total  shear  in  any  panel  is  divided  equally 
between  all  the  members  intersected  by  a  vertical  section  in  that 
panel,  which  is  equivalent  to  the  assumption  that  the  mean  stress, 
in  each  sloping  member  of  the  panel  in  question,  is  the  same. 

Take  the  length  of  panel  to  be  p. 

Ex.  2.  A  through  bridge-truss  of  the  Warren  type,  of  nine  panels  (in  which  the 
sloping  members  are  inclined  at  30°  to  the  vertical,  so  that  the  truss  is  made  up 
oj  equilateral  triangles).  In  such  a  truss  it  is  evident  that  the  maximum  stresses 
are  tension!)  in  the  members  .sloping  ilown  towards  the  centre  and  compressions 
in  the  members  which  slope  up  toirnru''  the  centre. 


r\u/\  uA^/  \ 
uVis\/i6\/  'V 


Fig.  629. 

A.  Dead-load  stresses  in  .sloping  members. 

Let  li  be  the  reaction  at  the  left  supjwrt.    Then 


liX»p=>idX4lip,    or    «=y36. 

Hence  the  shears  in  the  1st,  2d,  3d,  4th,  etc.,  panels  are  R—d,  R—2d, 
R— 3d,  etc.,  respectively;   i.e.,  jtSO,     27,  jrlS,  —9,  etc. 


DETERMINATION  OF  STRESSES. 


691 


(- 


Thus  the  shear  in  any  panel  is  the  product  of  a  constant  quantity  -5 

panel  dead  !oad\       .  ....         ,.,.„„.,„ 

"ruTof  oanels   /  ^      *  multipher,  which  is  36  in  the  first  panel,  and  which 

in  each  succeeding  pane!  is  diminished  by  the  total  number  of  panels. 

The  corresponding  stress  in  the  sloping  member  in  any  panel  is  the  shear 

in  that  panel  X sec  0,  d  in  the  present  ca.-«  being  30°,  i.e.,  i~  the  product  of  a 

^     ^  .,     d        ^  I     panel  dead  load  \        ,  , 

constant  quantity  gSecO  ["- -. r-XsecOl  and  a  multiplier  which 

is  30  in  the  1st  panel,  and  which  in  each  succeeding  panel  is  diminished  by 
the  number  of  panels  viz.,  9.  These  diagonal  stresses  may  be  tabulated  as 
follows : 


Col.  I. 

Col.   II. 

Col.  III. 

Col    IV. 

Col.  V. 

Col.  VI 

d 

-5  we  t-L. 

Total  Maxi- 

Member. 

Multiplier. 

-|8ec»-D. 

MultipUer. 

mum  Diaconal 

9 

Streaa. 

X.   1=    1,  2 

36 

36D 

8.4t-36 

36L 

360-)- sot 

2,  3-   3,  4 

27 

27D 

7.4    -28 

7»L 

27D  +  28I. 

4,  5-   5,  6 

18 

IGD 

6.3t-21 

2\L 

18»  +  21L 

«,  7-   7    8 

9 

9D 

5  3    -15 

15/. 

90+ 15^ 

8,  9-   9,10 

0 

0 

4  2i-10 

lOL 

lot 

10,11-11,12 

-9 

-9D 

3.2   -   6 

%L 

-9D+    6t 

12,13  =  13,14 

-18 

-18D 

2.1J-   3 

at 

-ISO-f   3t 

14,15-15,16 

-27 

-27D 

1.1    -    1 

L 

-27D+     L 

The  first  column  indicates  the  various  sloping  members.  Column  3  gives 
the  stresses  in  the  several  members  due  to  the  dead  loads,  and  these  are  the 

product  of  a  constant  quantity  5-  sec  0  at  the  head  of  Column  3,  by  the  corre- 
sponding multiplier  in  Column  2. 

B.  Live-load  Stresses  in  the  Sloping  Members. — The  maximum  live-load 
stresses  of  the  same  kind  as  those  due  to  the  dead  load  are  produced  when 
the  greater  segment  of  the  truss  on  one  side  of  any  given  panel  is  loaded. 
Thus  the  ma.ximum  live-load  stresses  in  the  sloping  members  of  the  first  panel 
are  due  to  the  concentration  of  I  at  each  of  the  pane-points  from  the  first 
to  the  eighth;  in  the  second  panel,  from  the  second  to  the  eighth;  in  the 
third  panel,  from  the  third  to  the  eighth;  etc.     When  all  the  panel-points  are 

loaded   the   reaction   at  the  left  support -g-X 8 X4i.     For  each  succeeding 

panel  one  load  leaves  the  truss  and  the  centre  of  gravity  of  the  remaining 
loads  move  one  half  panel  towards  the  right.     Thus  the  reactions  at  the  left 

support  are  g  X7X4;    gX6X3i;    g-X5X3;    gX4X2i;    ^X3X2;  |x2Xli; 

y  X 1  X 1 .    These  are  the  shears  which  develope  the  maximum  live-load  stresses 

in  the  sloping  members  of  the  1st,  2d,  3d,  etc.,  panels  respectively. 

TTnti!  the  middle  of  the  truss  is  reached  the  stresses  are  evidently  of  the 
Bame  kind  as  those  due  to  the  dead  load,  but  as  soon  as  the  centre  is  passed 


692 


THEORY  OF  STRUCTURES. 


the  livo  load  covers  Ic^s  than  one  half  of  the  bridge  anil  the  stresses  developed 
imnu'diately  in  front  of  the  load  are  of  the  opposite  kind  to  those  due  to  tlie 
dead  load.  As  loii<j,  however,  as  these  stresses  are  less  than  those  due  to  the 
(lead  load  they  wil  not  affect  the  des  gn  of  the  bridge,  but  if  they  are  greater 
fliaii  those  due  to  the  dead  load,  the  members  must  Ik?  trengthened  so  as 
safe'y  to  l)e!.r  stresses  of  the  opposite  kind,  or  else  additional  members,  eal  ed 
counter-braces,  must  be  iiitioduced  to  take  up  the  stresses  -a  question. 

As  before,  the  stresses  in  the  sloping  members  are  the  product  of  a  con- 
stant quantity  j^-Xsec  fl  and  a  multiplier  which  is  8.4i  in  the  first  panel,  and 
7.4,  C^Mi,  r)..3,  etc.,  in  the  2d,  3d,  4th,  etc.,  panels. 

These  results  are  tabulated  mi  the  above  table.  Thus  the  total  maximum 
stress  in  any  given  sloping  member  is  equal  to  the  algebraic  sum  of  the  two 
corresponding  stres.ses  in  Cols.  3  and  5.  If  the  total  maximum  stresses  in 
the  (ith,  7th,  and  8th  panels  are  negative,  no  strengthening  or  counter-braces 
ar  re(iui  ed,  as  the  stres.ses  are  less  in  amount  but  of  the  same  kind  as  those 
for  which  the  ordinary  bracing  is  designed.  If,  however,  any  of  these  stresses 
arp  po.^itirf,  then  counter-braces  must  be  introduced. 

C.  Chord  .S'/rc-,s(.s.— The  maximum  stre.s.ses  in  a  chord  occur  when  the 
live  load  covers  the  whole  bridge,  and  in  the  present  case  this  is  equivalent 
to  a.ssuming  that  a  load  d  +  l  \s  concentrated  at  each  panel-point.  Take 
P-d  +  L     Then 

/>        P       P       P     p 
the  f^hears  ar-  -30.  -L>7,  -18,  -!),  -0.  in  the  1st,  2d,  3d  4th,  and  5th  panels 

rr!S|icctively. 


Therefore 


-41  =  ,j  tan  0X36; 


B3  =  .ll  +  stress  transmitted  through  2d  diagonal 
+    "  "  "       3d 

p 

=  .41  +  ,^  tun  0(:u)  + 27); 

C5  =  B3+ stress  transmitted  through  4th  diagonal 
+    "  "  "        5th       " 

p 

=  /i3  +  -  tan  ocjr  +  IS). 

I'  P 

Smnlarly,     1)1  =  (  :>  +  ~  tan  IH\H  +  0) ,  and  K9  =  Dl  +  -  tan  (?(9  +0). 

&oagam,     A2=g  tan  0(36+ 3()),     A'4  =A'2 +^  tan  <?(27+27), 


.,     /' 


XO-Xl-r  ^i.iiiS'dS  t  IS),  A'S-A'0  +  ^  tan  0(9+9) 


DETERMINATION  OF  STRESSES. 
These  results  mav  be  tabulated  as  follows: 


693 


Col.  I. 

Col.  II. 

Col.  III. 

Col.  IV. 

p 

Total  Maxi- 

Member. 

Multipmr. 

„  tan  0-F. 

mum  Clioril 

Stpem. 

.41 

36 

36F 

36F 

B3 

36  +  27-63 

63F 

99f 

C5 

27  +  18-45 

45f 

144f 

D7 

18+  B  -27 

27F 

niF 

£•9 

9+0-9 

9F 

imF 

X2 

36 +  .16 -72 

72F 

72F 

X4 

27 +  27  =  .54 

MF 

i26F 

X6 

18+18-36 

36f 

\62F 

X& 

9+  9  -18 

18F 

ISOF 

Col.  Ill  gives  the  stress  tran.smitted  to  the  chord  through  the  sloping 

p 
members  and  is  the  product  of  the  constant  quantity  -5  tan  0  and  the  multi- 
plier in  Col.  II.    The  total  maximum  chord  stress  is  given  by  Col.  IV  and 
is  obtained  by  adding  to  the  stress  in  the  preced  ng  panel  the  stresses  trans- 
mitted through  the  sloping  members  at  a  panel-point. 

Ex.  3.  An  eight-panel  deck  of  the  Pratt  type,  with  web  members  sloping 
at  9  to  the  vertical.  These  members  are  designed  to  be  in  tension,  the  verticals 
being  in  compression. 

A1B«CSD«E»P«QTH 


XX        XX       X       X 

Fig.  630. 


A.  Dead-load  Stresses.    Let  /?  be  the  reaction  at  the  left  support.    Then 
ff8p-7f/-4p,    or    «=^28. 


Thus  the  dead-load  shears  in  the  1st,  2d,  3d,  etc.,  panels  are 

d 


R-d,     R-2,1,     R~3,l, 


d  d 

etc..    I.e.,    ^28,     -20, 


8l2. 


etc., 


the  stresses  in  the  corresponding  diagonals  being 


^8ecflx28,    ^sec«x20,    ^secexi2, 
000 


etc. 


Hence  the  dead-load  stress  in  a  sloping  member  is  the  product  of  a  con- 
^     ,  ,.      d        ,  /    pimel  (lead  lo.ul  ,\ 

stant  quantity  g-tecfl^=--—,y~^^j^Xsecflj   and    a    f.vtor  which  is  28 

for  the  first  pane'  and  is  diminished  by  the  number  of  panels,  viz.,  8  for  each 


ii'tj' 


694 


THEORY  OF  STtvCTURES. 


sucwoding  panel.     These  stresses  may  therefore  at   once   l)e  tabulated  as 
follows: 


Col.  I. 

Col.  II. 

Col.  III. 

Ceil.  IV. 

Col.  V. 

Col.  VI. 

,; 

•^«ofl  =  f.. 

Toial  Maxi- 

Member. 

Multiplier. 

-     scr  f-D. 

Multiplier. 

mum  Uiajronal 

Stress. 

Xa 

28 

2HI> 

7X4    =28 

2SZ, 

2Sl>  +  2S/, 

be 

20 

201} 

ox;<i--2i 

2U. 

20W  +  21A 

dt 

12 

12» 

fix 3    .    l."! 

15L 

120+1.5/, 

<& 

4 

4D 

4X2* -10 

lot 

40+10/, 

-4 

~4D 

.3X2    -   fi 

C,L 

-40+fiA 

Im 

-12 

-I20 

ZXIJ-  .-» 

3L 

-120  +  ,'Ji 

no 

-20 

-200 

1X1    =-    1 

L 

-20O  +  i 

Col.  I  indicates  the  nietnWr;  Col.  Ill  gives  the  .stress  in  the  meml)er,  and 
i  the  protluit  ol  the  constant  tjuantity  at  the  he:id  of  Col.  Ill  by  the  corre- 
sponding multij)lier  in  Col.  II. 

B  Live-load  stresses.  The  maximum  live-load  shear  in  any  panel,  of  the 
.tunr  k'rid  as  that  due  to  the  dead-load  shear,  occurs  when  the  live  load  covers 
the  greater  st>ginent  of  the  bridge  on  one  side  of  the  panel  in  (|uestion.  These 
niaxiimini  live-load  shears  for  the  Ist,  2d,  3d,  and  4th  panels  occur  when  L 
is  concentrated  at  the  panel-points  from  1  to  7,2  to  7, 3  to  7, 4  to  7,  n^s|>ectively. 

Oil  pa.ssing  the  centre  of  the  bridge,  that  is,  when  the  live  load  covers 
less  thiui  one  half  of  the  bridge,  the  live-load  shears  are  of  the  opjmsUe  kind 
to  tlKw  due  to  the  dead  load  an<l  therefore  dcvelope  .stresses  of  the  opposite 
kind  to  tlu).-«3  in  the  sloping  nieniUrs,  for  wh'ch  provision  must  therefore  lie 
matle  eiiher  by  strengthening  these  mcml)crs  or  by  introducing  counter-braces 
as  sliinvn  by  the  dotted  lines. 

There  is  a  diffcrrnt  end  reaction  for  the  ma.ximum  live-load  shear  in  each 
panel,  iind  if  !{,,  R,,  Ii„  etc.,  are  the  end  reactions  when  /  is  concentrated  at 
the  ])anel-point8  1  to  7,  2  to  7,  3  to  7,  etc.,  respectively,  then 


R,-»P'7h4p,    or 


«,-|7x4; 


R,-Hp~(U3ip,  or  /^--«x3i; 


R,-Hp'-5l-3p,    or    i?,--5X3; 


etc. 


etc. 


Thus  the  maximum  live-load  chcars  in  the  1st,  2d,  .3d,  eti'.,  panels  are 
Ri,  li,.  Rn,  etc.,  and  the  maximum  live-load  stresses  in  the  corresjjonding 
s  oping  meml)erH arc  R,  sec  0,  //.sec  0,  R, sec  0,  etc.,  rcsix>ctivcly, or 


-sec  0X7X4,     -ijecexflx3i,     -sec  0x3x3, 

o  o  a 


GtC. 


DETERMINATION  OF  STRESSES. 


695 


Hence  the  maximum  ILve-lcad  stresses  in  the  1st,  2d,  3d,  etc.,  panels  are 

I        „  /       panel  live  load    ^^       .\        , 
the  product  of  a  constant  quantity  g  «ec  <»("^^^^^  ^^  ^^^^^Xsec  ej  and 

li  multiplier  composed  of  two  factors,  one  of  which  is  the  number  of  loads  on 
tlic  truss,  while  the  other  is  the  distance  of  the  centre  of  gravity  of  these 
loads  in  nunil)er  of  panels  from  the  right  support. 

These  results  can  be  at  once  tabulated  as  already  shown,  Col.  V  giving 
tlip  maximum  live-load  stresses  in  the  sloping  members,  and  these  stresses 
are  obtained  by  multiplying  the  constant  quantity  at  the  head  of  Col.  V  by 
the  corrosjwnding  multiplier  in  Col.  IV. 

It  will  be  noted  that  for  each  succeeding  panel  one  live  load  leaves  the 
truss,  so  that  the  centre  of  gravity  of  the  remaining  live  loads  movc^  one 
hair  panel  nearer  to  the  right  support. 

C.  Dead-load  Stresses  in  VerticcJs.  The  dead-load  stresses  in  the  1st,  2d, 
3il,  etc.,  verticals  are  evidently  the  dead-load  shears  in  the  1st,  2d,  3d,  4th, 
etc.,  panels,  respectively,  these  shears  being  transmitted  through  the  sloping 
members  Xa,  be,  de,  }g,  etc.    The  values  of  these  stresses,  therefore,  which  are 

evidently  compressions,  are  -28,  --20,  -12,  g4,  . . .  ,  etc., and  the  compression 

on  the  middle  vertical  is  d,  the  dead  weight  concentrated  at  its  head. 

The  maximum  live-load  stresses  in  the  same  verticals  evidently  occur  when 
/  is  concentrated  at  the  panel-points  1  to  7,  2  to  7,  3  to  7,  4  to  7,  etc.,  the 

values  being  -  7X4,-  GX3J  -5  x3,  -4  X2J  etc.,  respectively. 

So  O  " 

The  max.  live-load  stress  upon  the  central  vertical  is  a  compression  due 
to  the  weight  I  concentrat  1  at  its  head.  The  total  maximum  stress  in 
any  vertical,  say  ej,  is  evidently  the  sum  of  the   corresponding  dead-  and 

live-load  stresses,  that  is,  gl2  +  -  4x2J. 

When  /  is  «-oncentrated  at  6  aiid  7  the  corresponding  stress  in  kl  is  a  ten- 
sion and 


■-2  xl§,  and  the  tension  in  mn  due  to  /  concentrated  at  7  is  — . 
8  ° 


it* 


Hence 


the  total  resultant  stress  in  kl 


-8^2  +  8^ 


i'J 


:    I 


and 


n,n- -^20  +  ^' 


Chord  Stresses.— The  stresses  in  the  chords  are  greatest  when  the  live  load 
covers  the  whole  bridge,  so  that  there  is  a  fMinel  load  of  d+l  (  -/')  at  each 
punel-|H»iiit.  Then,  remembering  that  the  stress  in  the  chord  is  due  to  the 
AwuT  transmitted  through  the  sloping  members  and  is  equal  to  the  produrt 
of  thi.-*  shear  by  tan  0.  and  also  remembering  thai  the  shears  in  the  Ist,  2d, 
3d,  and  4th  panels,  etc.,  arc 


P 


^20 
8*' 


I'- 


8 


4. 


etc.,  respectively 


■ve  have 


THEORY  OF  STRUCTURES 


i4a  =  ^  tan  0x28,      DC  -  Aa +^  t&n  d  X20, 

Cc-Bc  +  ^  tan  0X12,     Z)9-Ce  +  ^  tan  0x4, 
s  o 

und        A'6  =  ^  tan  0X28,    Xd=X6+|  tan  0x20,    X/-Xd  +  ^  tan  0X12, 
These  results  mav  be  tabulated  as  follows: 


Col.  I. 

Col.  11. 

Co.  HI. 

Col.  IV. 

p 

Total  Maxi- 

Member. 

Multiplier. 

^  t«n  t'F. 

mum  Chord 

8tren. 

Aa 

28 

28f 

28f 

Be 

20 

20f 

4SF 

r. 

12 

12f 

eoF 

^ 

4 

-»/• 

64f 

28 

28*" 

28i' 

Xd 

20 

•MF 

*SF 

XI 

12 

12F 

eof 

Note. — If  this  truss  is  inverted,  it  becomes  a  truss  of  the  Howe  type,  the 
sloping  members  being  now  in  compression  and  the  verticals  in  tension.  The 
magnitude  of  the  stresses  remains  tlie  same  as  above. 

(Iraphkat  Mfthtni. — The  stresses  obtained  in  the  above  tables  may  be 
dcterniitipd  in  a  very  simple  manner  graphically.  Fig.  631  gives  the  stress 
diagram  for  the  dead  loads  on  the  truss,  in  which  XA,  the  reaction  at  the 
U>ft  sup|)ort,=:}if/,  and  .l/<-flC'-C'Z>-etc.,  -d,  the  dead  load  concentrated 
at  each  panel-point. 


ABO 
OCT 


e 

c 

^. 

t 

\, 

\. 

k\\ 

\, 

\ 

Nr  ^ 

.»     \ 

6                 X? 

, /^                            1? 

AAMm 


•    "I    » 


Kio    ftJl 


/     4      6      X 


Fio.  632. 


From  the  same  diagram  can  In;  obtained  the  max  mum  live-load  stresses 
in  the  chortls  by  multiplying  the  corrcsp«)nding  dead-load  stresses  by  the 

ratio       !i<*  the  stresses  an-  greatest  when  the  live  loa<l  covers  the  whole  bridge. 

.Xgaiii.  it  has  iilteaily  U-en  shown  that  thore  isadiffen-nt  end  reaction  corrc- 
8|Hiridiiig  to  the  niaxiinuni  stress  in  caih  diagonal.  Suppose  that  the  only 
force  acting  U!M)n  the  truss  is  >\  vertical  rcnctinn  of  lOQO  units  ftt  the  left  *u|)- 
;H"-t      Fig.  \jii2  is  llic  r<.rrts|oniliiig  »ires.s  dagram  and  it  shows  that 


H' 


DETERMINATION  OF  STRESSEH. 


007 


(a)  the  stress  in  each  diagonal  dtie  to  the  assumed  reaction  —1000  sec  0; 
(6)    "      "     "     "     vertical     "    '    "        "  "      -1000. 

Now  the  actual  reaction  at  the  left  support  is 

I 

-7x4    when  I  b  at  panel-points  1  to  7 


^6X3J     "      "  " 


^5X3       "      "  " 


14X2J     "      "  " 


^3X2       "      "  " 


12X1J      "       "  " 


2  to  7 

3  to  7 
4to7 
6  to  7 
6to7 


8 


<(      II  II 


panel-point  7. 


It  is  also  evident  that 

th«  actual  ttreii  in  a  mcmbtr         actual  reaction 


Hence 


the  ttrcN  due  to  aiiumed  reaction    atnuned  reaction 


the  maximum  stress  in  Xa  — :,„^  -    -28 --  sec  0X28, 


II  II 


«  II 


<i  II 


<<  It 


II  II 


II    II  I.    1000  sec  er,    /      ^^„, 

'"^— T5to-821''8"''*^21. 

^' 1000^8^^    8 '^*^*^' 

etc.,  etc. 

"*    10008^^    8^*' 

..     ..  ,d-I««>l20-i20. 
1000  8         8     ' 

u      ..    e/-^^il2-A2 
etc.,  etc. 


it 

t       » 

i'  ! 


608 


THEORY  OF  STRUCTURES. 


Generally  speaking,  in  the  case  of  trusses  with  horizontal  chords,  unless  the 
panels  are  of  unequal  length  or  unless  the  bridges  are  skew  or  are  otherwise 
specially  designed,  the  stresses  are  more  easily  and  rapidly  obtained  by  tabulating 
the  results  as  previously  described. 

Ex.  4.  A  len-pattd  doMi -intersection  lattice  {trellis)  deck-bridge,  with  mem- 
bi  rs  s!<>piiig  at  0  to  the  vertical. 

First,  assume  that  the  nieniljovs  are  rivctt'd  together  and  therefore  that 
in  any  panel  the  shear  is  ofiually  divided  between  the  tuo  diagonals  met  by 
a  vertical  section. 

I 


10    m 


Fio.  633. 
A.  Dead-load  Stresses  in  Web.     If  R  is  the  reaction  at  the  left  support, 


/M0p=!t<i-5/j,     or     /^  =  ,-,45, 


and  ttie  dead-load  shears  in  the  Ist,  2d,  3d,  . 
K,  R,-d„    I{,-2d,    l{,-3d,    ... 


,  panels  are 
respectively,  or 


ii«. 


lO' 


}i, 


10 


the  corresponding  stresses  in  each  of  the  diagonals  in  the  1st,  2d,  3d,  ... , 
panels  being 

His-'^''^^).  Hfo"'''^^-^)'  K^-''^^..)'  ••• 

Tb  s  the  dead-load  .stresB  in  a  diagonal  is  the  prcnlurt  of  a  constant  quantity 


1  (/         „  /     1     r>anel  dead   oad  „ . 

^cO(  =^^    '  Xsecfl) 

2  10  \     2  nutniK-r  of  panels 

l>t  pane'  and  is  diminished  by  the  numlx'r  of  imncls  in  each  succee<ling  jjaiiel. 
The.se  results  nmv  Ih'  labuluted  as  follows: 


fl|  and  a  multiplier  which  is  4.5  in  the 


li 
M 


if^^W  '^'^ 


DETERMINATION  OF  STRESSES. 


Col.  I  indicates  the  member;  Col.  Ill  is  the  dead-load  stress  on  the  member 
and  is  obtained  by  multiplying  the  constant  quantity  at  the  head  of  Col.  Ill 
by  the  corresponding  number  in  Col.  IV. 

It  may  be  noted  that  the  stresses  in  the  inclined  members  in  any  panel 
arc  necessarily  of  the  same  magnitude  but  are  opposite  in  character,  being 
tensions  when  the  members  slope  in  one  direction  and  compressions  when 
they  slope  in  the  opposite  direction.  It  may  also  be  noted  that  the  riveting 
divides  the  compression  members  into  two  equal  lengths,  so  that  the  ratio 
of  the  length  of  the  strut  to  its  least  radius  of  gyration  "is  diminished  one  half 
and  the  rigidity  is  therefore  increased  (Art.  7,  Chap.  VIII). 

B.  Live-load  Stresses  in  Web.  The  maximum  live-load  shear  in  a  panel 
occurs  when  the  live  load  covers  the  greater  segment  of  the  bridge  on  one 
side  of  the  panel.  Thus  for  the  maximum  stresses  in  the  diagonals  in  a  panel 
there  is  a  separate  end  reaction  equal  to  the  maximum  live-load  shear  in  the 
panel  in  question. 

These  reactions  are  evidently  equal  to  the  live-load  shears  just  in  front 
of  the  live  load. 

Let  R„  R„  K„  R„  etc.,  be  the  end  reaction  when  the  live  load  I  is  con- 
centrated on  each  panel-point  from  b  to  I,  3  to  I,  d  to  I,  5  to  /,  and  /  to  I, 
rcsfjectively.    Then 


fl,10p=9;-5p,      or    /?!■ 


/o9x5; 


Rt-lOp'SlUp,    or    ff,-r^8x4J; 


RtlOp-7l-4p,      or    ft,--7x4; 


etc. 


etc. 


.\gain,  on  passing  the  middle  of  the  bridge,  that  is,  when  the  live  load 
covers  a  smaller  segment  of  the  bridge,  the  live-loaO  shears  are  of  the  oppo- 
site kind  to  those  due  to  the  dead  load  and  therefore  develope  stresses  in  the 
sloping  members  of  an  opposite  kind  to  those  due  to  the  dead  load. 

Thas  the  msximiim  live-load  stresses  in  each  of  the  diagonals  in  the  lat, 
2d,  3d,  . . . ,  panels  arc 

Iff,  sec  9.     iff,  sec  9,     Jff,  sec  9 

rospectively,  and  therefore  the  maximum   live-load  stress  in  any  diagonal 

.      .  .  .  ■      ^    I         «/     •     panel  live  load     ,,         .\ 

IS  the  product  of  ft  constant  quantitv- —sec  01 -- ~      ,        ,         ,  XsecffI 

'  2  10x2  numiter  of  panels  / 

and  a  multiplier  composed  of  two  factors,  the  one  Iwing  the  numlier  of  loads 

oil  the  briditf  Mild  llu!  oilief  the  distanre  in  number  of  panels  of  the  centre  of 

gravity  of  these  loads  from  the  right  support. 

It  may  bo  observed  that  for  each  succeeding  panel  one  live  load  leaves 


700 


THEORY  OF  STRUCTURES. 


the  truss  and  the  centre  of  gravity  therefore  moves  one  half  panel  nearer 
the  right  support. 

Col.  V  of  the  preceding  table  gives  the  maximum  live-load  stress  in  any 
specified  diagonal.  It  is  the  product  of  the  constant  quantity  at  the  head 
of  Col.  V  by  the  corresponding  multiplier  in  Col.  IV.  The  total  maximum 
stress  i  any  diagonal  is  the  algebraic  sum  of  the  correspfjnding  stresses  due 
to  the  dead  and  live  loads.  For  example,  the  total  maximum  stress  in 
o=6aD  +  loL.  The  total  maximum  stresses  in  67,  78,  89,  and  9,10  are 
-5D  +  10L,  -13D  +  6L    -2.')D  +  L,  and  -35/)+/,,  respectively. 

If  any  one  of  these  last  four  stresses  is  jKisitive,  it  indicates  that  the  live 
load  developes  a  stress  in  the  member  of  an  opposite  kind  to  that  due  to  the  dead 
oad  and  the  membe  must  therefore  be  designed  to  meet  this  change  of  stress. 
In  some  cases  the  members  of  the  truss  receive  a  greater  sectional  area  for 
this  purpose  while  in  others  any  additional  member,  called  a  counter-brace, 
is  introduced  to  take  up  the  stress  in  question.  If  any  of  the  stresses  are' 
negative,  they  are  less  in  amount  but  of  the  same  kirid  as  those  for  which  the 
members  are  designed. 

C  Chord  Stresses.  These  stresses  are  greatest  when  the  live  load  covers 
the  whole  bridge,  so  that  at  each  panel-point  there  is  concentrated  a  load 
d+l  =  P. 

The  shears  in  the  1st,  2d,  3d,  4th,  etc.,  panels  are  45  — ,  35—.  25—    15—. 

10  10  10  10 
etc.,  respectively,  and  in  any  given  panel  one  half  of  the  shear  is  transmitted 
to  the  chord  through  each  of  the  sloping  members.  The  corresf ending  in- 
crement of  chord  stress  is  the  product  of  a  constant  quantity  ^  —  tan  5  and 

a  multiplier  which  is  4.'}  for  the  first  panel,  and  for  each  succeeding  panel  is 
diminished  by  the  number  of  panels. 
Hence 

1    P 

lb  (  =  -a2)  =stress  transmitted  through  12  (or  ab) tan  *X45; 

63  (  =  -2c)  =lb  +stross  transmitted  through  ab 

+    "  "  "       be 

1   P 

"16+.,-:- tan  0(45+35); 

3rf  (  -  -r4)  =W  +Mn'ss  transmitted  through  23 
+    "  "  "       34 


So, 


1   P 

-fe3+.,  :-tanfl(35+25). 


1   P 


,/5  ( -  _ 4r)  -.1//  +^  jjj  tan  (9(26  +  16); 
I'fc.  etc. 


m 


DETERMINATION  OF  STRESSES. 


7Qt 


These  results  may  be  tabulated  as  follows: 


Col.  I. 


Member. 


16- -a2 
fc3--2c 
3d--f4 
rf.5=-4e 
5/--e6 


Col.  11. 


Col.  Illi 


1   P 
Multiplier.         5    -  tan  B  —  F. 


45 
45  +  35-80 
35  +  25-fiO 
25+15-40 
15+   5-20 


45f 

80f 
COf 
40F 
20F 


Co.  IV. 

Total  Maxi- 
mum Chord 
Stress. 


45f 
125F 
185F 
225/? 
245f 


It  is  evident  that  the  panel  lengths  in  the  upper  chord  are  in  compression 
and  those  in  the  lower  chord  in  tension. 

Second.  Let  the  truss  be  of  the  pin-connected  type.  Each  system,  abed 
. .  and  1234  .  .  . ,  must  he  regarded  as  being  entirely  independent  of  the  other 
and  as  being  strained  only  by  the  loads  at  the  panel-points  belonging  to  the  par- 
ticular syKlem  under  conrideration. 

A.  Dead-load  Stresses  in  Wrb  Members.  Consider,  first,  the  system 
aM  . .  Im.  A  load  I  is  concentrated  at  each  of  the  panel-points  b,  d,  /,  h,  I, 
and  the  corresponding  reaction  R  at  the  left  support  is  given  by 


R'lOp-odop,    or    R' 


10 


The  corresponding  shears  from  1  to  6,  &  to  d,  d  to  f,  }  to  h,  h  to  /,  and  / 
to  11  are 

d  , 


10 


10^"'     10^'        10^'        10^^' 


d  d 

so  that  the  dead-load  stresses  are  tt, sec  Ox 25  in  oft,  —  sec  0 X 15  in  6c  and  cd, 

—  sec  0  X5  in  de  and  e/,  —  —  sec  0  x  5  in  /(/  and  gh,  ——sec  0X15  in  hk  and  kl. 

Xvil,  consider  the  system   1234.  .  .  10,11.     The    oad   /  is  now    comci, 
tratod  at  each  of  the  paiiel-iKiints  3,  5,  7,  9,  and  the  corresiwnding  reactinn 
It"  at  the  left  support  is  givi-n  by 


R"-lOp''4d-ryp,    or    /?"-— -20. 


The  correspond  ng  shears  from  1  to  3,  3  lo  5,  5  to  7,  7  to  9.  and  9  to  11  are 
—  0,    -jT^'O,  res{)ectively,  bo  that  the  dead-load  stresses  are 


d  d 

lo"'   lo'"'   10"'       10 


-  sec  6X20  in  12  and  23,  —  sec  8:<  10  in  31  and  45,  0  in  56  and  67, aw  0 


10 

■:  10  in  78  and  89. 


10 


702 


THEORY  OF  STRUCTURES. 


These  results  may  be  tabulated  as 

folio  w.s : 

Col.  I. 

Col.  II. 

Col  in. 

Col.  IV. 

Col.  V. 

Col.  VI. 

Member. 

Multiplier. 

Yo^'-D- 

Multiplier. 

4«c9=.i. 

Total.  Maxi- 
mum Liiasonal 
Stress. 

ab 

25 

230 

5 . 5  =^  25 

25/. 

23/) +  25/, 

12-23 

20 

20/) 

4.5-20 

20/. 

20/) +  20/, 

bc~cd 

15 

15fl 

4  4=16 

16/. 

15/>+16t 

34  =  45 

10 

10/.I 

3.4=.  12 

12/, 

10/) +  12/, 

dr  =  fl 

5 

5D 

3  3=   9 

9/. 

5/1+   9L 

56  =  67 

0 

0 

2.3=  a 

6/. 

6/, 

t(i''a>> 

-5 

-oD 

2.2=   4 

4/-, 

-5/>+   4/, 

7«-=89 

-10 

-  IO/> 

12=   2 

2/, 

-I0/)+   2/, 

hk  =  lcl 

-15 

-15/) 

11-    1 

L 

-15D+     L 

Co  .  I  designates  the  member  and  Col.  Ill  gives  the  corresponding  dead- 
load  stress.  It  is  the  product  of  the  constant  quantity  at  the  head  of  Col.  Ill 
by  the  multiplier  in  Col.  II. 

B.  Live-load  Stresses  in  Web  Members.  Consider,  first,  the  system  abed  .  .  Im. 
The  maximum  live-load  shears  from  I  to  b,  b  to  d,  d  to  /,  /  to  h,  and  A  to  Z 
occur  when  I  is  concentrated  at  the  panel-points  (of  the  system  in  question) 
b  to  l,d  to  /,  }  to  I,  h  to  I,  and  at  I,  respectively.  If  R^,  R„  R„  R„  /?,  are 
the  reactions  at  the  left  support  for  these  several  concentrations,  then 

/i',-10p=5/-5p,    or    «,=r^.5x5; 


R,lOp^4l-4p     or     /?,-— 4x4; 


RlOp=3l-3p,    or    /e,=-3x3; 


R,lOp=2l-2p,    or    /?.=-2X2; 


R,-lOp-  Ip,      or    /?,  =  |-lxl. 

Similarly  if  /?,',  /?/,  R/,  R,'  are  the  reactions  at  the  left  support  for  the 
system  1234  ...  10,11 , 

«,'  =  /q  4  X o,  R,'  =  ,Jj  3  X  4,  R/  =  ,-Jj 2  x  3,  and    «/  - ^^1  X  2. 

Hence  the  maximum  live-load  stros.ses  are 


■!■■ 
I  hi 


—  sec  0X5X5  in  ab,  rj:'**'  9X4X4  in  6c  and  c(/, —  sec  0X3x3  int/eand  c/, 


DETERMINATION  OF  STRESSES. 


703 


—  sec  0X2X2  in  fg  and  gh  and  —  sec  0X1X1  in  ^and  kl; 


I  I 

—  sec  0X4X5  in  12  and  23,        — 


sec  0X3X4  in  34  and  45, 


—  sec  0X2X3  in  56 and  67, and  —sec  0X1X2  in  78  and  89. 

These  results  are  shown  in  the  precedinj;  table,  and  Col.  V  gives  the  maxi- 
mum live-load  stresses  in  the  diagonals.    This  stress  is  the  product  of  the 

constant  quantity  —  sec  0  at  the  head  of  Col.  V  and  a  multipUer  composed 

of  two  factors  of  which  one  is  the  number  of  loads  on  the  truss,  while  the 
other  is  the  distance  of  the  centre  of  gravity  of  the  loads  in  panels  from  the 
right  support. 

The  total  maximum  stress  in  any  member  is  given  in  Col.  VI,  and  is  the 
algebraic  sum  of  the  corresponding  stresses  in  Cols.  Ill  and  V.  For  example, 
the  total  maximum  stress  in  de—5D+9L. 

Again,  the  total  maximum  stresses  in  fg  ( —gh),  78  ( =89),  hk  (—kt)  are 

-5D+4L,     -10D+2L,    and     -15D  +  L,    respectively. 

If  any  one  of  these  results  is  positive,  it  indicates  that  the  stress  of  the  opposite 
kind  to  that  of  the  dead  load  is  developed  and  that  therefore  provision  must 
be  made  for  this  in  designing  the  members  affected.  If  the  results  are  posi- 
tive, it  shows  that  the  members  ore  simply  subjected  to  the  same  kind  of 
stress,  but  less  in  amount,  as  those  for  which  they  are  usually  designed.  Again, 
in  the  two  systems,  members  sloping  in  one  direction  are  in  tension  and  those 
sloping  in  the  opposite  direction  are  in  compression. 

C.  Chord  Stresses. — The  stresses  in  the  chord  lengths  are  greater  when  the 
live  load  covers  the  whole  girder,  so  that  a  load  of  d+l—P  is  concentrated 
at  each  panel-point.  Then,  calling  t„  <,,  <,,  ...  the  tensions  jn  a2,  2c,  ci, 
etc.,  and  c„  c,,  c,,  . . .  the  compressions  in  16,  63,  3d,  . . . 

p 

t\  —stress  transmitted  through  ah=-rz^  tan  0  X25, 


+ 

+ 

«.-<4  + 

+ 


12 

23 --tan  0(20 +20), 

he 

P 

cd-— tan  0(15  +  15), 

34 

P 
45-— tan  0(10  +  10), 

de 

f/-- tan  0(5 +5), 


h , 


704 
and 


THEORY  OF  STRUCTURES. 


c,  =  stress  transmitted  through  12  =—  tan  9  X20; 


+ 
c,-=c,+ 


+ 
+ 


ah 

6c=-  tan  5(25  +  15), 

23 

34=- tan  9(20 +  10), 

cd 
p 

dc=-  tan  0(15+5), 


56  =  - tan  0(10+0). 


These  results  may  be  tabulated  as  follows: 


Col.  I. 

Col.  II. 

Col.  III. 

Col.  IV. 

p 

Total  Maxi- 

Member. 

Multiplier. 

ib'"""-^- 

mum  Chord 

Stres-s. 

a2 

25 

25P 

25f 

2c 

20  +  20-40 

4QF 

esF 

r4 

15 +  !.">  =  . SO 

30f 

9aF 

4« 

10  +  10  «  20 

20f 

115F 

Hi 

5+   .'j=10 

iOF 

125f 

16 

20 

20F 

20F 

63 

25+15  =  40 

40F 

CiOF 

3d 

20+ 10 -.10 

mF 

OOF 

rf.5 

15+    5  =  20 

20F 

UOF 

5/ 

10+   0-10 

lOF 

120F 

C<1.  I.  indicate"  the  chord  member;  Col.  Ill  indicates  the  stress  trans- 
mitted to  any  give  emlxT  through  the  sloping  members,  and  is  the  product 
of  a  constant  quantity  at  the  head  of  Col.  Ill  by  the  corrcs|)onding  multi- 
pliers in  Col.  II  Col.  I\'  gives  the  total  maximum  chord  stress  in  any  given 
paiicl,  and  it.s  value  is  nhtaincd  by  adding  to  the  stress  in  the  [)receding  panel 
the  stresses  transmitted  through  the  diagonals  meeting  at  the  common  panel- 
[Mniii. 

Ex.  5.  An  cig'd-piiir'  thrivujh  hilticr  trusn  with  horizontal  chords  and  two 
series  of  diagonalx  inclined  in  opposite  directions  at  angle  6  to  the  vertical. 


a 

a 

a 

r 

r 

r 

c 

e 

»• 

a 

§§$<$$^8g^ 

f 

"•{ 

A 

b 

b' 

b" 

b" 

d 

d' 

d" 

[•id 

Fio.  634. 


An  objection  to  this  class  of  girder  is  the  number  of  the  joints. 
First,  consider  the  mcrnlx>rs  t"  1m;  riveted  togcthc. 


i 


DETERMINATION  OF  STRESSES. 


705 


Prefisely  the  same  method  of  analysis  for  the  determination  of  the  stresses 
ill  tlie  several  ucb  members  is  to  be  adopted  as  in  the  preceding  example.  In 
vrpry  panel  a  vertical  section  now  intersects  four  members,  and  therefore 

in  the  followinf?  t.ilili'  tli'  tcmslant  quantity  i   —  l„  see  Oj  "D  sit  the  liead  of 

Col.  Ill  for  tlie  dead  load  and  j  (-  sec  OJ  -  L  at  the  head  of  Col.  V  for  the  live 
load.    The  total  maximum  diagonal  stresses  are  given  by  Col.  VI  as  indicated : 


Col.  I. 

<'ol.  11. 

Col.  m. 

Cnl.  IV. 

C-ol.  V. 

Col.  VI. 
Total  Maxi- 

Member. 

Multiplier. 

D. 

Multiplier. 

L. 

mum  Diagonal 

Stress. 

ab  =  ati" 

2S 

28D 

7  4   -28 

28i 

2SD  +  2S/- 

All'"  =  a'b' 

20 

20/) 

fi.3i-2I 

21t 

20A)  +  21L 

bc  =  a"h" 

12 

120 

5..'t    -1.-) 

15t 

12/>+15t 

h'c'-a"'h"' 

4 

4D 

4  2t-10 

lot 

4D  +  10/. 

fc"  =  cd 

-4 

-4D 

.3  2   -   6 

6/- 

-4n+   6Z. 

b"'r'"-c'd' 

-12 

-12D 

2  M-   ,3 

.3t 

-120+  3L 

de   ^c"d" 

-20 

-20O 

1.1    -    1 

L 

-20O  +  L 

The  total  stress  in  a  diagonal  is  the  algebraic  sum  of  the  corresponding 
stresses  due  to  the  dead  and  live  loads.     For  example, 

the  total  maximum  stress  in  6c  (  —a"b")  «-12Z)  +  15L. 

Again,  the  total  maximum  diagonal  stresses  in  b"c"  (^cd),  6"'c"' (=</</'), 
and  </c  (  =c"rf")  are  -4D  +  6/.,  -1 2D +3/.,  and  - 20i)  +  L,  respectively. 

If  cither  of  these  is  negative,  the  result  indicates  that  the  live  load  pro- 
(liicos  a  stress  in  the  web  member  under  consideration  of  a  kind  opposite  to 
that  due  to  the  dead  load,  and  therefore  the  web  member  must  be  designed 
10  carry  this  additional  stress.  If,  however,  the  result  is  positive,  it  indicates 
tliat  the  stress  is  of  the  same  kind  but  less  in  amount  than  that  due  to  the 
(lead  load,  and  the  ordinary  bracing  is  therefore  quite  sufficient. 

It  is  also  evident  that  the  stresses  in  the  members  in  any  panel  sloping 
ill  opposite  (lircction.s  are  op|X)sitc  in  kind. 

For  the  greatest  stresses  in  the  chord  panel  lengths  the  live  load  I  is  con- 
centrated at  each  [lanel-point,  and  the  trthss  now  carries  what  is  a  uniformly 
distributed  load,  the  load  at  each  panel-jxiint  being  d  +  l.     Thus  the  con- 


.stant  quantity  at  th-  head  of  Col.  Ill  for  the  chords  is  now--—-  tanff- 
Hence  the  table  for  the  chord  strcs.ses  is  a.s  follows: 


P. 


Col.  I. 

Member. 

.46='  —a'n" 
6fc'=  —a"a"' 
b'h"  ^  —a"'c 
b"b'"--ci^ 

C-ol.  II. 
Multiplier. 

Col.  III. 

F. 

Col.  IV. 

Total  Maxi- 
mum <'hord 
Stres.«. 

28 
28  +  20-48 
2(1+  \i-:i2 
12+   4=16 

2liF 
4SF 
62F 
16f 

28#' 

7BF 

108** 

124F 

i'  .*}■ 


706 


THEORY  OF  STRUCTURES. 


Second,  let  the  truss  be  pin-connected.    There  are  now  four  sjrsteina  of 
bracing,  viz.,  abcde},  a'b'c'd'e',  aa"b"c"d"],  Aa"'b"'c"'d"'. 

The  following  table  gives  the  maxinjum  stresses  in  the  diagonals,  —  sec  fl  —D 

o 

being  the  constant  quantity  for  the  dead  load  at  the  head  of  Col.  Ill  and 
-  sec  e  =L  being  the  constant  quantity  for  the  live  load  at  the  head  of  Col.  V: 


Col.  I. 

Col.  11. 

Col.  III. 

Col.  IV. 

O.'.  V. 

Col.  VI. 
Total  Maxi- 

Member. 

Multiplier. 

D. 

Multiplier. 

L. 

mum  Diagonal 

Stress. 

ab 

10 

lOD 

10 

lOL 

lOD  +  lOL 

br~cd 

2 

2D 

3 

SL 

2D+  3L 

de-el 

-6 

-6D 

0 

0 

-6D 

a'b' 

8 

8D 

8 

SL 

80+  SL 

ftV-c'd' 

0 

0 

2 

2L 

2L 

d'e' 

-8 

-»D 

0 

0 

-»D 

aa"~a"b" 

6 

60 

6 

6L 

6D+  6L 

b"e"~c"d" 

-2 

-2D 

1 

L 

-2D+L 

d"l 

-10 

-lOD 

0 

0 

-iOD 

■•a"'  — a'"6'" 

4 

4D 

4 

4Z, 

4D+  41, 

t/"c"'-c'"d"' 

-4 

-47) 

0 

0 

-4D 

Th?  following  table  gives  the  maximum  chord  stresses,  the  constant  quan- 
tity for  the  combined  dead  and  Uve  load  at  the  head  of  Col.  Ill  being 

o 


Col.  I. 

Col.  II. 

Col.  III. 

Col.  IV. 
Total  Maxi- 

Member. 

MuItipUer. 

F. 

mum  Chord 
Stress. 

Ab 

4 

4F 

4F 

bb' 

10  +  2-12 

12F 

16f 

6'6" 

8  +  0-   8 

»F 

24F 

6"6"' 

6-2-   4 

4F 

2SF 

a'a" 

8 

SF 

SF 

a"a"' 

6-2-   4 

4F 

\2F 

a"'c 

4-4-   0 

0 

\2F 

cc' 

10  +  2-12 

12F 

24F 

Ex.  6.  The  bracing  of  a  lattice  girder  consists  of  a  single  system  of  triangles 
in  irht'rh  one  of  the  aides  is  a  strut  and  the  other  a  tie  inclined  to  the  horizontal 
at  mif/lfs  of  n  nnd  ,3  respectively;  in  order  to  give  the  strut  sufficient  rigidity  its 
section  is  made  k  times  that  indicated  by  theory,  the  coefficient  k  being  >  unity. 
Show  that  the  amount  of  material  in  the  struts  and  ties  is  a 
minimum  when  tan  a  =k  tan  /?. 

Let  /i=(lepth  of  tru.s.s.  Also,  let  .S  be  the  shear  between 
the  two  constTUtive  panel-points  A  and  B.  Take  AC  as  the 
strut  and  BC  as  the  tie.    Then 


i'la.  035. 


Stress  in  /1C=.S  cosec  a    and    in  BC-S  cosec  ,3. 


DETERMINATION  OF  STRESSES.  707 

Therefore,  total  amount  of  material  in  AC  and  BC 

kS  cosec  a  .^    S  coeec  ^ „„    S ,, 
-. At  + — at  -—(A  cosec  a-A  coeec  ct+coaecfi'hctmBc/l) 

-  —(A  cosec'  a  +  cosec'  ^  -    mimmum, 


80  that 


Again, 

Therefore 

Hence,  by  (I)  and  (II), 


Ifccosa.       C0B;9  ,^    ^ 
Bin'o  sin'^ 

ilB  -A(cot  a  +cot  /9)  -a  constant. 


...     (I) 


da         d/9 


sin'  a     sin'  ^ 


(11) 


tan  a  -k  tan  ^. 


Ex.  7.  Determine  the  maximum  stresses  in  the  members  of  a  through  lattice . 
truss  of  40  ft.  span  and  4  //.  depth,  tvith  tux)  systems  of  fr    ngles  (base -8  ft.), 
(a)  when    riveted   together;   (6)  when   pin-connected.    Dead   load  - 1    ton   per 
lineal  foot,  live  load"^  ton  per  lineal  foot. 

(a)  Riveted  rnws.— Panel  dead  load-1  ton;  panel  live  load -2  tons. 


Fio.  036. 


Diagonal  Stresses.  Constant  for  dead  load  --  (-7:  sec  45°)  - — -  ; 

2  \10  /      20 


"'live    load -i-(l  sec  45»)  - 


V2 

To' 


Chord  Stresses.  Constant  for  combined  dead  and  live  loads 

^  tan  45°) 


Table  of  Max.  Diagonal  Stresses. 


Col.  I. 
Member. 

Col.  11. 
Multiplier. 

Col.  III. 
20' 

Col.  IV. 
Multiplier. 

Col.  V. 

vT 
10  ■ 

Col.  VI. 

Totml  Hud- 
mum  DiaKonal 

Stress. 

d^'-D, 
d»--D» 

45 
35 
25 
15 
S 
-5 

2.25V2" 

1.7.'5V^2 

1,25VT 

.75V^2 

.25  ^^2 

-  .2S'^/2 

9X5   -45 
8X4i-36 
7X4    -28 
fix."»i-2l 
5X3    -15 
4X2i-10 

4..'iN/2" 
3.(.vr 

2.sv'i 

O.75V2" 
S.35V2 
4.05v^ 
2.K5N/2" 
1.75V2' 
.75vT 

I    !■ 


708 


THEORY  OF  STRUCTURES. 


The  stresses       '      i.inaiiunjj  diagonals  are  ncRativc. 
Table  of  Max.  Chord  Stres-ses. 


Col.  I. 
Member. 

Col.   II. 

Multiplier. 

Col.  III. 
3 

C..1.  KV. 

Total  Maxi- 
mum (^lionl 

St  reiw. 

fi  75 
IH  75 
27 .  75 
33  75 
3H  75 

'2~  -<'! 

1,'  ~C, 
li'~Cl 

45 
45 +  .35  »  SO 
.35  +  25  '  (iO 
25+15     40 
15+    5     2t) 

fi  75 
12  00 
»  00 
H  (X) 
3.00 

(6)   Pin-conncctcd  Truxa. — 

v'iT 

Diagonal  Strctises.  Con.staiit  for  dead  load  =-7^! 
"        "     live  load  =  —-—. 
Chord  Strcsscn.  Coii-stanl  for  combiiKHl  dead  and  live  loada 
-,|tan4.5°-f^. 


Table  of  .Max.  I 

Col    II.                   Col    III 

)iaj?onal   St  re 

Col.  IV. 

sses. 

Oil.  I. 

Col.  V. 

Col.  VI. 

Member. 

Multiplier 

v'2 
11) 

Multiplier. 

2V2 
10  - 

Total  Mui- 
mum  HionJ 

Streiw. 

rf,-  -Hi 

25 

ao 

15 
10 
5 

a.sVa' 

V2 

RX  2 

5X5-28 
4  y  5  -  20 
4X4- IB 
.SX4-I2 
3X3  •   9 

5v^a 

4X^2 

3.2Vi 
24V'2 
I.HV-2 

T-sVa" 
fiv'a 

4.7X^7 

s.4v^a 

2.3>/2 

dt'-O, 

0 
-5 

-.5W 

2x3-    fl 

ax2-  4 

I2v'a 

.((Va 

1.2V'2 

.3v^a 

The  .ttresrics  in  the  rem.'iiniiig  liiu^onali*  are  ne|{ative. 
Table  of  Max. 


r 

c«i.  1. 

Col     II. 

Member. 

Multiplier. 

«i 

20 

h 

a5  +  l5-40 

Ij 

Jii  ^  HI    :«» 

(4 

15  +    5     20 

t. 

l»¥    0     lU 

»! 

2.^ 

'l 

20  f  20     40 

f% 

15^15     :m 

ft 

10+111     20 

c* 

5  +    5  -  iii 

DETERMINATIOS  Ob   STRESSES. 


709 


Ex.  8.  Determine  the  stresses  in  the  xeveral  members  of  a  deck-truss  for  a 
(l<iublc-track  bridge  of  342  ft.  span,  33  ft.  depth,  and  uiih  eighteen  panels.  The 
panel  eiigint,  live  and  dead  UhuIs  art-  121, (XX),  (i5,(XI0,  and  40,(X)6  lbs.,  reaped- 
ivety,  per  Iriixx.     (Single  intersect  ion.) 


^ 


AlB»CJD;e^FGHK       t.M 


r? 


_M 


', «» 


»\ 


H        l>        U       H 


XXXXXXXXXXX 

I'Ui.  «37. 

19 
sec  e  -1.13393;    tan  »  -33 ;  0°  -29°  56'. 

121(XX) 
Diag.  Strestet.    Constant  for  engine  load  =    ",„- see  0  =  7757; 

for  train  load  -    .-jt—  see  0  -4167; 

40000 
for  dead  load  -     ■■■     sec  tf -2565. 

Chord  Stresses.  Constant  for  combined  engine  and  dead  loads 

121000 +  4aXX)         „     ,  ,^ 
- —     jjj        -  tan  9 -5150; 

for  combined  dead  and  train  loads 
40000 +ft-i(MX) 


w 


l.s 


tan  »-3359. 


id 


Diagonal  Strkhkek. 


1 

1 

1 

Total  Mad- 

M«mb«r. 

Multiplier. 

77«7. 

Muliiplirr 

41U7 

Miilliplirr. 

25IIS. 

mum  Diacn- 
lutlMlmMia 

l*f>UI«ill. 

, 

.Xa 

17 

1»I.N«W 

im 

IMMtJli 

IM 

.'«W.44!r~ 

1,1)0  l.irM 

br 

in 

VU.Wi 

\jn 

.■WKi.tmi 

1.15 

:«4«l.27.1 

07f),437 

4> 

\H 

llli..V>.% 

KKV 

4:t7.."l;l.'i 

117 

.Mm.lM 

8Aa.aM 

ii 

14 

KM.JWM 

01 

.•|7lt,MI7 

W 

iM.QSA 

741,7,10 

1:1 

1(I<I.M< 

*H 

:w5,i«ii 

Ml 

•J07.7tW 

H,'<:i.tua 

Im 

IJ 

(i:i,l)H. 

mi 

■2T\mj 

IM 

ini..ws 

r<M,n)\ 

IW 

1 1 

M^f27 

.Vi 

TM.DC, 

*i 

ll5.4aA 

4»jKn 

M 

to 

77  .W) 

4:. 

m7  ..■>!.'> 

27 

m3M 

im.st4v 

.-» 

V 

fio.Nl:» 

■m 

i.vi.ou    1 

9 

3H.IMA 

242.«IO 

r» 

H 

tu.il.%0 

■••s 

Iltlti7ll 

« 

-!K1.IIM 

1.%A.I)47 

nr 

7 

M.-jm 

I'l 

>T.V»7 

•IT 

-HBJM 

TIMX 

The  Mrewt  in  the  next   tiiiigoiml  ly  i»  iiegHlive,  and  tlw-n-fore  counter- 
braces  (or  additiiHial  iitretigt>M>iiiiig)  an*  n*<|uired  in  tlie  two  ivntre  panek 


710 


THEORY  OF  STRUCTURES. 


only,  althouf^h  it  is  usual  in  practice,  for  the  purpose  of  stiffening  the  truss, 
to  introduce  them  into  other  panels. 

Again,  the  maximum  .stresses  in  the  verticals  ab,  cd,  ef,  gh,  kl,  mn,  op,  qr, 
and  st  arc  the  vertical  coniiwnents  of  the  maximum  stresses  in  the  diagonals 
Xa,  be,  de,  /g,  hk,  Im,  no,  pq,  and  rs,  respectively.     Hence  the  maximum  stress 


ab  =  1 ,001 .026  cos  0  =04.i,4.Sfi  lbs. ; 
el  =  S.53,S»")  cos  0  -  740,000  lbs. ; 
kl  =  G;}3,63-2  cos  0  =  .549,1 00  lbs. ; 
op  -  4l'»,837  cos  e  -.37-',r.OO  lbs. ; 
st=  242,910  cos  0-210,r>00  lbs.; 
wz  -      72,551  cos  0  -  62,874  lbs. 


ed  -  970,247  cos  0  -  840,820  lbs. ; 
^A  =  741 ,730  cos  0  =642,800  lbs. ; 
mn  -.529,701  cos  0  -459,050  lbs. ; 
qr  -334,340  cos  0  =289,740  lbs. ; 
uv  - 1 55,647  cos  0  - 1 34,890  lbs. ; 


Chord  Stre&«es. 


1 

Hum  of 

Total  Mmxi- 

Mam  her. 

Multiplier. 

0150. 

Multiplier. 

3350. 

3d  and  5tb 

mum  Chnrtl 

I     CMlumnM. 

Streu  in  Lba, 

An  .  Xb 

17 

HI. MO 

1,18     ~ 

456.824 

.^44,374 

544.374 

Hr  -  X^ 

—  1 

-5.1.W 

138 

4.^6,834 

451.674 

006,048 

tr~X/ 

—  1 

-5,1.'.0 

lis 

.106,362 

.301,212 

1.387,260 

l>u  ■  Xh 
hk  -  XI 

—  t 

-5,150 

100 

SK.'i.yOO 

330,750 

1.718,010 

!        ~  1 

-.■i.l.TO 

83 

275,4:W 

270,288 

I,0H8,208 

Fm  -  Xn 

—  1 

-5.1.V) 

64 

214.076 

209,826 

2, 1X8,134 

Oo-Xp 

—  1 

-5.1.V) 

4« 

l.M.SM 

140  ..304 

2.,M7,488 

H'l-Xr 

—  1 

-5.1.V) 

W 

04.(K52 

88.002 

2,436  ,.390 

At 

~' 

-5,150 

10 

33,500 

38.440 

2.464.830 

These  results  can  easily  be  checketl  by  the  method  of  moments. 
For  instance,  the  reaction  /?  at  the  right  support 

-  rj^(j»6000)  + .^ . 

Taking  moments  almut  the  f(N)t  of  vertical  st, 

r-      oo     /    '*f^^*^     17vl050(IO\ 

A»X33-(       j^-  + :, 19x19-9x105000x4x19-81,329,500 


and 


A's-2,404,.530lbs. 


,\s  already  state<i  at  the  end  of  Kx.  2,  the  stresses  in  the  several  membem 
can  U'  nion-  easily  and  readily  obtained  by  tabulating  them  in  the  manner 
just  de.HcrilKMl,  itui  the  wiine  results  may  In>  found  graphicaUy  as  follows: 

Fig.  •i3S  gives  the  stres.«»es  in  the  several  menilH>rs  for  a  draii  Imd  of  40,000 
Uw.  (•((iiceiiiraled  al  every  pMiiel-|Miiiit.  The  greatest  .stressi>s  in  the  choni 
panel  h'ligths  occur  when  the  live  load  covers  the  whole  bridge,  mi  that  the 
total  loud  on  the  bndKe  is  then  ef|iiivalent  to  a  load  of  Itil.OOO  lbs.  at  the 
1st  |mn<l-ixiint  and  lO.'i.tNN)  lbs.  al  every  other  panel-|Mnnt.  Then,  »t  the 
left  su|>js>rt, 

the  reaction  /^--^lOlOOO-^.— -16x105000 xH^x  19 -945,389  Iba. 


DETERMINATION  OF  STRESSES 


711 


Taking     XA  -  &t5,3S9  lbs.,      ilB- 161,000   lbs.,    and     BC~CD- 
—105,000  lbs.,  Fig.  639  gives  the  Mai  maximum  chord  stresaa. 


Fio.  638 


Fio.  639. 


m      k 


AM 
BHKL 


_        "       J       S      7      5 

Fio.  wo 

For  the  maximum  diagonal  ttresxen  assume   that  the  only  force  acting 
upon  the  truss  is  a  vertical  reaction  of  1000  units  at  the  ieft  support.     Then 
Fig.  640,  in  which  XA  -1000  units,  is  the  corresponding  stress  diagram  and 
shows  that 
(a)  the  stress  in  each  diagonal  due  to  th»  asaumed  reaction 

- 1000  sec  fl  - 1 1539.3  uniU; 

(6)  the  streaa  in  each  vertical  due  to  the  aaaumed  reat^ion 

- 1000  uniU. 

The  total  aclutd  reactiima  at  the  left  support  are 

M5,3S9  lbs.  when  the  live  load  covers  I  to  17 


840,>W0  " 

2tol7; 

740,000  " 

3  tor; 

642,723  " 

4  to  17; 

64ft,aW  " 

5  to  17; 

mi.OOO  " 

6  to  17; 

372,.'M6  " 

7  to  17; 

2H»,723   " 

8  to  17; 

210,.'i00   " 

9to  17; 

134,SH9   " 

lot  J  17; 

62,889  " 

11  to  17. 

''t 


712 
1  Icnee 
or 


THEORY  OF  STRUCTURES. 

the  actual  stress  in  any  diagonal,  say  on      3725^6 
the  sin'ss  due  to  the  assumed  reaetion  1000    ' 

actual  stress  on  =372.')r)fi  -'y^mT"  ='*--'.'-^'-  ^^'^■'> 

the  actual  stress  in  any  diagonal,  say  c/)  _ 289723 
the  stress  due  to  assumed  reaction  1000  ' 

or  actual  stress  «/'  =289,723  lbs. 

Ex.9.  Thr  Schirrdtrr  Inisn  in  which  the  mini'  ■.  strrxx  in ^very  diagonal 
i,s  to  he  nil. 

Consider  the  eight -panel  truss,  Fig.  M\,  and  let  l>  and  /-  l)e  the  panel  dead 
and  live  loads  n-six-ctively.     I^t  a  be  the  panel  length.    Let  y„  j/„  y„  y,  be  the 


o. 


o. 


Oi         O     o       I 


<i     ;     a 


length.*  of  the  verticals.  Let  the  straight  portions  of  the  Ix)'-.  meet  the  hori- 
zontal through  (J  ill  Oi,  Oi.  0,,  and  take  0(),"X,,  ^W>,  =/„  (W7,  =.j,.  For 
initiimuni  stress  in  2.4,  /-  is  at  I  and  the  reaction  at  O  is  then 

The  stress  in  2.1  is  to  \tf  nil,  ;ind  then-fon',  taking  moments  almut  0^, 
-(3i/>  +  i/,)x, +{/)  +  /,)(x, +fl)=0,     or     -"'!'*  n^v-. 


Hence 


y,     J, +2fi     «(S/>+jr,) 
I/,  °  J-,  +<i      7(4/)  +  /.)" 


For  miiiimutn  stress  in  .'{/>,  /,  is  at  1  and  2,  and  the  n-action  at  O  is  then 

3i/> +  ',"/.. 
The  stit'ss  in  '.Vi  is  to  !»■  nil,  anil  tlien-fon",  taking  moments  alM)ut  Ot, 

X,      H{li  +  l.) 


■V.iil>  +  ','l.)x,+Jil)i-l.,(x,-\-ia)  -0,     or 

V,      X  -f  3fi      .'>(4/>  k- 1.) 
I/,  "/..  '  2«i'  2(*</)4-3/<)' 


a      4D-L 


Ffsr 


mmim'-Hi!  .".!«>«.■■.  m 


-U-   !. 


!!i!  H  nssd  ih*'  mart  ion  st  O  is  th^tt 


3j/>f  J/.. 


4.31 


DETERMINATION  OF  STRESSES.  713 

The  stress  in  4c  is  to  be  nil,  and  therefore,  taking  moments  about  0„ 

-(3iD  +  *L)x,+3(D  +  L)(x,+2a)=0,    or    ??  =24(D  +  Z,)^ 

a       2D  —  3L 


Hence 


y.  _^. +4a  ^4(HD+3L) 


Case  a.  If  L=iD,  a-j-=ac,  and  therefore  the  two  central  chord    panel 
lengths  of  the  bow  are  horizontal. 


Also, 


"  •    y.  ""«i'    y.  "49- 


y 


If  />>  ID,  r,  is  negative  and  the  depth  of  the  truss  might  be  diminished 
from  3  towards  the  centre. 

So  if  L>41),  J-,  is  negative  and  the  depth  might  diminish  from  2  towards 
the  centre. 

If  L>  MD,  X,  is  negative  and  the  depth  might  diminish  from  1  towards  the 
centre. 

It  is  inadvisable,  however,  to  construct  a  truss  in  such  a  manner. 
r.i.sf  b.  If  L-D,  x,"  -,^,  and  the  three  central  verticals  should  be  each 
made  of  the  same  length,  viz.,  y,.     Aim, 


^     2.5  y,     .54 

J; -22     «"d     --35. 


The  calculations  will  be  the  same  if  the  truss  is  inverted,  <nd  also  the 
<lepths  V,.  ;/.,  .  .  .  may  be  plotted  one  half  above  and  one  half  below  the  hori- 
zontal through  O. 

If  the  diagonals  in  Fig.  (i4l  are  made  to  slope  in  the  opptwite  direction, 
I  hey  are  tieci-ssnrilv  in  tension. 

i;x.  10.  I',iiili\  iruKu  in  ivhirh  earh  panel  length  of  the  bow  i«  to  be  subjected 
III  the  xnnie  Hlrrxs. 

Consider  an  cight-|>anel  truss,  Fig.  042,  each 
|.ancl  length  iK'ing  a.     I...f  /.,,  /.^,  ,,„  ;,,  jx.  the 


|.ancl  length  iK'ing  a.     I...f  /.,,  p^,  />„  ;,,  Jx.  the  p         .         -         . 

IHTiH-ndiciiliirs  frnm  2,  3,  4, .')  upon  the  ojJiioHiiP  J^""    t*  **  "j^^"?*  "  "j* 

chord  length,  iiirlined  to  the  vcrlicnl  at  angles  '  ->J/^       /|       A 

".  '      <>.     The  chord  SI H'sscs  an- gn-atest  when  ,^p  '*^\/  •*•    / 
til.       <■  liNid  covers  the  whole  truss.  *^I>^>^/_H 


l^'t  /'  (  -1)^1.)  Ir  the  panel  load;  and  let 
/•'  \n'  the  constant  for..-  in  e«<h  pHnd  length  of 
the  bow.     Tlie  reaction  at  1  -.\\l'.    Then 


Fio.  042. 


A\P 


Pi 


_ f' . •If''  in- Pa _ 3| P  3a -2P •  |o  _ SjP  4a-3P2a 


P, 


714  THEORY  OF  STRUCTURES. 

*  Pi~ P,~ P>~  Pt' P<*' 

defining  the  values  of  p,,  p„  p„  p,. 

Cnnslruclion. — Through  1  draw  the  tangent  \A  to  the  circle  having  2 
as  centre  and  p,  as  radius  to  meet  the  vertical  through  2  in  ^. 

Through  A  draw  the  tangent  Ah  to  the  circle  with  2  as  centre  and  p,  as 
radius  to  meet  the  vertical  through  3  in  h. 

Through  h  draw  the  tangent  be  to  the  circle  with  4  as  centre  and  p,  as 
radius  toVneet  the  vertical  through  4  in  c. 

Through  c  draw  the  tangent  cd  to  the  circle  with  5  as  centre  and  p^aa 
radius  to  meet  the  vertical  through  5  in  d. 

Again,  if  y„  y.-,  i/„  and  i/«  are  the  lengths  of  the  verticals, 

»'     ..„              P.            K-K     ,„„  a     iV-P.'. 
—  -tana-    . =     -tan/J  — ; 

a  Va'-p^*  o  P» 


K-h,    ^  ^h,'-p,'      A.-A,    ^_.     Vy-P,' 

—tan  r  — —tan  a  — . 

a  '  p,       '        a  p. 


If  in  any  panel,  e.g.,  the  3d,  Pt^a,  then 

o"2  VA,     a/" 


If  the  perpendiculars  are  increased  m  times,  then  F  is  diminished  m  times, 
but  remains  unchanged  if  P  is  increased  m  times. 

7.  Single-  and  Double-intersection  Trasses. — Fig.  643  represents 

the  simplest  form  of  single-intersection  (or  Pratt)  truss;  i.e.,  a  truss 


— 1 


Fio.  613. 

in  which  a  diagonal  oros.<ies  nne  panel  only.  It  may  l)e  eonstructed 
entirely  of  iron  or  steel,  or  may  have  tlie  chords  and  verticals 
of  wooil.  The  v»'rtieals  are  in  rompression  and  the  diagonals  in 
tension.  Th(>  angle-l docks  are  tlierefore  placed  alM)ve  the  top  and 
b<>lo\v  the  l)o(tom  c-liord.  ("ounter-hraces,  sljown  hy  the  dotted 
iliagonals,  are  introdvieed  to  withstand  tiie  effect  of  a  live  load. 

Fig.  611  shows  a  16-panel  deck  tru.ss  of  the  Baltimore  type,  having 
straight  chords  and  sulKlivided  panels.  Th»^  Petit  truss  is  similar, 
the   j)nly   difference  iM'ing   that  one  of  its  chords  is  inclined.    A 


■^i.:'k 


SINGLE.  AND  DOUBLE-INTERSECTION  TRUSSES.  716 


14-panel  through  truss  of  the  Petit  type  is  represented  by  Fig.  645. 
Both  classes  of  truss  are  standard  forms  for  bridges  of  long  span. 


i^ 

9, 

c 

P, 

it 

f  , 

0 

"i 

.^ 

h 

1  > 

«y 

•/ 

\': 

It/ 

\" 

y 

\b 

V 

\- 

ao/ 

\» 

^ 

\u 

"^ 

ixN 

(l 

A 

/. 

w\ 

(«* 

«^ 

6« 

a\ 

/*> 

«N 

/» 

«N 

**/ 

/* 

\ 

/ 

\ 

/u 

\ 

4 

-\ 

/» 

/ji 

\ 

/« 

\ 
\ 

K 

J 

( 

X 

X 

X 

X 

X 

Via.  644. 

Ex.  11.  The  stresses  in  a  Baltimore  or  Petit  truss  may  be  easily  obtained 
graphically.  Take,  for  example,  the  truss  rtpresenled  by  Fig.  644,  and  let  d 
and  I  be  the  panel  dead  and  live  loads  respectively. 


Fio.  645. 


Chord  Stresses.— The  stresses  in  the  chords  are  greatest   when  the  live 
load  /  is  concentrated  at  each  of  the  panel-points.    Talcing  p-'d+l.  Fig.  646, 


Fio.  646. 

is  the  corresponding  stress  diagram,  in  which  XA  -7Jp,  and  AB-BC-ae~p. 
Then 

Al^B2-^7iptunf>;    (0'D6-\2ipUu  6,    £.11 -^,I2-15Jp  tan  *; 
(7,17-//,18-161ptan  fl-A',23-L,24,  for  compression  chord; 


0  J  IF 


l": 


716 


THEORY  OF  STRUCTURES. 


and  X7=X8-7ptane;  A',13=X,14  =  12p  tan  fl;  X,19-X,20-15p  tan  «for 
tension  chord. 
The  corres})onding  chord  stresses  due  to-  the  dead  and  live  loads  separately 

may  be  obtained  bv  multiplying  these  results  by  the  ratio  -  and  -  respectively. 

P  P 

Diagmial  Strcssen. —  (a)  Stresses  due  to  dead  load. 

Xl=7J</serO;    A'3  =  7(isectf;    23  =i(/ sec  0=69-12,15-18,21 -22,23; 
45 -aid  sec  tf;    89=5<isee0;     10.11  =3irf  sec  fl;     14,15-3^8600; 

10,17  =  lJ(/secfl;    20,21  =d  sec  0. 
(6)  Stresses  due  to  live  load.    Fig.  047  is  the  stress  diagram  for  the  truss 


n.M.27 

i:.ii.ti 

11.W.I* 

«.«.* 

butt 

\ 

\ 

\ 

N 

t 

1 

tt.aM 

I6.M.III 

MU.!* 

<.3.« 

* 

Fio.  647. 

on  the  assumption  that  it  is  acted  upon  by  an  arbitrary  vertical  reaction 
at  0  only      Call  this  reaction  R,  so  that  AM  =//. 

On  the  same  scale  the  length  of  each  sloping  member  in  the  diagram  is 
R  sec  0,  and  R  is  the  length  of  each  vortical. 

For  the  live  load  /  concentrated  at  nil  the  panel-points,  at  all  the  panel- 
points  except  the  first,  at  all  the  panel-points  except  the  first  two,  three,  four, 
etc.,  the  corresiKinding  reactions  at  O  are: 

7il.  W.  VJ,  Ul,   ?5/,   ;!/,   HI,   r,l,  W,  Ul,  W,   W,   M  Al,  and  ^l. 

Hence  the  maxinium  live-load  tensile  stresses  in  the  main  diagimaU  are: 

Al=71/s.c0;        .Y3=',V/socfl;       i^^Uli^cO;         89-U/8ectf; 
10,11-; 5/ sec  fl;     14,15-|iiZ8ecO;     16,17 -tl/ sec  0;    20,21 -Hf  sec  *; 

and  for  the  c(>unUr.t: 

10,13=  |?/s«M.O;     16,19-fl/secO:     12,15-U/ »ec  0;    18,21 -f|Z  sec  0. 


Als 


4,7-,?o/8ec9;    6,9-,VsecO. 


Virfirnl  Strrxsrs.—  (,t)  Strcssi>s   due    to   dead    load.     It 


is   assumed  that 


two  thirds  of  the  dcail  |i)atl  is  concentrated  at  the  top  panel  and  one  third 
at  the  iMittom.     Then,  for  the  nuhrertirals. 


12  -  §./  - .10  - 1 1 .  12  -=  1 7. 1 S,  and  are  com 
78 -J(i- 13,14 -19,20,  and  are  ten-siona, 


pressions; 


BOWSTRING  TRUSS.  717 

For  the  main  verticals, 

34-6Jd;    9,10-4§d;    15.16-2|d;    21 ,22 -l§d;  and  are  aU  compressions. 

(6)  Stresses  due  to  live  load.    For  the  suhverticah, 

12 -Z-56 -11,12 -17,18,  and  are  all  compressions. 
78-0-13,14-19,20. 

For  the  main  verticals, 

34-V,'/;    9,10-IJ/;    15,16-H/;    21,22 -nz,  and  are  all  compressions. 
Again,  when  the  live  load  covers  less  than  half  the  bridge, 

the  total  resultant  stress  in  22,23  -(i?|  - Jrf)  sec  0; 

" "  26,27  =  (fJZ-li<0  sec  »; 

"      "  "  "    "  28,29 -(H;-2Jrf)  sec «; 

"      "  "  "     "  32,23 -(H/-3id)  sec  0; 

The  last  term  in  these  resultant  stresses  is  due  to  the  positiw  dead-load 
shear  and  must  be  neutralized  by  the  negative  shear  due  to  the  live  load 
iH'fore  the  counters  begin  to  act. 

8.  Bowstring  Truss.  —  The  frame  represented  by  Fig.  64S 
r«'st.s  ujx)n  aupiwrts  at  .4  and  B  and  consists  simply  of  a  curved 
luemlxr  ACB  with  its  ends  tied  together  by 
;i  horizontal  chord  AS.  There  in  no  inter- 
mediate bracing  and  it  is  assumed  that 
(Icforniation  is  prevented  by  the  stiffness  of 
the  bow. 


■K- 


■<» 


Fio.  648. 


I/«t  AB=-1,  and  let  the  chord  AB  carrj'  a  uniformly  distributcil 
Inad  of  intensity  w.     I^'t  k  l)e  the  central  depth  of  the  frame. 

Consider  the  equilibrium  of  one  half  of  the  frame  between  A 
and  a  vertical  section  MX  immediately  on  the  right  of  the  crown  C. 

U't  H  be  the  horizontal  thrust  at  C,  and  T  tlie  horizontal  tension 
in  liie  tie.    Then  ACD  is  kept  in  eciuilibrium  by  //,  T,  the  vertical 

n  act  ion  y  at  X,  and  the  load  is  ^  on  AD. 

Taking  moments  about  A, 


Hk' 


wP 

'  8* 


718  THEORY  OF  STRUCTURES. 

Taking  moments  about  C, 


Thus, 


Tk^'f^Hk. 


jj     wP 


siiul  it  is  evident  that  the  horizontal  component  of  the  thrust  at 
(■\ery  point  of  the  bow  is  a  constant  quantity. 

Again,  if  x,  y  are  the  co-ordinates  of  any  point  P  with  respect  to  C, 

the  bending  moment  at  P  =  Y\2~y)  ~I^(2~^)*~^^*~*^ 


'  8\k     P  )' 


X    4f 


which  is  nil  '^^  J  =  ~p,  >p>  if  the  axis  of  the  bow  is  a  parabola. 

Fig.  649  represents  an  iV-panel  bowstring  truss  of  span  I  and  cen- 
tral depth  k,  the  axis  of  the  curved  member  being  a  parabola. 


Fia.  649. 


Let  L  be  the  panel  live  load  and  take  o  (  -tt)  to  be  the  length 

of  a  panel. 

The  maximum  stress  in  the  nth  diagonal  is  a  tension  (=dii)  when 
L  is  concentrated  at  each  panel-point  from  n  +  l  to  N —  1,  and  a 
compression  (//„')  when  L  is  concentrated  at  each  panel-fx)int  from 
1  to  n. 

For  the  maximum  tension  consider  the  equilibrium  of  the  por- 
tion of  the  truss  b<'tw('«'n  .4  and  a  vertical  section  immediately  on 
the  left  of  n  +  I  (or  Q),  and  let  the  tangent  at  P,  the  head  of  the 
vertical  PQ,  meet  the  horizontal  through  A  ia  E  and  the  axis  of  the 


BOWSTRISO  TRUSS. 


71» 


parabola  in  T.    Let  x,  y  be  the  coordinates  of  P  with  respect  to 
the  vortex  C.    Then,  since 

X    Ay* 
EQ = QP  tan  PTC  =  (*  -i)|-  - 


Agaii, 


AE.EQ-(l-y)SlA, 


2x       2y  • 

a 


BE-AE+l'- 

I 


(i-r 


2j/     • 


Therefore 


and 


AE     2 


-y 


n  +  1 


i^Q      I  N-n-V 

2+y 


z 


JBJ?      2 


o+y 


.V-n-1 


i^Q      i  n  +  1 

2-y 

The  reaction  A^  at  ^  is  given  by 


RaI  -(N-n-  l)L~^a  -  R^Na, 


or 


ftx  = 


L  (N-n-l)(N-n) 


2  N 

Draw  i?F  at  right  angles  to  d„,  and  take  moments  about  E.    Then 
/.  (N-n-l)(N-n) 


d„EF  =  T, 


N 


-AE, 


.      L  (N-n-\)(N-n)  AE 

L  iN-n-l){N-n)AE 
"  2  "  N  EQ 


ciiaec 


L  (Ar-n)(w-f  1) 


iV 


eosec  tf 


»i 


On  being  the  inclination  of  the  nth  diagonal  to  the  horiiontal. 


720 


THEORY  OF  STRUCTURES. 


For  the  maximum  compression  (d„')  in  the  nth  diagonal,  con- 
sider tlie  equilibrium  of  the  portion  of  the  truss  between  MN  and  B, 
and  take  moments  again  alwut  E. 

The  reaction  Rg  at  B  is  given  by 

/2j.\a  =  HL— :^a        or        /i*  =  —  -^ — . 


Therefore 


d,'Wsinfl„  =  |'^W, 


or 


L  n(n  +  l)BE 


d„'  =  Y— ^  ^cosec0„  =  Y 


L  (X-n-\)n 


X 


eosec  d„ 


Hence,  if  every  panel-point  is  loaded,  the  resultant  stress  in  the 
nth  diagonal 

=  d„-d„'  =  ^^^^^l(-V-n)(n  +  l)-(.V-n-l)nj=^cosec0,. 


A' 


L 


The  vertical  com[)onent  of  this  stress  =  -;:7  and  it  follows  that  when 

a  truss  of  this  tyjn^  carries  a  uniformly  distributed  load,  ti.'>  vertical 
component  of  the  stress  in  any  diagonal  is  a  constant  quantity  equal 
to  one  half  nl  the  panel  load,  and  is  a  tension. 

Let  D  be  the  panel  dead  load.    Then 
the  maximum  tensile  stress  in  the  nth  diagonal 

r.     ,(A' -n) (n -4-1)1  cosec^n 
and  the  maximum  compressive  stress  in  the  nth  diagonal 


-  (.V-n-l)»  I  COST c  On 

~  Y  J       2 


If  this  last  result  is  negative,  it  indicates  a  stress  of  the  same 
kind  as  that  due  to  tlir  dead  load,  but  is  less  in  amount  and  is  con- 
seciueiitly  provided  for  by  the  ordinary  bracing. 

If.  however.  th<*  result  is  [Misitivp,  it  indicates  that  the  resultant 
stress  is  of  an  opi)Osite  kind  to  that  due  to  the  dead  load  and  must 


BOWSTRISG  TRUSS. 


721 


be  provided  for  by  giving  the  members  affected  an  additional  sec- 
tional area  or  by  the  introduction  of  counter-braces  shown  by  the 
dotted  lines,  which  will  evidently  be  in  tension  and  thus  every 
diagonal  becomes  a  tie.  In  precisely  the  same  manner  as  above 
it  may  be  easily  shown  that  the  stress  dn"  in  the  nth  "dotted" 
diagonal  is  given  by 

,     L(N-n)(n  +  l) 
"   ~2 N cosecf?,', 

»,'  being  the  inclination  of  the  diagonal  to  the  horizontal. 

Again,  the  maximum  live-load  stress,  V„,  in  the  nth  vertical,  is 
a  compression  and  is  developed  when  L  is  concentrated  at  each 
jmnel-point  from  ?i  +  l  to  A'-l.  It  is  evidently  the  vertical  com- 
ponent of  the  corresponding  stress  in  the  (n-l)th  diagonal.  This 
stress  can  be  determineil  by  considering  the  equilibrium  of  the  por- 
tion of  the  truss  between  A  and  a  vertical  section  immediately  on 
the  left  of  n.    It  Ls  at  once  found  that 


.         L(X-n-l){X-n)     n  Ln(N-n-l) 


cosec  dn-l. 


Hence 


4.      w        •    n         Ln(N-n-l) 
f»  =  d„_,  sm  <>„-i  =-  ^^ — .-, — ^, 


The  vertical  comjxinent  of  the  dead-load  stress  in  the  (n-l)th 
diagonal  is  a  tension  ^,  and  tiierefore  the  total  maximum  stress  in 

the  nth  vertical  due  to  both  dead  and  live  loads  =  ?-.  +  --  /)  =  ?•  -- 
The  stress^is  consequently  a  compression  or  tension  according 
as  i„>  or   <— . 

If  the  truss  now  under  consideration  is  inverted,  the  stresses 
are  reversed  in  kind  but  remain  the  same  in  magnitude  The 
lenticular  truss  (Fig.  650)  is  a  combination  of  the  two  forms  and 
the  nio..t  iniijortant  example  in  practice  of  such  a  combination  is 
the  l)owstnng  suspension  bridge  erected  at  Saltash  (Eng.).     The 


722 


THEORY  OF  STRUCTURES. 


Ik)w  is  a  wrought-iron  tube  of  oUipfical  section,  and  is  strengthened 
at  intervals  by  diaphragnL'^.  The  tic  consists  of  a  pair  of  chains. 
The  mathematical  analysis  of  the  stresses  in  the  several  •  nibers 
is  precisely  similar  to  that  of  the  simple  bowstring  truss. 


t^*---^^-'^V/if1 

0 

R 

^» 

"       n\ 

Fio. 

«       0 

6«). 

First  assume  that  there  is  no  intermediate  bracing  and  that 
deformation  is  prevented  by  giving  the  bow  sufhcient  rigidity.  If 
the  load  is  uniformly  distributed  and  of  intensity  w,  and  if  k  and  fc' 
are  the  central  tlepths  of  the  l)ow  and  tie  respectively,  then  the 
half  of  the  truss  betwwn  .1  and  a  vertical  section  A/A'  immediately 
on  the  right  of  the  crown  C  is  kept  in  equilibrium  by  the  horizontal 
thrust  //  at  C,  the  horizontal  tension  T  at  D,  the  vertical  reaction 

—  at  A,  and  the  imiformly  distributed  load  -ir  between  A  and  MN. 

Taking  moments  about  C  and  D, 


and  therefore 
Iffc  =  t', 


If  intermediate  bracing  is  intnMhjced,  as  shown  in  Fig.  650,  the 
tnaxinumi  tension  (d„)  in  the  /ith  diagonal  is  prchluced  wiien  the 
paiu'l  livi-  load  L  is  concentrated  at  each  |>anel-[M)int  from  n  +  l 
to  .V-1,  while  the  maxiiiuun  r()nipre.«<sive  stn>sH  (</,')  in  the  same 
diagonal  is  due  to  a  panel  load  L  concentrated  at  each  |Kinel-|x)int 
from  1  to  n. 


iii 


BOWSTRING  TRUSS. 

To  find  these  stresses,  consuler  as  l)efore  the  equiHbrium  of 
each  of  the  two  portions  of  the  truss  nmde  by  a  vertical  plane  iVAT 
immediately  on  the  left  of  the  (n  +  l)th  vertical  PQ.  The  two 
tangents  at  P  and  Q  meet  in  a  point  E  in  the  horizontal  through 

^1,  the  distance  ER  being  — ^ ,  where  x  is  the  horizont-1  distance- 
of  PQ  from  the  vertex  of  each  parabola. 

I^t  /)  =  perpendicular  from  E  upon  nth  diagonal. 

For  maximum  tension  in  nth  diagonal,  consider  equilibrium 
of  portion  of  truss  between  A  and  MN,  and  let  R  be  the  reactioa 
at  A.    Then 


RXa-iN-n-l)U^a,   or   fl,:^(-V-n)(^-n-l)^ 


Also, 
and 


AE-^ 


\2    V       (n+l)2a 


2x        N-2n-2' 


LnOV-n)(.V-n-lV            La  (■V-n)(.V-n- l)(n-H)» 
anP     ^  y  -XAb^Y  X(X-2n-2) • 

For  maximum  compression  in  nth  diagonal,  consider  the  equi- 
librium of  the  portion  of  the  truss  between  B  and  MX,  and  let  R* 
Im'  the  reaction  at  B. 


Then 


and 


R'Na  -  — Im,   or    ft'  -  -    '■  ^  ■•^. 


(^-r 


BE— ^.a 


(.V-n-l)» 
.V-2n-2' 


'II      r  J  /       D/  nil    /-onfn  +  Df.V-n-l)* 

iluT'fore         dn'p-R'BE"-^      ..,    '    . ;-ii- 

2       A0V-2n-2) 


1^ 


1 1 


Hence     p(d,-d,')- 


Aa(.V-n-l)(n-»J) 
2        A'-2n-2      ■ 


724 


THEORY  OF  STHUCTIRES 


liCt  0„  be  inclination  of  nth  diagonal  to  the  tangent  at  Q,  and 
let  «„  be  the  angle  between  EQ  and  the  vertical. 


Then 
Hut 


p  =  EQ  sin  Un  —  EH  sin  o„  sin  6„. 


ER''AE+{n  +  l)a^a 


(n  +  l)(.V-n-l). 
A'-2h-2 


Hence  dn-dn  =-^co8cc  ancoscc0„ 


Thus  the  vertical  comixinent  of  the  diagonal  stress 

Im  cos(a„-fl„)     La  .  /,    .  ,n 

2  mi  tt-  sm  0.     2 


If  a„  =  90",  this  coni|M)nent 


Im 


us  liefore. 


liowstring  tru«s«'s  of  the  ty|>e  just  descrilx>d  are  now  rarely 
us<'d.  hut  it  is  a  continon  practice  to  s|)ecify  that  the  |)an('i-|K)int» 
ill  one  of  the  '-liords  are  to  lie  in  the  urv  of  a  paralx)la,  the  |X)rtioii8 
of  ihr  chord  iM'twcM'ii  consecutive  panel-|M>ints  l>eing  straight.  Tin* 
ileteniiiMatit)!!  of  the  stress's  in  this  case  is  very  simple  and  is  h'st 
niadt'  i)y  the  graphical  method.  With  a  Irak's  of  this  type  the 
diajroiiais  arc  iinxlniiin'il  uiidiT  a  uniformly  flistrihutrd  load,  and 
only  ciinii'  into  play  when  a  live  load  crosw-s  the  bridge  Take, 
for  exam|)le,  a  ten-panel  truss,  Fig.  (Wil,  with  {kuicI  dead  ami  hvc 


i 


loads  of  />  and  /-  ref^jn'ctively.  The  U|>|>er  ends  of  the  verlicals 
are  in  the  arc  of  a  parabola  with  its  vertex  at  the  highest  jmuicI- 
jKiinl. 


BOWSTRISO  TRISS 


725 


Fig.  (\ry2  is  tho  dead-load  stress  diugruin  and  the  diagonal  Rtresscs 
are  evidently  tiil.  Tlie  maximum  stn'sses  in 
the  I'hortla  are  also  produonl  when  the  live 
load  covers  the  whole  bridge,  i.e.,  Is  uniformly 
distributed,  and  the  ntagnitude  of  thene  live-load 
stn'sses  may  be  found  from  the  dead-load-stress 
diagram  by  multiplying  the  corresponding  chord 

.stH'ss  by  the  ratio  j.  The  total  maximum  chord 


L  +  l) 
I) 


Fui.  652. 


stresses  may  be  found  by  using  the  ratio 

jia  the   multiplier. 

F'lr  each  diagonal  and  vertical  (excepting  the  vertical  next  a 
supiKtrt)  there  is  a  different  end  reaction.  A.ssunu'  that  a  constant 
reaction  of  UKM)  units  at  tlic  left  8up|)ort  is  tlie  only  load  on  thf* 
truss,  then  the  corres|K)nding  stn'ss  diagram  is  that  shown  by  Fig. 
♦)')3,  anil  the  stress  in  the  <liagoiials  an<l  verticals  due  to  the  assumed 
reactions  can  l)e  easily  .scaled  off. 

Again,    the   reactions  at   the   left    sup- 


9X5.  ,j^NX4i.  ]7)7X4.  j^Oxaj.  /^oX3, 
/;4x2i.4'ix2.;^2Xli^lXl, 


10 


10" 


10 


Fio   O.'h-? 


when  I,  is  conct'ntrated  at  the  [>anel-points  2 
to  9,  3  to  9,  4  to  9,  a  to  9,  tJ  to  9,  7  to  9.  8  to  9,  and  9,  respiM-tively. 
Then 

tiie  actual  stn-ss  in  a  nM'nil)er  the  actual  reaction 

the  stress  due  to  the  ijssunieil  reaction     the  aM8unK>d  reaction' 

.\gain,  it  can  Im-  cjisily  shown  aiialyticully  that  the  diagonal  stresn 
Ml  an  A-|»unel  trus<  uiuUt  a  uniforinly  distributed  |)anrl  load  is  nii. 

\A'i  /».  7  be  the  lengtlis  of  the  «tli  ami  (n  +  l)th  verticals,  anil 
l<-t  X  be  the  horizontal  distance  of  the  latter  from  the  vertex. 

I>'t  the  sloping  chord  Mieml)er  in  the  .'Uh  panel  meet  the  hori- 
loatul  througli  A   in  E,  and  take  AE^i. 


Vhen 


(j-»-l)o, 


lit! 


A 

i 


1^ 


726 


THEORY  OF  STRUCTURES. 


1-  „  )  =  Y2(n  +  l)(A'-n-l)i 
also 


4x  +  n\      4k 


Hence 
and 


/       4J  +  a\       4* 

z-\-(n  +  \)n     a     (n4-l){.V -n -1) 


z  +  /ta 


n{N-n) 


2=aTr 


n(n  +  l) 


S-2n-V 


r^ 


Fio.  6M. 
If  p  is  the  porpondicular  from  E  \i\^m  rf„, 

In  + 1 


d„/)  =  ffz  -  «f  ("  .7-  a  +  «) , 


whore  P  is  the  panel  load  and /J  is  the  reaction  at  A 
Hence 


.V  -1, 


_f.V-l    M(«  +  l)  /n  +  1  nn+1    \] 


-0, 


and  therefore 


rf„-0. 


g.  BoW/Cring  Girder  with  Isosceles  Bracing. — Diagimal  siressea 
due  III  the  dead  lirnti. 

l.et  />  1h'  the  dead  load  concentrated  at  each  panel-|K)int; 
^,.  /„'     Ik'  the  t«'n.sioris  in  the  diag-t.  jm,  qn,  at  n; 
0„,  t*,,'   Ik*  the  iiirlinations  of  the  diags.  to  tlie  horizontal; 
«,.,  ttn  1h'  lh«'  inclinations  of  the  tangtuits  at  p  and  9  to  the 
horizontal. 
Ia'I  the  taiigertts  iil  p  an<l  7  intcrsi-ct  BA  produced  in  K  an<i  K'^ 
re.s|iectively      Thon 


BOWSTHIXO    TRUSS 


727 


1       /  '       / 

tnEn  sin  ^"  =  g ^^y    *'"'    '"' '^''^  ■'^^"  ^"'  "^  S^ '^  V' 


where 


,.       1    I  4A'n-4n2  +  l  l  <  4.Vn-4n'  +  l 

^'•^i  iV -^V^2nTr     "^^     ^«=4iV    A'-2«-l 

Diagonal  stres/^  due  to  the  lire  load  (D.—Jjot  d„,  d,,'  lx>  the  diagonal 
stresses  at  the  nth  j)anei-|)oint.     I^t  i  Ix*  the  horizontal  distance  of 


p  from  tlie  \-ertex  C,  and  let  the  tangent  at  :>  meet  the  horizontal 
AB  in  E.    Then 


.\l80, 


OX  Hz 

i  =  j-(n-i)-^.    and    x~-~--{n^^)-; 


dn  is  the  marimum  tensile  xtre.-^s  when  L  is  concentrated  at  all  the 
|)anel-|)oint8  from  n  to  .V-1. 

The  correspomhng  n>action  at  .4-7^^      .~^'^;\-Z^L±l) 

Ck)n8idor  the  equilibrium  of  the  portion  of  the  truss  between  ^4 
and  a  verticul  section  inuuetliately  on  the  right  of  p.  Taking 
n.oments  about  E, 

d^  En  sm  0„  - ^^ aE. 


J,  I 


Tlierefore     dn  ^ 


L  (.V-n)(.V     n+D/l^coBec^, 


A' 


AE  +  nc, 

iV 


728  THEORY  OF  STRUCTURES. 

and  the  angle  0„  is  given  by 

d„  is  the  maximum  com]rressi}'e  stress  when  L  is  concentrated  at 
all  the  panel-points  from  1  to  i.  -1. 

Tu  1-  *•         .  D  •  Ln(n-l) 

The  corres|)on(ling  reaction  ut  B  is  no\v  =  —  — r; . 

J       A 

Consider  tlie  equilibrium  of  the  jwrtion  of  the  truss  between  B 

and   a    vertical   section   immediately   on   the   right   of   p.    Taking 

moments  about  E, 

L  n(n-l) 


Therefore 


dn  ■  En  sin  0„  =  '-^  '^^^BE. 
Ln(n-\)BEc(m'e0n 


d„  =  : 


'^         AE  +  H^. 


Again,  d„'  is  the  maximum  tenxile  stress  when  L  is  concentrated 
at  every  panel-point  from  1  to  n. 

Tu                    I-             »■        .  D     Ln(n  +  \) 
The  corre8|K)iuinig  n-action  at  «  =  —  — . . 

Consider  the  oquilihrium  of  the  iH)rtion  of  the  truss  between  B 
and  a  vertical  sccti(»ii  inniiidijttely  on  \hv  left  of  q.  I,et  the  tangent 
at  q  meet  the  horizontal  AH  in  E'.    Taking  moments  about  E', 

.  ,     L  n(n  V\)BE'  ciifecO^' 


«-^2       -V 

AE'  +  ni   ' 
A 

where 

(/-2j>* 

RF'     ^^^^^' 
^^'      Hx 

and 

x--^-{n  + 

Th*  angle  (?,'  is  given  by  tan  (9„ 


I 


(-T)- 


rfn'  is  the  maxim um   comjirrssi   •  strenif  when  L  is  concentrated 
at  all  tlie  panel-points  from  n  i-l  to  A'~l. 


BOWSTRING  TRUSS. 


72» 


Tu                    r             ,-        ,  A     ^  {N-n)(N-n-l) 
The  corresponding  reaction  &t  A=— -^ -. 

Consider  the  equilibrium  of  the  portion  of  the  truss  between 
.1  and  a  vertical  section  inunediately  on  the  left  of  q.  Taking 
moments  about  E', 

L(N-n)(N-n-l)  Xg^cosecg/ 

AE'+nf 

It  is  a  common  practice  to  require  that  the  panel-points  in  the 
curved  chord  lie  in  a  parabola  and  that  the  portions  of  the  chord 
between  consecutive  panel-points  are  straight.  The  maximum 
stresses  in  the  several  members  can  then  be  easily  determined  graph- 


»ii 


Fio.  (US6. 


Fio.  857. 


n 


Fio.  ttS. 


ir<dly.  Fig.  657  is  the  stress  diagram  for  the  dead  weight  upon 
the  truss  repn'senteti  by  Fig.  65G.  The  chord  strt'sses  are  greatest 
when  the  live  load  covers  the  whole  truss,  and  therefore  the  same 
diagram  gives  the  maximum  chord  stresses  due  to  the  live  load  or 


i 


m 


'30 


THEORY  OF  STRUCTURES. 


the  maximum  chord  stresses  due  to  the  combined  dead  and  live  loads 
by  multiplying  the  corresponding  stresses  obtained  from  the  dead- 

111-  L       ^  /.  +  /> 

load  diagram  by  j)  "f  ~n~'  respectively. 

Again,  assuming  that  a  single  vertical  reaction  XA,  at  the  left 
8upix)rt,  is  the  only  force  acting  upon  the  truss,  Fig.  658  is  the 
stress  diagram  from  which  the  maximum  stresses  in  the  diagonals 
can  at  once  be  found.    Thus 

length  of  ef 


actual  strenK  «■/  =  actual  corresjwnding  end  reaction  X 


length  of  XA' 


Ex.  1 2.  Given  a  through  right-panel  truss  oj  SO  //.  span  and  10  ft.  rite,  with 
a  pnrnholir  rumpresxiim  chord,  the  fmnel  live  and  drail  loads  firing  10  and  5  torn 
rtsiKiti ivly  and  the  hraring  isosceles  as  in  Fig.  fi.'W};  find  the  maximum  strestet 
in  the  3d  and  4th  diagonabt. 

L-lOtons;    A'=8;    n-2;    /-SOft.;    ifc-lOft. 

dt  is  a  maximum  tension  when  x-=2."j  ft.     Thru,  p^ig.  G-W, 

AE-4i{l.    and    RE -H4i  U.; 

Un  0,-11;    fl,-'iO°38';    coaec  fl, - 1 .2935. 

Therefore 

.,  ,       .     10  fix7   4iXl.29.3.5     „  ,^^ 

d,{max.  tens.)--  — —      '-.  ^  -6.24  tons 

Jo  4j  +.JU 


and 


.,                         10  2X1   WiXLLnW."!     .^„ 
dttmax.  comp.)  -—   — — — -5.58  tuns. 


dt  is  a  maximum  tenstion  when  x  -15  ft.    Then 

.4fc''-»,V  ft.,     «£'-4k«ft.; 
tane.'-U;     fl,'-.VJ=4S';     ro«e<- fl/-1.157. 


Therefore 


d»'(max.  tens.)  -  "t^   - 


10  2X3  Jf.'»X1.157 


',V  +  20 


■10.72  tons 


and 


.  ,,  10  0X5  »VX  1.157 

d»'(mox.  fomp.)  -  -    -  ;  -11.07  tons. 

—        "  ft   "^  -O 


TUHEE-HINGED  BRIDGE. 
For  the  dead  load  stresses 


731 


T, 


5  cos  59°  4«' 


and     r.-'V^™'-'^^' 


Therefore 


sin  110°  26'        "■     ■■     sin  110°  26 
r,  -2.684  tons       and      T,  -3.38  tons. 


xo.  Three-hinged  Bridge.— Fig.  659  represents  two  bridge-trusses 
hinged  to  the  abutments  at  A  and  B,  and  also  hinged  at  an  inter- 
mediate point  C.  An  objection  to  this  type  of  truss  is  the  large 
.Lfiection  due  to  the  hinge  at  C,  but  on  the  other  hand  it  has  the 
a.lvantage  of  eliminating  tem|)erature  stresses.  If  the  trusses  are 
inverted,  the  bridge  is  then  of  the  suspension  type.  The  stresses 
rciiuin  the  same  in  magnitude,  but  are  reversed  in  kind. 


/-:: 


K-t-H'uftQ 


- — / ^ _^ 


«<M 


Fro.  659. 


Any  weight  P  concentrated  as  in  Fig.  659  produces  a  reaction 
at  .4  which,  to  prevent  rotation  at  the  hinges  ,4  and  O,  must  pass 
along  AO,  intersecting  the  direction  of  P  in  M.  Then,  for 
iquilibrium,  the  reaction  at  B  must  pass  through  M.     Hence— 

Firxl.  Kvery  load  on  the  horizontal  member  EF  induces  a  stress 
III  the  sloping  member  AO,  but  has  no  effect  upon  the  horizontal 
member  CD  or  uixm  the  bracing  connecting  CD  with  AO. 

So  every  load  on  the  horizontal  member  CD  affects  only  the 
-ii»ping  member  BO. 

Second.  If  P  is  the  resultant  of  a  number  of  loaiU  on  EF,  then 
HM  is  the  direction  of  the  resultant  reaction  at  B,  and  the  loa<ls 
nil  the  left  of  A'  tend  to  pmiuce  rotation  from  left  to  right,  while 
ilKK-^e  on  the  right  of  K  tend  to  produced  rotation  from  right  to  left, 
K  U'ing  evidently  the  point  of  no  moment. 


■?l 


^'# 


73:i  THEORY  OF  STRUCTURES. 

For  equilibrium  of  left  portion,  taking  moments  about  A, 

Va-Hb^O. 
For  equilibrium  of  right  portion,  taking  moments  about  B, 
Vd  +  He-P/^0. 
Also,  r,-r=0    and     Ta  +  l'^P. 

These  four  equations  give  //,  V,  Ti,  and  Tj. 

\ote.—H  and  V  arc  horizontal  and  vertical  components  of  reac- 
tion at  0;  H  and  I'l  are  horizontal  and  vertical  components  of  reac- 
tion at  i4;  //  and  V^  are  horizontal  and  vertical  components  of 
H'action  at  B. 

V.\.  13.  To  find  point  in  CD  at  which  a  weiyht  Q  tniul  be  concentrated  to 
makr  the  B.M.  at  j  from  C  fi/ual  to  nil. 

The  resultant  reartion  at  .1  must  necessarily  pass  through  the  point  in 
fiuiwtion,  and  its  directidii  inters«'cts  the  nwtion  ulong  HO,  arising  from  Q, 
ill  a  [toint  N.    Then  Q  must  lie  vertically  below  \.     Also, 


f-r—y    and     -.-y — («+c)— ' 


V  +  c) 


an  equation  fciving  y. 

ICx.  14.  Thrrc-hingeii  hritigr-trugtrs  with  horizmtal  top  chords,  straight  slop- 
iny  lower  chords,  «/  the  dinunsions  and  loadeti  as  in  Fig.  6fiO. 

tan(/-=4;    cota-f;    cot/3-|;    cotr-2;    cot9-6. 

fkaiNoad  Stresses.~i>\nce  the  deatl  load  is  syminetrically  distributed,  the 


V-ll 


Pta.  mo. 


vertical    n-aetion  V  at  each  side  8U[>port  is  8  tons,  and  there  is  no  vertical 
■hear  iit  the  ccntrul  hiiiKc. 


TUREE-HISGED  BRIDGE 


733 


H^S  Iniw 


Calling  //  the  liorizontul  reaction  at  the  hinges  and  taking  moments  about 
a  side  hinge, 

-//xiai+SxJO-O,    or    ^-12  tons. 

Fig.  661,  the  stress  diagram  for  the  dead-load  stresses,  ran  now  be  drawn. 
The  live  load  is  three  times  the  dea<l  load,  and  henee,  when  the  live  load  covers 
the  whol(>  bridge,  the  corresponding 
stresfd's  in  the  several  members  arc 
t  liree  times  the  stresses  due  to  the  dead 
load. 

Method  of  Sections. — For  the  cKords 
and  dingimals  consider,  successively, 
the  e(|uilibrium  of  portions  of  the  truss 
ljetw(H>n  the  central  hinge  and  ver- 
tical sections  in  the  Ist,  2d,  3d,  and 
4th  (Mtnels. 

For  the  vrrtiralx  consider,  SU(  a-s- 
sively,  the  equilibrium  of  portions  of 
tlie  tru.s8  between  the  central    hinge 


Fio.  G61. 


and  sloping  sections  between  the  1st  and  2d,  2d  and  .3<l,.ld  and  4th  diagonals 
For  horizontal  chord  panel  stresses  take  moments  about  the  foot  of  a  vertical. 
"    sloping  "         "  "  "  "  "       "    top     "  'i 

"  stresses  in  diagonals  and  verticals  take  moments  about  the  point  S,  the 
|H)int  «)f  intersection  of  the  two  chords  produce*!.  If  x  is  the  horiiontal  dis- 
tance of  the  point  from  the  central  hinge, 

£  +  40 

T 


— -j-9    and 


■5  ft. 


Then,  denoting  tensions  by  T  ar  "  compressions  by  C.  the  stresses  in  tons 
are  given  by 


«/.  Hi-//  10-2(10+20)-!  X30        -  .30  and 

Ai.Y  1.5  co>»0 -A/  1§ +3  2-20  +  1x40     -180  " 

A7.   l.-.sin  .. --//IJ+a^-iS  +  lxS    -13.i  " 

r.VS§=//Oi| +2X10 -1X20  -  40  " 

.Y.»/  llJcosO-//  lJ+2-2  l.-i  +  l  X.30   -110  " 

MX  X>  sin  ,9-  -//IJ  +2-2-20  +  1  X.i  -  ft",  " 

/>/'.-.-//  3J -IX 10  -  30  " 

.YO  Si  •  cos  0  -     ffl| +2X10  +  1X20-  (50  " 

O/' 2.1    sinr  --//IJ +2x15  +  1  X.5  -   ir>  " 
f/Mi-O 

.YV  .-)cosfl-//lJ+lxlO  -  30  " 

V/M.'isina- -HU+lx.'i  --1.')  " 

.UA3.')--//1§ +3-2  25  +  1x5  -135  " 

O.V.25--//  lj+2  2  20  +  1x5  -  65  " 

«.»/'  15--//  1§+2X15  +  1X5  -15  " 

Z^/-' 5-1x5  -     5  " 


BL-2HC); 

XK  - 12  sec  0  -4  vTofC) ; 

A7.  -3  cosec  a  -V'85{r); 

tW-4|(r); 

XM  -  V  sec  0  -  V  vlo(C)  ; 
.WiV  -  Vcosec^-ilv/eKr); 

Di'-mc): 

Y0-Vsec(»-V>/l0(O; 
Ol'-i  cosec  r-}V5(T); 
A«=0;  _ 

A'y-(>8ecO-2v'lO(r'); 
QH-  -coseca-  -\/37(C); 
Ml.  '3?(0 ; 
OiV-2|{C); 

RF'HC). 


MICROCOPY   RESOLUTION   TEST   CHART 

lANSI  and  ISO  TEST  CHART  No    2i 


1.0 


I.I 


|"iM  ilia 

i  ;f:  iiiii^ 


1-25  IIIIII.4 


1.8 


1.6 


A     APPLIED  IIVMGE 


•-  '    p,is'    M.J-       "ee- 
'  "^;   ♦§?  -  0300  -  PhDn» 


734 


THEORY  OF  STRUCTURES. 


U.  Live  Lnaih.— Every  load  on  the  ri<;ht  truss  induces  a  stress  along 
ihc  ^lopiiijr  chord  of  tho  left  truss  and  ilu  ivlorc  affects  the  members  of  this 
diord  only.  For  all  other  niemlxTs  oi  the  left  truss  the  stresses  remain  the 
same  whether  the  right  tru.---  is  loailed  or  unloaded. 

The  panel  live  load  is  ■•>  tons,  or  three  times  the  panel  dead  load.  The 
values  of  \  and  H  for  any  distribution  if  the  load  may  be  easily  found  by 
the  nu'thod  already  described. 

.Maximum  Stres.sex  in  Sloping  Clioril.,.—The  moments  of  the  stresses  in 
XK,  XM,  XO,  and  XQ,  with  reference  to  the  points  1,  2,  3,  and  4  respect- 


"l 

3 

/  1 

3       -.in.      i 

a< 

_^^4<i^ 

v\ 

v\ 

lio   662. 

ively,  are  severally  equal  in  magnitude  to  the  moments  w-ith  reference  to 
the  same  jwints  of  the  resultant  reactions  through  a.  If  a  reaction  passes 
through  a  panel-iK)int  in  the  upfH-r  chord,  its  moment  with  respect  to  this 
j>onit  is  >((7,and  therefore  the  resulting  stress  induced  in  the  corresponding 
nienilK'r  of  the  sloping  chord  is  also  /((/. 

Let  ch  produced  meet  n\,  „2,  «.l,  n4  produced  in  the  points  n„  «„  „,  n, 
The  reaetKm  at  c,  due  to  a  load  vertically  Ik'Iow  «,  evidently  pas.ses  through 
2  and  has  no  moment  about  2,  so  that  the  corresponding  "stress  in  .Y.V  is 
niV.  Similarly  loads  at  1,  r,,  jin,!  r„  resix-ctivelv,  produce  no  corresponding 
stress«>s  in  A' A',  A'O,  and  XQ.  Hence 
all  loads  on  the  ri,M  of  I    tend  to  produce  rot.-ition  from  left  to  right  and 

induce  a ni/tn.y.iioii  in  A' A'; 
.ill  loacls  on  the  riuht  of  r,  tend  to  produce  rotation   from  left  to  right  and 

induce  row/jrc.v.v/i)//  in  AM/  ; 
iill  lomls  on  the  »•/,//,/  of  ,,  ten.l  to  produce  rotation  from  left  to  right  and 

induce  rom/irc^sinn  in  A'O ; 
all  loads  on  the  ri,,l,l  of  r,  tend  to  produci'  rotation  from  h-ft  to  right  and 

induce  rt»n pension  in  XQ; 
.ill  loa.ls   on   the   /.//  of  1   tend   to   produce   rotation   from  left  to  right  and 

induce  h iisiun  in  A' A'; 

all  loads   .>n   the   hjl  of  r,  lend  to   produce  rotation  from  left  to  right  and 

iiKliice  /(//.v/V(N  in  X.M : 
all  loads    on  the  l,J(  of  ,v  tend   to  i>r.Kluce  rotation  fnm.  left  to  right  and 

induce  tntsion  in  A7>; 
all  loa.ls  on  the  1,/t  of  r,  tend  to  pnKJuce  rotation  from  left  to  right  and 

induce  <«7(,v(om  in  XQ. 


KK1 


THREE-HINGED  BRIDGE. 
Table  of  Maximum  Stresses  in  Slopino  Chord. 


735 


Panel-points 

Member. 

Loaded  fi>r 
Max.  Stress. 

Maximum  Stresses. 

XK 

1  to  10 

V-O;      //=.36; 
A^Ai.l5ros_9-/y.U-(-4.tJ.20 

and 

JfA.'  =  12V^10  tons'- max.  comp. 

XK 

0 

V  =  0,     Il-O; 
no  tension  in  XK 

X.\t 

3  to  10 

y  =  i      //-33»: 
Jr.W.mcos9  =  W.lJ+l'.30  +  2.6.15  +  3X30 

and 

Xff  -  W^IO  tons  =  max.  comp. 

XM 

1  to  2 

V-~i.     //-2l; 
JT-UIU  cos  » >/.  U-  K.30 

and 

JT.V-JjVlO  tons -max.  teru. 

XO 

4  to  10 

V-21,     //  =  29t; 
jrO.Si  cos  #-//.lJ+r. 20  +  6X10  +  3X20 

and 

XO-  IS'  "^10  tons -max.  romp. 

XO 

1  to  3 

V--2t;     «-6f; 
XO.Sieond H.H-V.io 

and 

ATO-fjVlO  tons- max.  frnt. 

XQ 

Sto  10 

V-44;     //-22i: 
XQ. 5  cos  9^- /A  If +r.  10  +  3X10 

and 

XC- V^IO  tons- max.  eomp. 

XQ 

1  to  4 

V-.-*y.     //-LSii 
XQUfonf l/H-V  10 

and 

XQ-i^lO  tons -max.  Uiu. 

Hence 

total  maximum  stress  XK-4V10+  12Vio-  IGN/fotons; 
"      -Y-lZ-VvIo+Vi'v^iO-Vr'^iO    " 

"       "    XO  -yVio+w^io-W^/io   " 
"    XQ  -  2vTo+  V>/io-VVi6~  " 

These  stresses  are  all  rompressions  and  there  are  never  any  tensile  stresses 
in  any  member  of  the  sloping  chords. 

Maximum  Liir4oa(l  Stresses  in  Members  of  Horizontal  Chord.— The  loads 
;if  2,  3,  4  combininl,  severally,  with  the  forrcs|M)iuiing  reactions  along  cb, 
tend  to  caii.sc  a  rotation  from  right  to  left  with  rc8j)ect  to  the  points  p,  q,  r, 
rtsiH-ctivdy,  and  this  tendency  must  l)e  equiUbrated  by  a  tendency  of  the 
stresses  in  the  members  12,  23,  34  to  cause  rotation  from  left  to  right,  with 
rcsjH'ct  to  the  same  points.  The  8tros.ses  in  these  members  are  therefore 
Mhv.iys  compressive  and  are  great  est  when  the  live  load  covers  the  left  truss. 

These  stresses,  again,  are  unaffected,  ns  already  shown,  by  loads  on  the 
richt  truss  which  produce  reactions  along  ah  onlv.  It  may  therefore  lie 
assumed  that  the  truss  is  loaded  or  unloaded.  .Assuming  that  the  live  load 
covers  the  whole  bridge,  the  ma.ximum  compres.sive  stresses  in  12,  23,  34  are 
three  times  the  corresponding  stresses  due  to  the  deml  load.     Thus 

max.  RL'.iX2i  -  7|  tons; 
"     r,V -3x41-141    " 
"     /)f-3X6  -18     " 
"     EH  -0 


i)  < 

i 

:! 

jlfi 

'1' 

'1 

'■ 

^1 

'■ 

« 

j  ■ 

1 

i .  1 

- 1 

11 

■;    I 

■         !      ! 

'hi 

I 


41 


736 


THEORY  OF  STRUCTURES. 


As  a  verification,  find  CN  on  the  assumption  that  the  right  truss  is  un- 
loaded.   Then 

r-40+//13J=4f)-20=480,    or    3V  +  H=3(i; 
r-40-W13J=0,  or    3r-Af=0. 

i'liorefore  V'^Gtons    and    //  =  lStons. 

llence  CJV-SJ  =  r-'iO+Z/Gj -()10-3-20  =  120, 

and  C\  =  \',''=UAtona. 

Maximum  f.ivr-lond  SIressc.t  in  Diayoiialx.— Assume  right  truss  unloaded, 
as  the  loads  on  this  truss  produce  reactions  along  ab  only  and  do  not  affect 
the  diagonal  stresses. 

Let  the  horizontal  and  slo])ing  chords  be  produced  to  meet  in  the  (wint 
S,  Fig.  (tOO,  which  is  .">  ft.  from  the  panel-point  5.  Then  the  stresses  in  the 
several  diagonals  arc  obtained  by  taking  moments  about  S.  These  stresses 
are  tabulated  as  follows: 

Tahle  of  Maximum  Live-load  Sthesse.s  in  Diagonals. 


Member. 

/cZ 

KL 
MN 

M.\ 

OP 

OF 

QR 

QR 


Panel-points  , 
Loaded  ft>r  ' 
Max.  StreM.  ; 


Maximum  StresKS. 


2  to  5 

1 

3  to  5 

2 

4  to  5 

2  to  3 

5 
2  to  4 


r-6;     //-18; 
A7..4.')«iL.i-=3.6.2J  +  3X5-//.lf-V.5 
and      KL  •«  fv's.'i  ton.*  -  max.  ten: 

no  compre?*^ioii  in  KL. 

.W.V  3.")  ^\nj  "  i  t.  21)  +  3  X  .j  -  // .  IJ  -  K.  5 
and       .WA'^IA^til  loll^=  max   /en*. 
V-1.     II     2\. 
.W.V  35  sin  J.  .  //    U-K  I'.i 
and       .*/.'»■  =  AV(il  iiin»«m!ix.  romp. 
V-3},     //      llj, 
or  2.isinj;  =  ti   l.")  +  3  -S-H.ll-V.S 
anil      O/'^'fi'iV.')  tons-^nmx   Irnt. 
1-2J.     //-lif- 
OP  2h  ciiij  -//!§  + I    .5 

and       OP  -  la^^  ton*  -  mnx.  romp. 
I'     U       //     4i- 
««  l.'.  Kin  ,1-3  .S-//.H-V'.5 
and      y«  =  0. 

V-4J:     //-13J. 
(,/ft  1.1 -lin^t'  //   1J+1'.5 
and      yR'-3'^'^37  ton»  ■  max   romp. 


Hence  the  total  maximum  .strc.s.sc.s  in  tons  are 

MN-H^'in  +yv'(Tr=  Vc'^-'i^.  "  tcn.sion; 
OP-  |Jv  .T  +  fv'5  -JiV.T,  u  tension; 


THREE-niSGED  BRIDGE.  737 

and  OP=^y^  -  'vT  =tV  ^5]  a  compression; 
(3«=3  v/37-(  -n/37)  =4>/37,  a  compression. 

Maximum  Live-load  Stresses  in  Verticals.— It  is  assumed  that  there  are 
no  loads  on  the  right  truss,  as  such  loads  produce  reactions  along  ab  only  and 
do  not  therefore  affect  the  stresses  in  the  verticals.  Thus,  from  the  rieht 
truss,  I  ••«)-//•  13J=0,  or // =31'. 

The  sloping  planes  (Method  of  Sections,  p.  733)  divide  the  loads  on  the  left 
truss  into  two  groups.  The  group  on  the  right  combined  with  the  correspond- 
inj:  reaction  along  ch  tends  to  produce,  with  respect  to  the  point  S,  a  rotation 
ro,n  right  to  left  which  must  be  equilibrate.!  by  an  equal  tendency  to  rotation 
Jroin  left  to  right  due  to  a  compres.sive  stress  in  the  vertical  cut  by  the  sloping 
plane.  So  the  stress  in  this  vertical  developed  by  the  group  on  the  left  must 
be  a  tension. 


Table  of  Maximum  Vertical  Stresses. 


Maximum  Stresaea. 


AK 


NO 
SO 


NO 
NO 


V-6;     //-18; 
45-4.8.25-//. IJ 
— 12  tons  -  max.  comp. 
0 
V-6;     //-18; 
35-3  6.25  +  3.5-//  1J-K.5 
-!!♦  tona  — max.  comp. 
0 

„=    „*'-5*:    //-iSf: 

25-2  6  20  +  3  5-//.  I}- K.6 
— 8  A  «on«  -  max  comp. 
V-J;     //-2tr 
.25-//  1J+K.5 

—  A  tons  =  max  ltn». 
V-3i.     //-IIJ; 

15-6  15+.I5-//  IJ_r.6 
-4i  tona -max  eomp. 

I5-//.U+V5 

—  li  tona  —  max.  ten*. 
V-n.     //-4»j 

5-3  5  *' 

-3  tuna  — max   comp. 

,5»0     ^'"'**'     """*  = 
-0. 


Hence  the  total  maximum  stresses  are 

AK-A  +12    =16  tons,  a  compression  ; 
A.l/-3f +1H-1.5?    "  " 

A^0-2|+  S,'„=IOT'n  " 

PQ~\  +  41-  :,i  "  «' 

andPy-1-lj--     J    •<     a  tension; 
RF-\   +3-4      " 

Rrap/itca/  J/WAod.-It  is  much  better,  however,  to  use  the  eraphical  method 
1"  (he  determination  of  the  stresses  in  trusses  of  this  type.     The  following 


1 

r 

I 

1;^ 

w  1 

1 

t 
! 

1 

} 

w. 

738 


THEORY  OF  STRUCTURES. 


Fio.  663. 


Fio.  664. 


Fig.  665. 


oeFR 


Fig.  666. 


NP      V       COEFR 

Fio.  667 


Fig.  668. 


Fig.  660. 


TIIREE-HINGED  BRIDGE.  739 

lionlonoad:'""'  ^"^  '''  ^'""^  correspon.fing  to  the  different  distribu- 


Member. 


Fig.  664. 


3(  at  5. 


Fig.  66S. 


«at4. 


XK 
XM 
XO 

XQ 
BL 
CN 
DP 
KR 
KL 
.U.V 
OP 

QK 
AK 
I..\f 
XO 

RF 


v-u, 

H~4i. 


-jVio 
-}vTb 


V~2i, 
«-6f. 


Fig.  666. 


6<at3. 


0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
0 
-3 


--V 

0 

AVS5 

0 


//-4i. 


-Av^lO 
4vTo 

-2t 

-I' 

0 

-;v'37 
-3 
-3» 
-51 

+  1 
0 


Fig.  667.      Fig.  660. 
6t  at2, 


H'h. 


ivTo 

-3» 

-3{ 

-3 

0 

-AvT 

-i>/37 


3<  at  1. 


1=0. 
//-O. 


H 


0 
0 
0 
0 
0 
0 
0 
0 
0 
-3 
0 
0 
0 
0 


Fig.  668. 

Loads 
6  to  10. 


Fig.  669. 


V'  =  6. 
//  =  18. 


-6V10 
-ev'io 

-evTo 
-eVio 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 

0 


Loads 
1  to  10. 


V  = 


24, 
36. 


-12vTo 
-  V>/lO 

-V^ia 

-7f 
-141 
-18 
0 

»v^ 
JlVei 
IvT 

-3^37- 
-12 
-11» 
-7t 
-3 
-3 


II.  ITiree-hinged  Bridge  (arched  or  suspension)  with  the  panel- 
points  of  the  curved  chord  in  the  arc  of  a  parabola  having  its 
vertex  at  the  central  hinge. 


Fio.  670. 

I^t  the  whole  bridge  be  covered  with  a  uniformly  distributed 
l..a,l  of  intensity  t..  There  is  no  shear  at  the  central  hinge  but 
merely  a  horizontal  reaction  H.  ' 

Let  h  be  the  stress  at  any  point  P  in  the  horizontal  chord  ver- 
tically above  the  point  Q  in  the  bow,  whose  vertical  and  horizontal 
c  .stances  from  the  central  hinge  are  x  and  y  respectively,  a  beinjr 
the  central  depth  of  the  bridge.  * 

Take  moments  about  Q.    Then 

A(a+x)-//x-^. 


m 


( 


. ) 


ill 


\- 

i  ,■■ 

'■■ 

I-. 
: 

ii  ' 
ti 
ij    1 

! 

I' 

I ! 

i!; 


iiiij 


740 

But 

Therefore 


THEORY  OF  STRUCTURES. 


f 


--  rx    and 


Hk  = 


wP      w  P 


If  the  trusses  are  inverted,  the  bridge  is  one  of  the  ordinary 
suspension  type. 

Thus  under  a  uniformly  didributed  load  h  is  nil  and  there  is  rw 
stress  in  the  horizontal  chord. 

Ex.  14.  .1  three-hinged  eight-panel  tru.is  of  80  //.  upan,  15  ft.  deep  at  the 
supports,  1§  ft.  deep  tit  the  centre,  and  with  the  lower-chord  panel-points  lying  in  a 
parabola  having  iti  vertex  at  the  central  hinge.  The  panel  dead  and  live  loads 
are  2  and  6  tons  rcspeetivehj,  Fig.  671. 

Then  56  =  lift.,    4r=5ft.,    37=2ift.,    2p=9ift.,    and    lo-15ft. 

Also,  cota=^%,    cot,5  =  ,»F,    cotr-1,    cot5=xV. 

Stress  diagram  for  a  uniformhj  distributed  loati. 


H-u 


Fio.  671. 

Fig.  672  is  the  stress  diagram  for  a  uniformly  distributed  panel  dead 
load  of  2  tons.  The  stresses  in  similar  menil)ers  for  a  panel  live  load  of 
6  tons  are,  of  cours*-,  three  times  the  corresponding  stresses  in  the  diagram,  if 
the  live  load  is  uniforiniy  distributed. 


The  diagram  also  shows  that  for  uniformly  distributed  loads  the  stresses 
in  the  members  of  the  horizontal  chord  and  in  the  diagonals  are  always  nil. 


>i 


THREE-IIINQED  BRIDGE. 


741 


Method  of  Sections— Maximum  live-had  strcxi^rs  in  horizontal  chord  Loads 
at  the  points  m„  m„  m,  which  are  vertically  below  the  points  n„  n„  «,  in  which 
the  members  ap,  pq,  qr,  produced,  intersect  the  line  of  reaction  cb,  develope 
no  stress  m  the  members  BL,  CN,  DP,  respectively.    Thus 


^^   ~-<^ 


10 

— 1 


Fio.  673. 


For  the  left  truss  every  load  on  the  left  of  m„  combined  with  the  corre- 
sponding reaction  along  cb,  tends  to  prwluce  a  rotation  from  right  to  left  with 
r('s,*ct  to  the  point  p,  and  must  be  equilibrated  by  an  equal  tendency  to 
rotation  from  left  to  right,  with  respect  to  the  same  point  p,  due  to  a  com- 
iwssion  m  BL,  and  this  compression  is  greatest  when  the  live  load  is  con- 
.ct.trated  on  the  panel-p,jints  1,  2,  3.  So,  everv  load  from  m,  to  5  com- 
bined with  the  corresponding  reaction  along  cb,  tends  to  produce  a  rotation 
(mm  left  to  right  with  respect  to  the  ix)int  p,  and  developes  a  tension  in  BL 

tor  the  right  truss  every  load  from  6  to  10  causes  a  reaction  from  6  to  o 
and  tends  to  produce  a  rotation  from  left  to  right  with  respect  to  the  point 
//.  A  tension  is  therefore  developed  in  BL  which  is  greatest  when  everv 
panol-point  from  4  to  10  is  loade<l. 

Again,  the  compression  in  C'A'  is  greatest  when  the  live  load  is  concen- 
trated on  panel-points  from  1  to  4. 


Table  o,-  Stresses  Developed   by  a   Uniiyirmly  Pistribi'ted  Load. 


Panel-puintii 

■ — . 

Mfmb«r. 

Loaded  for 
Mux.  Stresii. 

Maximum  Streinen. 

HL 

1  lo3 

V-2J:     //-G#: 
fl/..9i-V  30  +  //  7i-0X10 

HL 

4  tu  10 

and 

«/. -Otl  'onB-max.  comn. 
V-2i:     //-29f; 

(•\ 

1  to  4 

and 

«/.  fli-V'.IO-//  7i  +  n'30  +  3X30 
BL-{>\%  tons -max.  Irnt. 
r-4}:     //-i;U; 

r.y  5»r.20+//  3i-oxio 

(  S 

5  to  10 

and 

f'.V-=  1.5  tons -max.  citmn. 
V'il,.     //-22I; 
fA;  .'.-V.20-//  3i  +  3x20 

DP 

1  to  4 

and 

(  .\  -  u>  tons -max.  Irnt. 

/>/'2»=i;-„iV//"r'='»= 

f,P 

3  lu  10 

and 

l>P-'-2-i\  i<iri-<-max.  mmp. 
1  •.  4J;     //>.224- 
nP  21,^  r  ID-//  J+3X10 

1 

and 

/'/'-2a J  tons  "max.  Irnt. 

If     i.       > 


■*?«! 


■t  s  J  ! 


Hu 

t 

IH 

.  ,  . 

i  I^^S 

^^^^n 

\ 

H^H 

r      *,■ 

Hk 

jl: 

^^Hh 

1 

^^^Hv 

742  THEORY  OF  STRUCTURES 

Maximum  live-load  strrnnfs  in  curved  chord. 

cot  a  =^,  cosec a  =  f'jViaS; 

cot  ,9  -  Ti ,  cospc  ,?  =  1 1 ; 

cot  r=i>  ooscc  r="i^'7; 

cot  3=j\  cosec  3=i^\^li5. 


Fig.  674. 

For  the  right  truss  every  load  from  6  to  10  causes  a  rpaction  from  6  to  a 
-which  tends  to  i)r(Hluce  a  lejt'handcd  rotation  with  respect  to  the  points  1, 
2,  3,  4,  and  ').  This  must  be  equilibrated  by  a  right-handed  rotation  due  to 
comprtssive  forces  develoiied  in  the  menilxTs  A'A',  XM,  XO,  and  XQ. 

Apain,  v,,  n.^,  n,,  n*  are  the  points  in  which  cb  produced  is  intersected  by 
«1 ,  a2,  a3,  «4  produced,  and  loads  at  1 ,  w„  m,,  m„  vertically  below  the  points, 
produce  no  stress  in  the  memljers  ,YA',  XM,  XO,  and  XQ,  respectively.  Hence 
for  the  left  truss  loads  on  the  right  of  1  develope  compressive  stresses  in  A'A', 
and  the  stres.ses  in  XM,  XO,  XQ  are  compressive  or  tensile  according  as  the 
loads  are  on  the  rif^ht  or  left  of  Ws,  Wj,  in„  resixjctively. 

(KSSKS     IN      MeMBEUS      of     t'VBVED     CHORD. 


Taulk    of 

Maximi  M 

I.IVh 

-LOAD 

~" 

MemlMT. 

I'liiiel-iMiitits 
l..iu<leil  for 
Mux.  Strei»». 

_^ — 







XK 

2  to  10 

XK 
XM 

0 
3  to  10 

and 

XM 

iu,2 

aiid 

xn 

4  to  10 

aiid 

XO 

i 

!       11.1 3 

and 

XQ 

^  to  10 

and 

XiJ 

i  t„4 

j 

and 
an.l 

Maximum  Stresses. 


V-0;     II     ■M, 
XK .  \ry  siiwr- 4 . B. 20  +  // ,  1 J 
X K  =•  3'^^ I'J-i  tcMi.f  'max.  coinp. 
0 

r-j.    //-:):)}; 
,Y.U  nt  sill  ?    2  (1  i.->  +  n  X  .-JO  +  // .  U  + 1 .  30 
.^.l/  =  -4Ug  liins  —  max.  comi>. 

I      };     II -2i. 
.YU  0*  sin,?  '  V  .30~//.lJ 
X.M  '-  213  tons  =  niiix.  lint. 
V     2t;     //-29J; 
XO  r,  Mil  r  J'  •  10  +  3X20  +  //. H  +  V. 20 

XO  -=  "iV^  '  i'  '"n»  -  niax.  tomp. 

y  2J;  II  -m. 
AY)  .I  sin  r_  T  20-//.1J 
XO  '  fJN'^iy  tuns  "  max.  /rn«. 

l„4»;     //-22*: 
.Y(^.2J  sin_i_;:ixl0  +  //   U+  »M0 
XQ  --  '/^'l '•"'  Ions     max.  comp. 

V   u;  ii-m. 

.Y(^2i  sin  I     y. 10-11   Ij 
jrQ-iv'i45  tons-max.  tn: 


mm 


THREE-HIXGED  BRIDGE 


743 


Maximum  live-load  stresses  in  diagonals.     Consider  the  equilibrium  of  a 
portion  of  the  left  truss  between  vertical  sections  y,y,  and  yy  in  the  first  panel 


Fig.  675. 

and  immediately  on  the  right  of  the  central  hinge.  The  diagonal  stress  KF^  is 
found  by  taking  moments  about  the  point  m,  in  which  ap  produced  intersects 
tilt'  horizontal  chord. 

Every  load  on  the  left  of  ?«,  combined  with  the  corresponding  reaction 
nlong  cb  gives  a  resultant  which  lies  above  the  point  to,  and  which  therefore 
tciuis  to  produce  a  rotation  from  left  to  right  with  respect  to  m,.  This  must 
be  equilibrated  by  an  equal  tendency  to  rotation  from  right  to  left,  with  respect 
to  the  same  point,  and  must  therefore  be  produced  by  a  tension  in  KL. 

Every  load  from  w,  to  5,  combined  with  the  corresponding  reaction  along 
el),  gives  a  resultant  which  lies  below  w,  and  therefore  tends  to  produce,  with 
r('s])oct  to  this  point,  a  rotation  from  right  to  left,  developing  a  compression 
in  KL. 

A1.SO,  every  load  on  the  right  truss  causes  a  reaction  from  6  to  a  which 
lies  below  the  point  ?«,,  and  the  resulting  stress  in  KL  is  therefore  a  com- 
(ircssion. 

Again,  let  pq,  qr,  rb  be  produced  to  meet  the  horizontal  chord  in  the  points 
"',.  .1,  and  m„  and  let  x,,  x„  x^,  and  x,  lie  the  distances  of  m„  to,,  5,  and  to,, 
r(s|jectively,  from  1.     Then 


X,         15 


J, -10      »i'     X.-20     o'    X3-30     2i' 


Mini 


x,  =25?ft.,       xi=32{t.,        a:,=40ft.,        a-3  =  G0ft. 


Consider  the  cquihbrium  of  portions  of  the  truss  between  yy  and  vertical 
>'(li()ns  y,y.„  y^„  y,y,  in  the  2d,  3d,  and  4th  panels,  respectively. 

I'A-cry  load  on  the  right  truss  produces  a  reaction  along  ba  which  falls 
'■'hm-  the  points  toj  and  5,  and  above  the  jxHut  m,,  so  that  the  tendency  to 
relation  with  resjMjct  to  the  points  w,  and  5  is  from  left  to  right,  while  with 
I  s|H>ct  to  m,  it  is  from  right  to  left.  Thus  the  corresponding  stresses  de- 
vcio|)ed  in  MN  and  QP  mast  be  eompressions,  while  the  stress  in  GH  must  be 
a  liiisi/m. 

Again,  every  load  on  the  left  of  the  jjoints  m.^,  5,  tn,,  combined  respectively 
with  the  corresponding  reactions  along  cb,  give  resultants  which  fall  above 


11, 


744 


THEORY  OF  STRUCTURES. 


the  points  ma,  5,  and  mj,  and  therefore  the  tendency  to  rotation  with  respect 
to  ihf  "  points  is  from  right  to  left  and  the  eorresponding  stresses  developed 
in  MN.  OF,  and  Qli  are  compressions. 

80,  also,  the  stresses  develojK'd  in  the  diagonals  MN ,  OP,  and  QR,  re- 
six-ctively,  by  loads  Ijetween  m.^  and  ii  are  tensile. 

Finally,  when  th  •  whole  bridge  is  loaded  equably,  the  diagonal  stresses 
are  nil.  Hence,  i:.  any  diagonal,  the  maximum  live-load  compression  must 
be  equal  in  magniluf.e  to  the  maximum  tensile  load. 

Graphical  Method. — The  stresses  in  the  several  members  may  be  obtained 
more  easily  by  drawing  the  stress  diagram  for  each  distribution  of  the  loud 
and  then  superposing  the  results  as  follows: 


Fig.  677. 

Fig.  f.78. 

Fig.  679. 

Fig.  080. 

Fig.  681. 

Fig.  0. 

Fig.  682. 

Member. 

3/  at  5. 

(Uat  4. 

6<  at  3. 

6/  at  2. 

Loads 
6  to  10. 

3(  at  1. 

Loads 
1  to  10. 

V-U, 

//  =  6}.' 

//-4J. 

V=J. 

V  =  6. 

K-=0. 

V-18. 

W-2i. 

//-18. 

H-0. 

H-36. 

XK 

-J>/i93 

-  A  v'Iro 

-fv/l93 

-AV193 

-jVl93 

0 

-3V193 

X.\f 

-BA 

-71J 

-2H 

2lil 

-24,«, 

0 

-39 

XO 

-V-vT? 

-?4^17 

llv/17 

AVi7 

-V^17 

0 

-9-*/l7 

M 

f>/l45 

-2,', 

-ttr 

-3V145 
4}? 

0 
0 

-3V145 
0 

CN 

3 

-It 

-9 

-4 

12 

0 

0 

DP 

4i 

-lli 

-7i 

-3 

18 

0 

0 

ER 

0 

0 

0 

) 

0 

0 

0 

KL 

- .',  V2b5 

-  ,».V2B5 

iW265 

ft^v'265 

~  A  V265 

0 

0 

MS 

-if^5 

H>/5 

W^s 

A"^5 

-  H'^S 

0 

0 

OP 

-i^u 

jiv'n 

-1^17 

-AVi7 

-i^yn 

0 

0 

QR 

iv'37 

-r" 

-8^17 

-IV37 

3v'37 

0 

0 

AK 

-a 

-3U 

*i 

-3 

0 

Lit 

.\o 

I 

i^i 

-W 

-f 

i\' 

0 
0 

0 
0 

'r$ 

— 

H 

-3 

0 

0 

~3 

0 

0 

0 

0 

0 

0 

B8C       3D<      E8 


Fio.  676. 


Fir.,  f.77. 


Fui.   (.".S 


THREE-HINGED  BRIDGE. 


745 


V  DEFR 

FiQ.  67ft 


K 

\ 

M 

vy''^— 

-»v 

^-^ 

0 

+^ 

^ 

.^-"^••CD 

(t»i 

:=&( 

Xl^ 

s:^— — 

Fig.  681. 


Fig.  682. 


Ex.  16.  Show  how  to  determine  the  maximum  stresseii  in  the  verticals  and 
diagonals  of  the  trusses  for  the  SauU  Ste.  Marie  Bridge,  Fig.  624,  Ex.  1. 

.\8  already  stated,  the  maximum  stresses  in  the  several  members  are  best 
obtained  graphically.  Fig.  683  is  the  stress 
diagram  when  the  only  force  on.  the  truss  is 
a  vertical  reaction  at  A  of  100,000  lbs. -A'l'. 
Taking  Xy  =  100,  so  that  each  unit  is  1000 
lbs.,  the  strfses  in  the  other  members  are 
indicated  on  tlie  figure. 

The  load  distribution  for  which  the  stress 
rd  is  greatest  is  Case  3,  the  actual  reaction  at  A  being  124,230  lbs.    Then 


[ind 


So 


actual  stress  «/-tVj(124,230)  -90,688  lbs., 
"      de  =  ,V6(124,230)  - 100,627  lbs. 
"      /y-i»o"iT(95,020) -64,614  lbs., 
etc.,  etc. 


12.  Wind-pressure. — Numerous  experiments  to  determine  the 
pressure  and  velocity  of  the  wind  have  been  made  by  means  of 
feathers,  cloud-shadows,  anemometers  of  \arious  kinds,  wind- 
pauges,  pendulum,  tube,  and  spring  instruments.  The  results, 
cither  through  errors  of  observation,  errors  of  construction,  or  for 
other  occult  reasons,  are  almost  wholly  unreliable  and  give  the 


.« 


^ijl 


I 


ft 


74G 


THEORY  OF  STRUCTURES. 


engineer  no  accurate  information  upon  whicli  to  base  his  calcula- 
tions as  to  the  effect  of  wind  ujKjn  a  structure.  Theoretical  inves- 
tigations on  the  subject  are  equally  unsatisfactory.  The  formulse 
expressing  the  relations  between  the  sjx'ed  of  the  anemometer,  the 
velocity  of  the  wind  and  its  pressure,  are  of  a  purely  empirical 
character,  and  are  only  applicable  to  a  sjiecific  series  of  recorded 
obs(>r  vat  ions. 

Smeaton  inferred  from  Rouse's  exjx'riments  that  the  average 
pressure  in  pounds  ix?r  siiuare  foot  =  (velocity  in  miles  per  hour)^ 
-^200,  or 

200 
According  to  Dines  the  formula  should  be 

2000 
Tlie  AVind-pre.'^sure  Commission  (Kng.)  recommended  the  formula 

100' 

as  giving  with  tolerable  accuracy  the  mnximuhi  pressure. 

Stokes  considered  that  tlu  liinl  wind  velocities  should  be  about 
four  fifths  of  the  values  recorded  by  anemometers,  so  that  a  velocity 
of  04  miles  jmt  hour  recorded  as  correspon<ling  to  a  maximum  pres- 
sure of  40.0  ll)s.  |M'r  sciuare  foot  (the  average  of  fnr  observed  pres- 
sures) would  Ik'  reduced  to  r)1.2  miles  \n>r  hour.  Tiie  (irermje  pres- 
sure corresiH)niling  (o  ')1.2  miles  |ht  hour  would  In*  13.1  lbs.  jxt 
scjuare  foot  according  to  Smeaton 's  rule  and  only  9.18  lbs.  according 
to    Dines. 

.\gain,  certain  exiM-riments  at  tJm'nwich  indicated  that  the 
pressure  was  increased  by  the  stifTness  of  the  copjK-r  wire  connecting 
tlie  recording  ]M'ncil  witli  the  pres.sure  plate,  and  a  flexible  brass 
chain  was  therefore  substituted  for  tlie  wire.  Tims  modified,  a 
pressure  of  20  lbs.  ])er  square  foot  was  registered  as  correspf»nding 
to  a  velocity  of  ((4  miles  |H'r  hour,  whereas  witii  the  cop|x'r  wire  a 
pressure  of  A%  lbs.  ])er  square  foot  had  U-en  registered  with  a 
velocity  of  on\\  53  miles  jxt  iiour. 


EMPIRICAL   REGULA  TIONS. 


747 


These  facts  tend  to  show  tliat  the  actual  pressure  is  much  less 
than  that  given  by  a  recording  instrument,  and  that  the  very  high 
pressures,  as,  e.g.,  80  lbs.  per  square  foot  and  even  more,  must  be 
ilue  to  gusts  or  squalls  having  a  jnirely  local  effect.  This  opinion 
.seems  to  be  confirmed  by  Sir  B.  Baker's  experiments  at  the  Forth 
Bridge,  which  also  indicate  that  the  pressure  per  square  foot 
diminishes  as  the  area  acted  upon  increases.  Xo  engineering  struc- 
ture could  withstand  a  pressure  to  80  lbs.  per  square  foot  of  surface, 
and  a  pressure  of  28  lbs.  to  32  lbs.  would  overturn  carriages,  drive 
trains  from  the  track,  and  stop  all  traffic. 

It  is,  of  course,  well  known  that  wind-forces  sufficiently  powerful 
to  uproot  huge  trees  and  to  demolish  the  strongest  buildings  are 
occasionally  develojwd  by  whirlwinds,  tornadoes,  and  cyclones, 
but  these  must  be  classed  as  acta  Dei  and  can  scarcely  be  considered 
by  an  engineer  in  his  calculations. 

Numerous  ob-^ervations  as  to  the  effect  of  wind  upon  structures 
in  ditTerent  localities  must  yet  Ik-  made  In-fore  any  useful  and  reliable 
rules  can  be  enunciated.  In  the  case  of  existing  bri(lg<>s  the  elonga- 
tion of  the  wind-braces  during  a  storm  can  easily  l)e  measured  within 
iT,*iST,  of  an  inch.  Investigations  should  Ik'  made  as  to  the  action 
of  the  wind  ujion  surfaces  of  different  forms  and  uiHin  sheltered 
surfaces,  as,  e.g.,  uinm  the  .surfaces  iH'hiiul  the  windward  face  in 
hridge-tru.sses.  Again,  it  is  (juite  iiossible,  if  not  prol)al)le,  that 
many  of  the  recorded  upsets  have  \yrpn  due  to  a  cnwhineii  lifting 
and  side  action,  rerjuiring  a  nuieh  less  flank  prcs.sure  than  would  be 
iK'ce.ssary  if  tliere  were  no  upwanl  force,  and  hence  further  light 
should  1m'  obtained  on  this  point. 

Under  any  circumstances,  the  wind-stre.s.ses  should  Ix*  as  small 
as  jK)ssible,  compatible  with  safety,  seeing  how  largely  they  influence 
the  sections  of  the  several  members,  especially  in  bridges  of  long  span* 

13.  Empirical  Regulations. 

Wind-prcusurc  Comwissinn  Rules.  — Vor  railway  bridges  and 
viaducts  a.ssume  a  maxini'im  pres.sure  of  .'>()  lbs.  iH'r.s(]uare  foot  uimjii 
an  area  to  1m'  estimated  as  follows; 

A.  In  r/».sp-girder  bridges  or  viaducts  the  area 
=-area  of  windward  face  of  girder 

+  an'a  of  train  surface  above  the  top  of  the  same  girder. 
^  li.r  iiioDt  roi-ont  researches  sw  pit\^T  by  Siaiiton,  Proc.  Inrt.  of  C.  E.(Eng.)  Vol. !.(,. 


ft; 


\\ 


H 


748 


THEORY  OF  STRUCTURES. 


B.  In  o;)en-girclcr  bridges  or  viaducts  the  area  for  the  windward 
girder 
=  area  of  windward  face,  assumed  close,  between  rails  and 

top  of  train 
+ calculated  area  of  windward  surface  above  the  top  of  the 

train 
+ calculated  area  of  windward  surface  below  the  rails. 
For  the  leeward  girder  or  girders  the  area 

=  calculated  area  of  surface  of  one  girder  above  the  top 
of  the  train  and  below  the  level  of  the  rails,  the  pres- 
sure being  28,  42,  or  56  lbs.  per  square  foot,  according 
as  this  area  <^S,  >iS  and  <  J5,  or  >}S,  where  S 
is  the  total  area  within  the  outline  of  the  girder.    The 
assumed  factor  of  safety  is  to  be  4. 
American  Specifications— (a)  The  lateral  bracing  in  the  plane 
of  the  roadway  is  to  be  designed  so  as  to  bear  a  pressure  of  30  lbs. 
per  square  foot  ujjon  the  vertical  surface  of  one  truss  and  upon  the 
surface  of  a  train  averaging  12  sci.  ft.  per  lineal  foot,  i.e.,  360  lbs. 
per  lineal  foot;    this  latter  is  to  be  regarded  as  a  lire  load.     The 
lateral  bracing  in  the  plane  of  the  otiier  chord  is  to  be  designed  so  as 
to  bear  a  pressure  of  50  lbs.  p(»r  square  foot  upon  twice  the  vertical 
surface  of  one  truss. 

(6)  The  fx)rtal,  vertical,  and  horizontal  bracing  is  to  be  propor- 
tioned for  a  pressure  of  30  lbs.  jx-r  scjuare  foot  upon  ttnce  the  ver- 
tical surface  of  one  truss  and  upon  the  surface  of  a  train  averaging 
10  s.,.  ft.  jHT  lineal  foot,  i.e.,  300  lbs.  {xt  lineal  foot,  the  latter  being 
treated  as  a  lire  load. 

(c)  Live  load  in  plane  of  roadway  due  to  wind-pre.ssure  =  300  lbs. 
jH'r  lineal  foot. 
Fixed  load  in  plane  of  roadway  due  to  wind-pres.sure  =  150  lbs. 

|K'r  lineal  foot. 
Fixed  load  in  plane  of  other  chord  due  to  wind-presure  =  150 
lbs.  |K'r  lineal  foot. 
Lateral   linwi uf/.-Vousuhr  a  trus.s-bri(lge  with  parallel  chords 
and  panels  of  length  p.     U-t  A  Ik-  the  area  of  the  vertical  surface 
of  one  tru.ss. 

According  to  (a)  the  lateral  bracing  in  the  j)lane  of  the  roadway 
is  siibj,.(.te.l  to  (1)  a  pjinrl  live  Imd  of  360/;  ll)s.  and  (2)  a  panel 


EMPIRICAL  REGULATIONS. 


749 


fixed  load  of  30^1  lbs., while  in  the  plane  of  the  other  chord  it  is  sub- 
jected to  a  panel  fixed  load  of 

50X2.4  =  100.4  lbs. 

Thus,  if  the  figure  represent  the  bracing  in  the  plane  of  the  road- 
way of  a  ten-panel  truss,  and  if  the  wind  blow  upon  the  side  AB, 


Fio.  684. 

the  maximum  horizontal  force  for  which  any  diagonal,  e.g.,  CD,  is 
to  be  designed  is 

=  45i4  lbs.  due  to  the  horizontal  force  of  30 A  lbs.  at 

each  panel-point 
+  756/)  lbs.  due  to  the  horizontal  force  of  360p  lbs.  at 

each  panel-point  between  C  and  B. 

The  dotted  lines  show  the  bracing  required  when  the  wind  blows 
on  the  opposite  side. 

It  is  sometimes  maintained  that  the  wind-forces, in  the  plane  of 
the  upper  chords  of  a  through-bridge  or  the  lower  chords  of  a  deck- 
bridge  are  transmittetl  to  the  floor-bracing  through  the  posts.  This 
can  hardly  be  correct  in  the  case  of  long  posts,  as  they  do  not  possess 
sufficient  sMffness.  It  has,  however,  Ix'en  jwinted  out  that,  in 
through-bridges,  the  cumulative  effect  of  the  wind-pressure  at  the 
ends  of  the  bridge  might  produee  a  .serious  lK>nding  action  in  the 
end  }W)sts  Thi.-t  action  would  have  to  Ix*  resisted  by  additional 
plating  on  the  end  jxwts  l)elow  the  portals,  or  by  an  increase  of 
their  sectional  area. 

Under  wind-pressure  the  floor-lM'ams  act  as  posts;  hence,  if 
the  wind-i)racing  is  attached  to  the  top  or  compression  flange  of  a 
fl<)or-l)eam,  the  flange's  sectional  area  must  be  proportionately 
increased.  If  the  bracing  is  attached  to  the  lower  or  tension  flange, 
the  stresses  in  the  latter  will  be  diminished. 


? 


It  i 


w 


750 


THEORY  OF  STRUCTURES. 


14.  Chords.— The  wind-pressure  transmitted  through  the  floor- 
bracing  increases  the  stresses  in  the  several  members,  or  panel  lengths, 
of  the  leewanl  chord,  the  greatest  increments  being  due  to  a  hori- 
zontal force  of  (8G0/>  +  30.4)  lbs.  at  each  of  the  panel-points  in  AB. 
The  corresponding  chord  stresses  in  the  ten-panel  truss-bridge 
referreil  to  above  are: 

C,=0; 

C2  =  4i  (360p  +  30.4 )  tan  ^  lbs. ; 

C3  =  C2  +  3.J  (3G0/>  +  30.4 )  tan  <?  =  8(360p  ^  30/i )  tan  0  lbs. ; 

C4  -  Ta  +  2 J  (3G0p  +  30.4 )  tan  0  =  10 J  (3(K)/)  +  .30.4 )  tar  '^  lbs.  ; 

C5  =  C4  + 1  i (360/;  +  30.4 )  tan  0  =  12(360p  +  30.4 )  tan  ./ 1 1  -,., 

90° -fl  being  the  angle  between  a  diagonal  and  a  chord. 

.Again,  the  wind-pressure  tends  to  capsize  a  train  and  throws 

an  additional  pressure  of  P^  lbs.  per  lineal  foot  upon  the  leeward 

rail,  P  being  the  pressure  in  pounds  per  lineal  foot  on  the  train 
surface,  y  the  vertical  distance  Ix'tvveen  the  line  of  action  of  P  and 
the  top  of  the  rails,  and  G  the  gauge  of  the  rails. 
Thus,  the  total  pressure  on  leeward  rail 

/  w  y  \ 

and  the  total  pressure  on  windward  rail 

("'      y  \ 
:7  ~^*(j)  ^l)s-  per  lineal  foot, 

w  being  the  weight  of  the  train  in  pounds  per  lineal  foot. 

Hence  the  total  vertical  pressure  at  a  panel-iwint  of  the  lee- 
ward truss 


-G 

2S 


W  ,  w  a      Iw  y  \ 


H- 


S  being  the  distance  l)etween  the  trusses. 


CANTILEVER  TRUSSES. 


751 


15.  Cantilever  Trusses. — A  cantilever  is  a  structure  supported 
at  one  end  only,  and  a  bridge  of  which  such  a  structure  forms  part 
uiuy  be  called  a  cantilever  bridge.    Two  cantilevers  may  project. 


BRlDaE.OVER  ST.  LAWRENCE  AT  NIAGARA. 

Fio.  6S& 


from  the  supports  so  as  to  meet,  or  a  gap  may  be  left  between  them^ 
which  may  be  bridged  by  an  independent  girder  resting  upon  or 
hinged  to  the  ends  of  the  cantilevers.  The  form  of  the  cantilever 
is  subject  to  considerable  variation. 


8UKKUR  BRIDGE 

Fio.  686. 


forth  bridge. 
Fio.  687. 


Figs.  688  to  693  represent  the  simplest  forms  of  a  cantilever 
frame.    If  the  member  AB  has  to  support  a  unifornUy  distributed 


Fio.  688. 


Fio.  689. 


Fio.  090. 


Fio.  601. 


load  as  well  as  a  concentrated  load  at  B,  intermediate  stay?  may 
Ix!  introduced  as  shown  by  the  full  or  by  the  dotted  lines  in  Figs, 
090  and  691.    Should  a  live  load  travel  over  AB,  each  stay  must  be 


h 


752 


THEORY  OF  STRUCTURES. 


designed  to  bear  with  safety  the  maximum  stress  to  which  it  may 
be  subjected. 

Figs.  692  and  693  show  cantilever  trusses  with  parallel  chords. 
If  the  truss  is  of  the  double-intersection  type,  Fig.  693,  the  stresses 


NWKKK 


Fio.  682. 


F E 


Fio.  693. 


Fio.  694. 


•fs7N7fv?^s>-^ 


Fig.  695. 


in  the  members  terminating  in  B  become  indeterminate.  They 
may  be  mjide  determinate  by  introducing  a  short  link  BD,  Fig.  694. 
TI1U.S,  if,  in  DB  produced,  BG  be  taken  to  represent  the  resultant 
stress  along  the  link,  antl  if  the  parallelogram  HK  be  completed, 
BK  will  represent  the  stress  along  BE,  and  BH  that  along  BF. 

This  link  device  ha.s  been  employed  to  equalize  the  pressure 
on  the  turntable  TT  of  a  swing-bridge  (Fig.  695).  An  "equalizer  " 
or  a  "rocker-link"  BD,  Fig.  696,  conveys  the  stresses  transmittal 
through  the  members  of  the  truss  terminating  in  D  to  the  centre 
posts  BT. 

Theoretically,  therefore,  the  pressure  over  TT  will  be  evenly 
distributed,  whatever  the  loaditig  !iuiy  be,  if  the  direction  of  BD 
bisects   the  angle   TBT  and  if  friction   is  neglected. 

The  joint  between  the  central  span  and  the  cantilever  requires 
the  most  careful  consideration  and  should  fulfil  the  following  con- 
ditions : 

(rt)  The  two  cantilevers  should  Ix'  free  to  expand  and  contract 
under  changes  of  tenifx'rature. 

(b)  T!ie  central  span  should  have  a  longitudinal  supjxjrt  which 


CANTILEVER  TRUSSES. 


753 


Nvill  enable  it  to  withstand  the  effect  of  the  braking  of  a  train  or 
the  pressure  of  a  wind  blowing  longitudinally. 

(<•)  The  wind-pressure  on  the  central  span  should  bear  equally 
on  the  two  cantilevers. 

_(d)  The  connections  at  both  ends  should  have  sufficient  lateral 
rigidity  to  check  undue  lateral  vibration.  Conditions  (a)  and  (c) 
would  be  fulfilled  by  supporting  tiie  central  span  like  an  onlinarv 
bridge-truss  upon  a  rocker  bolted  do^-n  at  one  end  and  upon  a  rocker 
nsting  on  expansion  rollers  at  the  other.  This,  however  would 
nut  satisfy  condition  (6).  It  is  preferable  to  support  the  span  bv 
means  of  rollers  or  links  at  both  ends,  and  to  secure  it  to  one  canti- 
I.  yer  only  on  the  central  line  of  the  bridge  with  a  large  vertical  pin 
adapted  to  transmit  all  the  lateral  shearing  force.  A  similar  pin 
at  the  other  end,  free  to  move  in  an  elongated  hole,  or  some  equiva- 
l.nt  arrangement,  as,  e.g.,  a  sleeve-joint  bearing  laterally  and 
with  rollers  in  the  seat,  is  a  satisfactory  method  of  transmitting 
the  .shearing  force  at  that  end  also.  (If  there  is  an  end  post,  it 
may  be  made  to  act  like  a  hinge  so  as  to  allow  for  expansion,  etc  ) 
llK'  points  of  contrary  flexure  of  the  whole  bridge  under  wind-pres- 
sure are  thus  fixed,  and  dl  uncertainty  as  to  winil-stresses  removed 

W  lu-re  other  spans  have  to  te  built  adjacent  to  a  large  can- 
tilever span,  it  should  not  iyo  hastily  assumed  that  it  is  necessarily 
l.<-st  to  counterbalance  the  cantilexer  by  a  contiguous  cantileveV 
■n  the  opposite  direction.  If  it  is  possible  to  obtain  good  founda- 
tions ami  It  piers  are  not  exjx'nsive,  it  might  be  cheaper  to  build 
a  number  of  short  independent  shle  spans  and  to  secure  the  can- 
II  .'ver  to  an  independent  anchorage.  If  this  is  <Ione.  care  must  be 
taken  to  give  the  abutment  sufficient  stability  to  take  up  the  unbal- 
anced thrust  along  the  lower  boom  of  the  cantile\er. 

Suppose  that  the  cantilever  is  anchored  back  bv  means  of  a 
single  back-stay. 

I^t  ir  =  weight  neces.sary  to  resist  the  pull  of  the  back-stay 
A  =  depth  of  end  post  of  cantilevtr;  ' 

«  =  horizontal  distance  l«>tw."en  foot  of  post  and  anchorage- 
M  =  ben(hng  moment  at  abutment  =  ir2. 
If  it  is  now  a^ssumed  that  the  sectional  areas  of  the  post  and  back- 
stay are  proportioned  to  the  stresses  they  have  to  bear  ^vhirh  is 
n.xcr  the  case  in  practice),  the  quantity  of  material  in  these  members 


i^  I 


ii 

i   i    1 
fl       ■ 


")4 


THEORY  OF  STRUCTURES. 


must  be  proportional  to 

..Z2  4-/(2 


ir 


h 


which  is  a  minimum  when  z='v2h. 

If  a  horizontal  mombor  is  introduced  between  the  feet  of  the 
back-stay  and  the  post,  the  quantity  of  material  becomes  propor- 
tional to 

,2  +  ^2  ^2  ?2j./,2 

which  is  a  minimum  when  z  =  h,  i.e.,  when  the  back-stay  slopes  at 
an  angle  of  45°.  By  making  the  angle  between  the  back -stay  and 
the  horizontal  a  little  less  than  45°,  a  certain  amount  of  material 
may  be  saved  in  the  joints  of  the  back-stays  and  also  in  the  anchors, 
which  more  than  compensates  for  the  increased  weight  of  the 
anchors  themselves. 

{Note. — In  these  calculations  it  is  assumed  that  the  top 
chord  is  horizontal,  and  that  the  feet  of  the  post  and  back-stay 
are  in  the  same  horizontal  plane.  This  is  rarely  the  case  in  prac- 
tice.) 

According  to  the  above  the  weight  of  material  necessary  for 
the  back-stay  is  directly  proportional  to  the  bending  moment  at 
the  abutment  antl  inversely  proportional  to  the  depth  of  the  can- 
tilever, other  things  IxMng  e(iual.  A  double  cantilever  has,  in  general, 
no  anchorage  of  any  groat  iniportauco. 

If  the  span  is  very  great,  a  cantilever  bridge  usually  requires 
less  material  than  any  other  rigid  structure  of  ecjual  strength,  oven 
though  anc!if)rago  may  have  to  be  provided.  If  two  large  sj)iuis 
are  to  bo  built,  a  double  cantilever,  ro(iuiring  no  anchorage,  may 
effect  a  very  considerable  saving  in  material,  altliough  a  double 
pier,  of  sufliciont  width  for  stability  under  all  conditions  of  loading, 
will  bo  necessary. 

Again,  wliere  false-works  are  costly  or  impossible,  the  property 
of  tho  oantilovor  tliat  it  can  lie  made  to  support  itself  during  erection 
give«  it  ;\!!  iiiinieDso  :u!v:intam..  If  tlie  design  of  the  cantilever  is 
such  that  it  can  bo  built  (jut  rapidly  and  cheaply,  it  will  often  be 


CANTILEVER   TRUSSES. 


755 


the  most  economicaU.ame  in  the  end,  even  if  the  total  quantity  of 
material  is  not  so  small  as  that  required  for  some  other  type  of 
bridge.  In  all  engineering  work  quantity  of  material  is  only  one  of 
tlie  elements  of  cost,  and  this  should  1k>  carefully  borne  in  mind 
wlien  designing  a  cantilever  bridge  because  a  want  of  regard  to 
the  method  of  erection  may  easily  add  to  its  cost  an  amount  much 
greater  than  can  be  saved  by  economizing  material. 

In  ordinary  bridge-trusses  the  amount  of  the  web  metal  is 
greatest  at  the  ends  and  least  at  the  centre,  while  the  amount  of 
tlie  chord  metal  is  least  at  the  ends  and  greatest  at  the  centre.  Thus 
the  assumption  of  a  uniformly  distributed  dead  load  for  such  bridges 
IS,  generally  speaking,  sufficiently  accurate  for  practical  purposes 
In  the  case  of  cantilever  bridges,  however,  the  circumstances  are 
entirely  different.  In  these  the  amount  of  the  metal  both  in  the 
web  and  in  the  chords  is  greatest  at  the  support  and  least  at  the 
<'nd.  For  example,  the  weight  of  the  cantilevers  (exclusive  of  the 
weight  of  platform,  viz.,  i  ton  per  lineal  foot)  for  the  Indus  Bridge 
ix'r  lineal  foot,  varies  from  6i  tons  at  the  supports  to  1  ton  at  the 
outer  ends.  Hence  the  hypothesis  of  a  uniformly  distributed  dead 
Joad  for  such  structures  cannot  hold  good. 

The  weight  of  a  cantilever  for  a  given  span  may  be  approxi- 
mately calculated  in  the  following  manner: 

Determine  the  stresses  in  the  several  members,  panel  by  panel— 

(A)  P'or  a  load  consisting  of 

(1)  a  given  unit  weight,  say  100  tons,  at  the  outer  end; 

(2)  the  corresponding  dead  weight. 

(B)  For  a  load  consisting  of 

(1)  the  specified  live  load; 

(2)  the  corresponding  panel  dead  weight. 

Thus  the  whole  weight  of  a  panel  will  l,e  th<'  sum  of  the  weights 
'I'duced  in  (.\)  and  (H),  and  the  total  weight  of  the  cantilever  will 
1"'  the  sum  of  the  several  panel  weigiits. 

Tliis  process  ovidi-ntly  gives  at  the  same  time  the  weights  of 
-  intilevers  of  one,  two,  three,  etc.,  panel  le.igths,  the  loads  remain- 
ini;  the  same. 

The  panel  dead  weights  referred  to  in  (A)  and  (B)  must,  in  the 
frst  place,  be  assumed.  This  ran  },e  d..i..'  will,  a  large  degree  of 
;HTuracy,  as  the  dead  weight  mast  necessarily  gradually  increase 


i 
i  11 


-  1 


ill 


m 


756 


THEORY  OF  STRUCTURES. 


towards  the  support,  anil  any  error  in  a  particular  panel  may  be 
easily  rectified  by  subsequent  calculations. 

Again,  the  preceding  remarks  indicate  a  method  of  finding  the 
most  economical  cantilever  length  in  any  given  case. 

Take  e.g.,  an  ojx'ning  spanned  by  two  equal  cantilevers  and 
an  intermediate  girder.  Having  selected  the  tyiH>  of  bridge  to  be 
employed  for  the  intermediate  span,  estimate,  either  from  existing 
bridges  or  otherwise,  the  weights  of  indejx-ndent  bridg(>s  of  the 
same  tyjx"  and  of  different  sjjans.  Sketch  a  skeleton  diagram  of  the 
cantilever,  extending  over  one  half  >of  the  whole  span,  and  apply 
to  it  the  processes  referred  to  in  (A)  and  (H). 

If  L  is  the  length  of  the  cantilever  and  /'  that  of  a  panel,  the 
following  table,  in  which  the  inte/ mediate  span  increases  by  two 
panel  lengths  at  a  time,  may  lie  prepared: 


—  aj 

, , 

.  j:  ji 

■    C  «    ■    ft) 

1  c  »  «    . 

II 

B  5 

'5  £ 

i 

ilii 

■srco-50 

^ 

s 

rs 

■^ 

is 

£s 

sf 

^ 

0 

/. 

?P 

L-2P 

iP 

L-AP 

tiP 

L-i\P 

8P 

L-HP 

etc. 

rtc. 

12  3  4  .5  0  7  '       8 

Weight  in  col.  3  =  one  half  of  the  weight  of  the  intermediate  girder 

+  ont'  half  of  the   live  load  it   carries  if  uniformly 

distributed.     (The    ])roportion    will    be    greater 

than  one  half  for  arbitrarily  distributed  loads, 

and   may   be   etusily   determined   in   the   usual 

niiumer.) 

Col.  5  gives  the  weights  obtained  ps  in  A. 

weiiilit  on  end  of  cantilever 
Col.  6=  col.   5x j.jjj — . 

Col.  7  gives  the  weights  obtained  as  in  B. 
Col.  S  =  e()l.  2  + col.  ()  +  col.  7. 

It  is  impftrtnnt  to  bear  in  mind  that  an  increa.se  in  the  weielit 
of  the  central  span  necessitates  a  corresponding  increase  in  the 


I     -k. 


CANTILEVER  TRUSSES. 


757 


weights  of  the  cantilevers.  Hence,  in  order  that  the  weight  of 
tlio  structure  may  be  a  minimum,  the  best  material  with  the  highest 
practicable  working  unit  stress  should  be  employed  for  the  centre 
span. 

The  table  must  of  course  be  modified  to  meet  the  requirements 
of  different  sites.  Thus,  if  anchorage  is  needed,  a  column  may  be 
added  for  the  weights  of  the  back-stays,  etc. 

Curve  of  Cantilever  Boom.— Consider  a  cantilever  with  one  horizon- 


Fio.  697. 


tal  boom  OA,  and  let  x,  y  he  the  coordinates  of  any  point  P  in  the 
(»ther  boom,  0  being  the  origin  of  coordinates  and  A  the  abutment 
end  of  the  cantilever. 

I/'t  W  be  the  portion  of  the  '^ight  of  an  independent  span  sup- 
ported at  0. 

Let  w  Ik"  the  intensity  of  the  led  at  the  vertical  section  through  P. 
Assume  (1)  that  there  are  no  diagonal  strains,  and,  hence,  that 
the  web  consists  of  vertical  members  only; 

(2)  that  the  stress  H  in  the  horizontal  boom  is  con- 

stant, and  therefore  the  bending  moment  at  P 
^Hy; 

(3)  that  the  whole  load  is  transmitted  through  the  ver- 

tical members  of  the  wob. 

JH  k  be  £uch  a  factor  that  kTl  is  the  weight  of  a  member  of 
I'tiflth  /,  subjected  to  a  stress  T. 

(Xote—lf  I  is  in  feet  and  T  in  tons,  then  k  for  steel  is  about 
.ii(K);},  allowance  being  made  for  loss  of  section  or  increase  of  weight 
at  connections.) 

V  consists  of  two  parts,  viz.,  a  connlani  part  p,  due  to  the  weight 
of  the  platform,  wind-bracing,  etc.,  which  is  assumetl  to  be  uniformly 


75S 


THEORY  OF  STRICTURES. 


(listril)uto(l;  and  a  variahle  port,  duo  to  the  weight  of  the  cantilever, 
which  may  be  obtained  as  follows: 

Weight  of  element  dx  of  horizontal  \)oon\  ^kHdx. 
"      "    web  correspondins;  to  r?x  =kmjdx. 

"      "    element  of  curved  boom  corresponding  to  dx 


Mth- 


Hence  the  variable  intensity  of  weight 

■"ind  w  =  p  +  kH  +  kwy  +  kHl^j  . 

Again,  if  M  is  the  Ix'iiding  moment  and  S  the  shearing  force  at 
the  vertical  section  through  P,  then 

'^-K^dS^  d^ 

dx^     dx  "dx^- 

Therefore    //g  =  p  +  kH  +  kinj  +  kH  (^)  ' 
Integrating  twice, 


y 


Hy^A+Bx  +  (p  +  2kH)-  +  A;H| 
A  and  B  being  constants  of  integration 


111 


When  x=0, 
Thus 


2/  =  0    and    H%-W. 
A=0    and    B  =  W. 


','{ ' 


CANTILEVER  TRUSSES 


Hence  Hy  =  Wx  +  ip  +  2kH)^  +  kH^ 


is  the  equation  to  the  curve  of  the  boom,  and  represents  an  ellipse 
with  its  nuijor  axis  vertical,  and  with  the  lengths  of  the  two  axes 


in  u  ratio  equal  to 


'p  +  2kHy 


kH 


The  depth  of  the  longest  cantilever  is  determined  by  the  vertical 
tangent  at  the  end  of  the  minor  axis,  and  corresponds  to  the  value 

of  y  given  by  making  ^=0  in  the  preceding  equation,  which  gives 
1 

For  a  given  value  of  H  the  curve  of  the  boom  is  independent 
of  the  span.  Again,  for  a  given  length  of  cantilever  with  a  bo"m 
of  this  elliptic  form,  a  value  of  H  may  be  found  which  will  make 
the  total  weight  a  minimum,  and  which  will  therefore  give  the  most 
economical  depth.  Such  an  investigation,  however,  can  only  be 
of  interest  to  mathematicians,  as  the  hypotheses  are  far  from  being 
even  approximately  true  in  practice,  and  the  resulting  depth  would 
be  obviously  too  great. 

/assumption  (1)  no  longer  holds  when  a  live  load  has  to  be 
considered.  Diagonal  bracings  must  then  be  introduced,  which 
l)ocome  heavier  as  the  depth  increases,  in  consequence  of  their 
increased  length.  The  diagonal  bracings  are  also  largely  affected 
by  the  length  of  the  panels.  If  the  panels  are  short,  and  if  a 
great  depth  of  cantilever,  diminishing  rapidly  away  from  the 
abutment,  is  usetl,  the  angles  of  the  diagonal  bracing,  near  the 
ajjutment,  will  be  unfavorable  to  economy.  This  difficulty  may 
1)0  avoided  by  adopting  a  double  system  of  triangulation  over  the 
deeper  part  of  the  cantilever  only,  or  even  a  treble  system  for  some 
distance  in  a  large  span.  Th'?  objections  justly  urged  against  mul- 
tiple systems  of  triangulation  in  trusses  lose  most  of  their  force  in 
large  cantilevers.  In  the  first  place,  the  method  of  erection  by 
building  out  insures  that  each  diagonal  shall 'take  its  proper  share 
of  the  dead  load;  and  in  the  second  place,  it  should  be  remembered 
tiiat  only  in  large  spans  could  a  double  system  have  anything  to 
recommend  it,  and  then  only  near  the  abutment  where  the  stresses 


;  '» 


760 


THEORY  OF  STRUCTURES. 


;ire  groatost:  in  such  cas(>s  tho  moving  load  only  produces  a  small 
[ortion  of  the  entire  stress  in  the  web.  In  practice  a  compromise 
lias  to  Ik*  made  Ix'tween  different  reciuirements,  and  the  depth 
must  l)e  kept  witliin  such  limits  as  will  admit  of  reasonable  propor- 
tions in  other  resjx^cts,  while  the  diagonal  ties  or  struts  may  Ije 
allowed  to  vary  in  inclination,  to  some  extent,  from  one  panel  to 
another. 

Apiin,  in  fixing  the  panel  length,  care  must  be  taken  that  there 
is  no  undue  excess  of  platform  wnght,  as  this  will  produce  a  corre- 
sjMJnding  increase  in  the  weight  of  the  cantilever. 

An  excessive  depth  of  cantilever  generally  caases  an  increase  in 
the  cost  of  erection. 

lioth  theory  and  practice,  however,  indicate  that  it  will  be  more 
advantageous  to  choose  a  greater  depth  for  a  cantilever  than  for 
an  ordinary  girder  bridge. 

An  ordinary  proportion  for  a  large  girder  bridge  would  lx»  one 
i.iiitli  to  one  seventh  of  the  span,  and  if  for  the  girder  were  substi- 
tuted two  cantilevers  meeting  in  the  midiUe  of  the  span,  the  depth 
might  v>ith  advantage  be  considerably  increased  beyond  this  pro- 
portion at  the  abutment,  if  it  l)e  reduced  to  nil  where  the  cantilevers 
meet.  When  a  central  span  is  introduced,  resting  ujwn  the  ends  of 
tiie  two  cantilevers,  the  concentrated  load  on  the  end  gives  an  ad<li- 
tional  reason  for  still  further  increasing  the  depth  at  the  abutment 
proportloiiaUn  (n  the  louith  of  the  earitilere).  The  greatest  economical 
depth  has  pr-ibahly  In-en  reached  in  the  Indus  bridge,  in  which  the 
depth  at  the  abutment -.54 X length  of  cantilever,  i'robably  the 
jiroportion  of  one  third  of  the  length  of  th(>  cantilever  would  l)e 
ample,  excepr  where  the  anchorage  causes  a  considerable  part  of  the 
wliol''  wcijihl,  l)ut  eacii  case  must  Im'  considered  on  its  own  merits. 
'1  lif  reduction  of  defl'ftion  obtained  by  increasing  the  depth  is  also 
;iii   appn'i'i;d)le  consideration. 

If  a  depth  1m'  chosen  nf)t  widely  dilTerent  from  that  which  makes 
t!ic  (|u;nitity  nf  inaterial  a  minimum,  the  lyeight  will  l)e  oi\ly  slightly 
iuc'eased,  while  it  is  possible  that  great  stnietm-al  ailvantages  may 
Ix'  gtiined  ill  (ither  diri'dions.  In  recommending  a  great  depth 
t"<'r  a  cantilever  at  its  abutment,  it  is  assumed  that  the  depth  will 
!«•  eoritiiuioiisly  reduced  frmn  the  al»utm<'nt  outwards.  If  the 
IiKid  were  (•(iiiliiiMdiN!','  i!i-i 'ilin'".!    if   i«  li\-  Tir>  tiu'ans  certain  tha.t 


STATICAL  DEFLECTION. 


761 


a  cantilovpr  of  uiiiloim  depth  would  require  more  material  than 
one  of  varying  depth,  but  it  has  already  been  pointed  out  to  what 
•  xteiit  the  weight  of  the  structure  itself  necessarily  varies,  and  if 
the  concentrateil  loa<l  at  the  end  were  separately  considered,  the 
economical  truss  would  be  a  simple  triangular  frame  of  very  great 
depth.  From  economic  considerations,  it  would  be  well  to  reduce 
the  depth  of  the  cantilever  at  the  outer  end  to  nil,  but  in  many 
cases  it  is  thought  advisable  to  maintain  a  depth  at  this  point  equal 
to  that  at  the  end  of  the  central  span,  so  that  the  latter  may  be 
built  out  '.vithout  false-works,  under  the  same  system  of  erection 
as  is  pursued  in  the  case  of  the  cantilever.  The  post  at  the  ends 
of  the  central  span  and  cantilever  is  sometimes  hinged  to  allow  for 
expansion. 

Deflection.— A  serious  objection  urged  against  cantilever  bridges 
is  the  excessive  and  irregular  deflection  to  which  they  are 
sometimes  subject.  They  usually  deflect  more  than  onlinary  truss- 
bridges,  anil  the  deflection  is  proportionately  increa-sed  under  sud- 
iletdy  applied  loads.  In  the  endeavor  to  recover  its  nornml  jiosition, 
the  cantilever  springs  back  with  increased  force  ami,  owing  to  the 
small  resistance  offered  by  the  weight  an*,  stiffness  at  the  outer  end, 
there  may  result,  esiiecially  in  light  bridges,  a  kicking  movement. 
It  must,  however,  Ik'  l)orne  in  mind  that  the  deflectioii,  of  which 
the  importance  in  connection  with  iron  bridges  has  always  Ix^n 
ncogiiized,  is  not  in  itself  necessarily  an  evil,  except  in  so  far  as  it 
is  an  indication  or  a  cau.^e  of  over-strain. 

i6.  The  Statical  Deflection,  due  to  a  quiescent  load,  must  be 
distinguished  from  what  might  Ix'  called  the  dynamical  deflection, 
i.e.,  tho  additional  deflection  due  to  a  loail  in  motion.  The  former 
should  not  exceed  the  deflection  corres|)ondiiig  to  the  statical  stresses 
Inr  which  the  bridge  is  desigred.  The  amount  of  the  dynamical 
lr(iccti(.n  deiM'iids  both  upon  the  nature  of  the  loads  and  uj)on 
I  III-  manner  in  which  they  are  applied,  nor  are  there  sutlicient  data 
t'p  ({(tcrmine  its  value  even  approximat<dy.  It  certainly  largely 
increases  the  statical  stresses  and  produces  other  ill  effects  of  which 
little   is  known. 

Hitherto  the  question  as  to  the  deflection  of  framed  structiireH 
lias  received  iiut  meagn'  attention,  and  formula'  deduced  for  solid 
SiFuiTs    huve    iwiMi    employed    with    inisicatUiig    results.     It    would 


762 


THEORY  OF  STRUCTURES. 


sopm  to  bo  more  scioiitific  and  correct  to  treat  eacli  member  sepa- 
rately iiiul  to  consider  it«  individual  defornuition. 

17.  Rollers. — One  end  of  a  briilgj'  usually  rests  upon  nests  6f 
turned  wrougiit-iron  or  steel  friction  rollers  running  between  planed 
surfaces.  The  dianunT  of  a  roller  should  not  Ix*  less  than  2  ins., 
and  tlie  pressure  upon  it  in  ])ounds  per  lineal  inch  should  not  exceed 
olM)  N  (I  if  made  of  wrought  iron,  or  GOO  ^d  if  made  of  steel,  d  being 
the  diameter  in  inches. 

18.  Eye-bars. — In  Ijigland  it  has  l)een  the  practice  to  roll  bar.s 
ha\ing  enlarged  ends,  and  to  forge  the  eyes  under  hydraulic  pres- 
sure with  suitably  shajK'd  (hes.  In  America  both  hammer-forged 
and  hydraulic-forged  eye-bars  are  made,  the  latter  being  called 
ircldlcss  njc-hnrs.  Careful  matiiematical  and  exjx'rimental  inves- 
tigations have  Ihm'ii  carried  out  to  tletermine  the  j)rojxT  ilimensions 
of  the  link-head  and  i)in,  but  owing  to  the  very  complex  character 
of  the  stres.'ies  develoix-d  in  tlie  metal  around  the  eye.  an  accurate 
matiiematical  .solution  is  im[)os.sible. 

Let  d  be  the  width  and  t  the  thickness  of  tlie  .>-■  nk  of  the  eye- 
bar  rej)resented  in  Fig.  699  Let 
S  he  the  width  of  the  metal  at 
the  sides  of  the  eye,  and  //  the 
width  at  the  end.  Let  I)  W  the 
diameter  of  the  jiin. 

The    projxirtions   of   the   head 
are  governed  by  the  general  coii- 
t''<i   690.  dition   that   each  and  every  jiart 

should  Im"  at  least  as  strong  as  the  shank. 

When  the  bar  is  subjecteil  to  a  tensile  stress  the  pin  is  tightly 
embraced,  and  failure  may  arise  from  any  one  of  the  'ollowing  eause.s: 
Ui)  Tii:'  }>in  intiy  he  shorn  throiKjh. 

Hence,  if  the  jnii  is  in  double  shear,  its  sectional  area  should 
Ix'  at  least  one  half  that  of  tiie  snank. 

It  may  hap|M'ii  that  the  pin  is  iM'ut,  but  that  fracture  is  pre- 
vented by  the  closing  up  of  the  j/n'ces  Ix'tween  the  jtin-head  and 
nut;  the  elhcifney,  liowrvtr,  «if  the  cdnnection  is  destroyed,  a»  the 
bars  are  no  longer  free  to  turn  on  the  pin. 

In  practice  I)  for  Hat  bars  varies  from  ]d  to  |r/,  but  usually 
lies  iM^iwrrn  ](/  and  J(/. 


EYE-BARS. 


763 


The  diameter  of  the  pin  for  the  end  of  a  round  bar  is  generally 
made  equal  to  l\  times  the  diameter  of  the  bar. 

The  pin  should  he  turned  so  as  to  fit  the  eye  accurately,  but  the 
l)est  practice  allows  a  difference  of  from  ^V  to  rto  of  an  inch  in  the 
diameters  of  the  pin  and  eye.. 

(/>)  The  link  may  tear  across  MX. 

On  account  of  the  perforation  of  the  head,  the  direct  pull  on 
tlu-  shank  is  bent  out  of  the  straight  and  distributed  over  the  sections 
S.  There  is  no  reason  for  ti..  assumption  that  the  distribution  is 
uniform,  and  it  is  obviously  probable  that  the  intensity  of  stress 
is  greatest  in  the  metal  next  the  hole.  Hence  the  sectional  area 
of  the  metal  across  MX  nmst  be  at  least  equal  to  that  of  the  shank, 
and  in  practice  is  always  greater. 

S  usually  varies  from  .55^  to  .625rf. 

The  sectional  area  through  the  sides  of  the  eye  in  the  head  of  a 
round  bar  varies  from  U  times  to  twice  that  of  the  bar. 

(f)  The  pin  may  be  torn  through  the  head. 

Theoretiailly  the  sectional  area  of  the  metal  across  PQ  should 
he  one  half  that  of  the  shank.  The  metal  in  front  of  the  pin,  how- 
ever, nuiy  l)e  likened  to  a  uniformly  loaded  girder  with  both  ends 
fixed,  anil  is  subjected  to  a  bending  as  well  us  to  a  shearing  action. 
Hence  the  minimum  value  of  //  luus  Imh'ii  fixed  at  id,  and  if  //  is 
made  ecjual  to  d,  both  kinds  of  action  will  b<'  amply  i>rovided  for. 

(d)  The  bearing  surface  may  be  insufficient. 

If  such  Im-  the  case,  the  intensity  of  the  pressure  ujMin  the  Ix^ar- 
iiig  surface  is  excessive,  the  eye  becomes  oval,  the  metal  is  upset, 
and  a  fracture  takes  place.  Or  again,  as  the  hole  elongates,  the 
inetul  in  the  sections  S  next  the  hole  will  Ik-  drawn  out,  and  a  crack 
will  conmienee,  extending  outwards  until  fracture  is   produced. 

Ill  practice  adeciuate  In-aring  .surface  may  Ix'  obtained  by 
thickening  the  head  so  as  to  confine  the  maximum  intensity  of  the 
|)ressure  within  a  given  limit. 

(e)  The  head  may  l)€  torn  through  the  shoulder  at  XY. 
Hence  XY  is  made  e(|ual  to  d. 

The  radius  of  curvature  //  of  the  shoulder  varies  from  1 W  to  7.6d. 


li 


'II 


(I 

Xote. — The  thicknesn  of  the  slmnk  f*liouKl  in-  -  ,  or  ^d  at  least. 


764 


THEORY  OF  STRUCTURES. 


The  following  table  gives  the  eye-bar  proportions  common  in 
American  practice: 


Value  of  d. 

Value  of  D. 

Value  of  S. 

Wel<lle»9 
Bars. 

Hammered 
Bars. 

1    00 
1.00 
1  .00 
1.110 
I.IM) 
1   00 
1.00 
1   00 

.67 
.75 
1.00 
1.2.5 
1.33 
1.50 
1.75 
2.00 

1.5 

1.5 

1.5 

16 

1.7 

1.85 

2.00 

2.25 

1.33 
1.33 
1..50 
1.50 

1.67 
1.67 
1.75 

Also,  in  weldless  bars,  H  =S;  in  hammered  bars  H  ^d. 

Steel  Eyf-barx. — Hydraulic-forged  steel  eye-bars  are  now  being 
largely  made.  The  steel  has  an  ultimate  tenacity  of  from  60,000 
to  ()S,()00  lb.s.  jK-r  .scjuare  inch,  an  elastic  limit  of  not  less  than  50  per 
cent,  and  an  elongation  of  from  17  to  20  jxt  cent  in  a  length  equal 
to  (en  times  the  least  transverse  dimension. 

The  Phd'iiix  Rridge  Company  antl  '  Edge  Moor  Iron  Company 
give  the  tables  on  page  7{\iy  of  steel  f.c-bar  projKirtions. 

19.  Rivets.  -A  riret  is  an  iron  or  steel  ttfiank,  slightly  tapered 
at  one  end  (the  tail),  and  .surmounted  at  tlie  other  by  a  cup  or  pan- 
shaped  hedd  (Fig.  7(K)).  It  is  u-sed  to  join  steel  or  iron  plates,  bars, 
etc.  For  this  piirjtosc  the  riv(>t  is  generally  heated  to  a  cherry-red, 
the  shank  or  spindle  is  pa.'^sed  through  the  hole  prepared  for  it,  and 


iSl 


Fio    700.        Fui    701. 


Vw    702. 


2=L 


Fio.  703. 


Fio.  704. 


the  tail  is  made  into  a  button,  or  jyoint.  The  hollow  cui>-t(H)l  gives 
to  llic  point  a  nearly  heinisplierical  shai)e.  and  forms  what  is  called 
a  smip-riirt  fFig.  701 1.  Sii:i|>-rivcts.  partly  for  the  sake  of  ai)i)ear- 
aiHT,  lire  coiniiioiily  u.'^cd  in  girdcr-wnrk.  but  they  are  not  so  light  as 
rri/NVy(/-pniiit(>tl  rivets  (.v7(;//-rivets),  which  are  hammered  into  shajx' 
uniii  almost  cold  (Fig.  702). 


RIVETS. 


786 


PhcBtiix  Bridge  Co. 

Edge  Mooi  Iron  Co. 

Wilth 

Diameter 

Diameter 

Width 

Diameter 

Diameter 

\finimum 

Excess  of 
Sectional  Area 

of 

of 

of 

of 

of 

of 

Thickness 

of  Head  aioDS 

Barrf. 

Pin-hole. 

Head. 

Bard. 

Pin-hole. 

Head. 

of  Bar. 

PP  over  Sec- 

tion of  Bar. 

•.i 

2A.2U 

„       1 
'       1 

2 

U 

-«i 

1 

33% 

3 

:JA,  3  1 

8      ! 

2 

2} 

5i 

1 

33 

4 

3^6 

g 

2i 

2J 

Hi 

1 

33 

4 

3{ii.lA.4tt 

10 

2i 

3i 

Ci 

1 

33 

5 

:Jfii,  4A 

11 

3 

2J 

01 

, 

33 

5 

■iH.  .-ft 

12 

3 

4 

8 

33 

5 

•■]*.  l-ft 

13 

4 

n 

9} 

33 

6 

,^ *.^^  , 

13 

4 

.^1 

lOJ 

33 

6 

•=5 A,  .'^A,  rn 

14 

.5 

4i 

11} 

1 

37 

6 

lift.  •■4i,  fiH 

1.5 

.5 

."•i 

12i 

37 

7 

.,  -^ 

1.5 

6 

•■>} 

1.3} 

■ 

37 

7 

■M.<-.A.<-« 

16 

(i 

*'>i 

14} 

■ 

37 

7 

<•'«.  7A.  7H 

17 

7 

■^ 

1.5} 

40 

H 

^A 

17 

7 

7i 

17 

. 

40 

8 

'Hi.  (ifi.  7A 

18 

8 

5 

17 

1 

40 

S 

7H.  xi 

1!» 

8 

6 

18 

1 

40 

X 

S2.  t>i 

20 

9 

7A.7H 

20 

<» 

Xi    Si 

2! 

10 

»i 

22 

10 

■•^J.  93 

23 

10 

10, lOj 

24 

III  li.Xh  the  I»h(rni\  and  KdRc  .Moor  liars  the  t liickncs-s  of  the  head  is  the  same 
as  tluit  of  the  body  of  the  liar,  or  does  not  e.xceed  if  bv  more  than  A  inch. 

Wli(>ti  a  smooth  surface  is  required,  the  rivt^ts  are  rovnhrsunk 
(Fif?.  70;i).  The  countersinking  is  drilled  and  may  extend  through 
the  plate,  or  a  shoulder  may  be  left' at  the  inner  edge. 

Cnld-rivetiufi  is  adopted  for  the  small  rivets  in  boiler-work  and 
also  wherever  heating  is  impracticable,  but  tightly  driven  turned 
bolts  are  .sometimes  substituted  for  the  rivets.  In  all  such  ca.s(>s 
tiie  material  of  the  rivets  or  bolts  should  be  of  sujierior  quality. 

I,oo.se  rivets  are  easily  discovered  by  tapping,  and,  if  very  loose, 
should  1h  at  once  replaced.  It  mu.^.t  1k'  borne  in  mind,  however, 
that  expansions  and  contractions  of  a  complicated  character  inva- 
riably accompany  linl-rircliiuj,  and  it  cannot  Ih*  suppo.sed  that  the 
rivets  will  1k<  ))erfectly  tight.  Indeed,  it  is  doubtful  whether  a  rivet 
has  any  hold  in  a  straight  drilled  hole,  except  at  the  enils. 

Hivefing  is  aci-omplished  either  by  imnfl  or  machine,  the  latter 
Iwing  far  the  more  effective.  A  machine  will  s(|ueez(>  a  rivet, 
at    almost    any  temix>raturc,  into   a   mast  irregular  hole,  but   the 


766 


THEORY  OF  STRUCTURES. 


exigencies  of  practical  conditions  often  prevent  its  use,  except  for 
ordinary  work,  and  its  advantages  can  rarely  be  obtained  where 
they  would  be  most  appreciated,  as,  e.g.,  in  the  riveting  up  of  con- 
nections. 

Dimensions  of  Rivets.— The  diameter  (d)  of  a  rivet  in  ordinary 
girder  work  varies  from  f  in.  to  1  in.,  and  rarely  exceeds  1|  ins.  The 
thickness  of  a  plate  in  ordinary  girder-work  should  never  be  less 
than  {  in.,  and  a  thickness  of  I  in.  is  preferable. 

According  to  Fnirbairn,  d  should  be  about  2t\it<\  in.,  and  should 
be  about  \\t  if  />i  in. 

According  to  Ummn,  d  should  lie  between  (|<  +  i'j)  in.  and 
(I'  +  f)  in.  when  t  varies  from  {  in.  to  1  in. 

When  the  rivets  join  several  plates,  d  =  (-+-j  in.,  T  being  the 

total  plate  thickness. 

According  to  French  practice,  the  diameter  of  a  rivet-head  =»  If d, 
and  the  length  of  the  rivet  from  the  head  =  7'+l§d. 

According  to  Rankine,  the  size  of  the  head  =  3d,  and  the  length 
of  the  rivet  from  the  head  =  7'  +  2irf. 

The  tliameter  of  the  rivet-hole  exceeds  that  of  the  shank  by 
from  sV  to  J  in.,  so  as  to  allow  for  the  expansion  of  the  latter  when 
hot. 

There  seems  to  be  no  objection  to  the  use  of  long  rivets  pro- 
vided that  they  are  properly  heated  and  secured. 

Strength  of  Punched  and  RiMed  Plates. — Experiment  shows  that 
the  tenacity  of  iron  and  steel  plates  is  considerably  diminished  by 
punching.  This  deterioration  in  tenacity  seems  to  be  due  to  a 
molecular  change  in  a  narrow  annulus  of  the  metal  around  the  hole. 
The  removal  of  the  annulus  largely  neutralizes  the  effect  of  the 
punching,  and  therefore  the  holes  are  sometimes  punched  \  in. 
h'ss  in  diameter  than  the  rivets  and  are  subse(|Uently  rinu-red  or 
drilled  out  to  the  full  size.  The  original  strength  may  also  In-  almost 
entirely  restored  hy  aiiiii-aling,  and.  gererally,  in  steel-work,  either 
this  process  is  adopted  or  the  annulus  referred  to  above  is  removed. 

runciiing  dotvs  not  sensil)ly  aiTeet  the  strength  of  Landore- 
SiemeiLs  uiiannealed  plates,  and  only  slightly  diminishes  the  strength 
of  thin  steel  plates,  but  causes  a  considerable  loss  of  tenacity 
in  thick  steel  plates;  the  lo.ss,  however,  is  less  than  for  iron 
plates. 


RIVETED  JOIN  'S. 


767v 


The  harder  the  material  the  greater  is  the  loss  of  tenacity. 
Iron  seems  to  suffer  more  from  punching  when  the  holes 
are  near  the  edge  than  when  removetl  to  some  distance  from  it, 
wliile  mild  steel  suffers  less  when  the  hole  is  one  diameter  from 
the  edge  than  when  it  is  so  far  that  there  is  no  bulging  at  the 
edge. 

The  injury  caused  by  punching  may  be  avoided  by  drilling  the 
lioles.  In  important  girder-work  and  whenever  great  accuracy  of 
worknmnship  is  required,  a  uniform  pitch  may  be  insured  and  the 
full  strength  of  the  metal  retained  by  the  use  of  multiple  drills. 
Drilling  is  a  necessity  for  first-class  work  when  the  diameter  of  the 
holes  is  less  than  the  thickness  of  the  plate,  and  also  when  several 
plates  are  piled.  It  is  impossible  to  punch  plates,  bars,  angles,  etc., 
in  spite  of  all  expedients,  in  such  a  manner  that  the  holes  in  any 
two  exactly  correspond,  and  the  irregularity  lx>comcs  intensified  in 
a  pile,  the  passage  of  the  rivet  often  being  completely  blocked.  A 
drill,  or  rimer,  is  then  dri  on  through  the  hole  by  main  force,  crack- 
ing and  bending  the  plates  in  its  passage,  and  separating  them  one 
from  another. 

The  holes  may  be  punched  for  ordinary  work,  and  in  plates 
of  which  the  thickness  is  less  than  the  diameter  of  the  rivets. 
Whenever  the  metal  is  of  an  inferior  quality  the  holes  should  be 
drilled. 

20.  Riveted  Joints.— In  /ap-joints  (Figs.  705  and  708)  the 
plates  overlap  anil  are  riveted  together  by  one  or  more  rows  of 
rivets  which  are  said  to  be  in  single  shear,  as  each  rivet  has  to  be 
sheared  through  one  section  only. 

In  fixh-  (or  butt-)  joints  (Figs.  706  and  707)  the  rivets  are  in  double 
■■^hear,  i.e.,  must  be  each  sheared  through  two  sections.    Thus  they 


^^^ 


Fm.  "(».5. 


S1& 


Fio.  706. 


Fio.  707. 


Fia.  708. 


are  not  subjected  to  the  one-sided  pull  to  which  rivets  in  single  shear 
are  liable. 

In  //.v/(-j()ints  the  ends  of  the  plates  meet,  and  the  plates  are 


,768 


THEORY  OF  STRUCTURES. 


riveted  to  a  single  cover  (Fig.  706),  or  to  two  covers  (Fig.  707),  by 
means  of  one  or  more  rows  of  rivets  on  each  side  of  the  joint. 

A  fish-joint  is  projx>rly  termed  a  butt-joint  when  the  plates  are 
in  compre.ssion.  The  jjlates  shoulii  butt  evenly  against  one  another, 
although  they  seldom  do  so  in  practice.  Indeed,  the  mere  process 
of  riveting  draws  the  plates  slightly  apart,  leaving  a  gap  which 
is  often  concealed  by  calking.  A  nmch  better  method  is  to  fill 
up  the  space  with  some  such  hard  substance  as  cast  zinc,  but  the 
best  method,  if  the  work  will  allow  of  the  increased  cost,  is  to  form 
a  jump-joint,  i.e.,  to  plane  the  ends  of  the  plates  carefully,  and  then 
bring  them  into  close  contact,  when  a  short  cover  with  one  or  two 
rows  of  rivets  will  suffice  to  hold  them  in  position. 

The  ri^■eting  is  .said  to  l)o  single,  double,  triple,  etc.,  according 
as  the  joint  is  secured  by  one,  two,  three,  or  more  rows  of  ri\'ets. 


o  o  o 

o  o  o 

O  0  o 

O  0  o 

o  o  o 

O  O   0 

o  o 


CHAIN 

Fio.  709. 


ZIQZAQ 

FiQ.  710. 


Double,  trijile,  etc.,  riveting  may  be  chain  (Fig.  709)  or  zif/zafj 
(Fig.  710).  In  Die  former  case  the  livets  form  .straight  lines  longi- 
tudinally and  transversely,  while  in  the  latter  the  rivets  in  each 
row  divide  the  space  between  the  riv<>ts  in  adjacent  rows.  Kxixri- 
nients  indicate  that  chain  is  somewhat  stronger  tiian  zigzag  riveting. 

Figs.  711  to  713  show  forms  f  joint  "sually  adopted  for  bridge- 
work.     In  boiler-work  tiie  ri\ets  are  necessarily  very  close  together, 


yo 

o 

o 

ON 

o  o 

0 

o 

0  o 

o  o 

o 

o 

o  o 

\o 

-^^ 

") 

oy 

°0° 


Fin.  712. 


Fio.  713. 


and  if  the  strength  of  the  solid  plate  be  assumed  to  l)e   100.  the 
strength  of  a  single-riveted  joint  hardly  exceeds  M.  while  double- 


RIVETED  JOINTS. 


769 


ri\pting  will  only  increase  it  to  60  or  70.  Fairbairn  proposed  to 
make  the  joint  and  unpunchcd  plate  equally  strong  by  increasing 
the  thichiess  of  the  punched  portion  of  the  plate,  but  this  is  some- 
wluit  difficult  in  practice. 

The  stresses  develojKHl  in  a  riveted  joint  are  of  a  most  complex 
character  and  can  hardly  be  subjected  to  exact  mathematical 
analysis.  For  example,  the  distribution  of  stress  will  be  necessarily 
irregular  (a)  if  the  pull  upon  the  joint  is  one-sideil;  (b)  when  local 
action  exists,  or  the  plates  stretch,  or  internal  strains  are  in  the 
metal  before  punching;  (c)  if  there  is  a  lack  of  .synmietry  in  the 
arrangement  of  the  rivets,  so  that  one  rivet  is  more  se -erely  strained 
than  another;    (d)  when  the  workmanship  is  defective. 

The  joint  may  fail  in  any  one  of  the  following  ways: 

(1)  The  rivets  may  shear, 

(2)  The  rivets  may  be  forced  into  and  crush  the  plate. 

(3)  The  rivets  may  be  torn  out  of  the  plate. 

(4)  The  plate  may  tear  in  a  direction  transverse  to  that  of  the 
stress. 

The  resistance  to  rupture  should  be  the  same  in  each  of  the  four 
cases,  and  always  as  great  as  possible. 

The  shearing  and  tensile  strengths  of  plate-iron  are  very  nearly 
•  qual.  Thus  iron  with  a  tenacity  of  20  tons  per  square  inch  has 
a  .shearing  strength  of  18  to  20  tons  yter  square  inch.  Rivet-iron 
is  usually  somewhat  stronger  than  plate-iron. 

Again,  the  shearing  strength  of  steel  per  square  inch  varies  from 
.•il)out  24  tons  for  steel,  with  a  tenacity  of  about  30  tons,  to  about 
Xi  tons  for  steel,  with  a  tenacity  of  about  50  tons;  an  average  value 
for  rivet-steel  with  a  tenacity  of  30  tons  being  24  tons. 

Hence,  if  4  be  a  factor  of  safety,  the  working  coefficients  become 

For  vvTought  iron  j  "'  ^T  ^'  ''^"'"^  '"'''  "'  ''''^^'■'  ^"^ 
f  5    "      "       "        "     "  tension. 

For  steel i  ^  *°"^  l^^  square  inch  in  shear,  and 

^  'i  tension. 

Allowance,  however,  must  ho.  made  for  irregularity  in  the  distribu- 

linti  of  stress  and  *"or  defective  workmanship,  and  in  riveting  wrought- 

iicti  platts  together  it  is  a  conmion  practice  to  make  the  aggregate 

-'■'•tion  of  the  rivets  at  kiust  equal  to  aiid  sumelimes  20  per  cent 

greater  than  the  net  section  of  the  plate  through  the  rivet-holes. 


770  THEORY  OF  STRUCTURES 

Hence,  the  working  coefficients  are  reduced  to 

4  or  4J  tons  per  square  inch  for  wrought  iron, 


and 


5or5i    " 


steel, 


according  to  the  character  of  the  joint. 

There  is  very  Httle  rehable  information  respecting  the  indenta- 
tion of  plates  by  rivets  and  bolts,  and  it  is  most  uncertain  to  what 
extent  tlie  tenacity  of  the  plates  is  affected  by  such  indentation. 
Further  experiments  are  required  to  show  the  effect  of  the  crushing 
pressure  upon  the  bearing  area  (i.e.,  the  diameter  of  the  rivet  multiplied 
by  the  thickness  of  the  plate),  although  a  few  indicate  that  the  shear- 
ing strength  of  the  rivet  diminishes  after  the  intensity  of  the  bearing 
pressure  exceeds  a  certain  maximum  limit. 

21.  Theoretical  Deductions. 

Let  S  be  the  total  stress  at  a  riveted  joint; 

/  >  lt>  /3.  /-  b<*  the  safe  tensile,  shearing,  compressive,  and  bearing  unit 
stresses,  respectively; 

t  be  the  thickness  of  a  plate,  and  w  its  width; 

JV  be  the  total  number  of  rivets  on  one  side  of  a  joint; 

n  be  the  total  number  of  rivets  in  one  row; 

p  be  the  pitch  of  the  rivets,  i.e.,  the  distance  centre  to  centre; 

d  be  the  diameter  of  the  rivets; 

X  be  the  distance  between  the  centre  line  of  the  nearest  row  of  rivets 
and  the  edge  of  the  plate. 

Value  of  X. — It  has  been  found  that  the  minimum  safe  value  of  x  is  rf, 
and  this  in  most  cases  gives  a  sufficient  overlap  ( -2x),  while  x  — Jd  is  a  maxi- 
mum limit  which  amply  provides  for  the  bending  and  shearing  to  which  the 
joint  may  ix-  subjected.     Thus  the  overlap  gill  vary  from  2d  to  3d. 

X  may  Im-  supixiscd  to  consist  of  a  length  x,  to  resist  the  shearing  action, 
and  a  lcn<ilh  x,,  to  resist  the  bending  action.  It  is  impossible  to  determine 
thcorctirally  the  exact  value  of  Xi,  as  the  straining  at  the  joint  is  very  com- 
iilcx,  but  the  metal  in  front  of  each  rivet  (the  rivets  at  the  ends  of  the  joint 

cx-coptpd')  may  Ix-  likened  to  a  uniformly  loaded  Ijeam  of  length  d,  depth  Xt-:^, 

;v.[i\  breadth  t,  with  both  ends  fixed.     Its  moment  of  resistance  is  therefore 

^ilx,-'-]  ,  j  iH'ing  the  maximum  unit  stress  due  to  the  bending.    Also,  if 

/'  i-<  the  load  upon  the  rivet,  the  mean  of  the  bending  moments  at  the  end 

P 
and  centre  is  rrd. 


RIVETED  JOINTS.  771 

Hence,  approximately, 

It  will  be  assumed  that  the  shearing  strength  of  the  rivet  is  equal  to  the 
strength  of  a  beam  to  resist  cross-breaking. 

Single-riveted  lap  and  single-cover  joints  (Figs.  705  and  706). 

~h-(p-d)tf,-dtf,;  2x,</,-^/,;      therefore  z,  - 1 ^ (1) 

^  It  /       d\'    nd'  J     ,     I     .,  r 

Sdi'^'-V    'Tf'-'     therefore  „J.^^.^s.'^_k ^2) 

As  already  pointed  out,  these  joints  are  weakened  by  the  bending  action 
developed,  and  possibly  also  by  the  concentration  of  the  stress  towards  the 
mner  faces  of  the  plates. 

Single-riveted  double-cover  joints  (Fig.  707). 

^\U-{V-d)th'-dt},;  therefore       2x,tf,~2~f,.     x-J-.    ...     (3) 

3-'J(---')'-.^f/.;.wo.  x.-f.L^. ,„ 

These  joints  are  much  stronger  than  joints  with  single  covers  Also 
equation  3)  shows  that  the  bearing  unit  stress  in  a  double-cover  joint  is  twice' 
as  great  (theoreticaUy)  as  in  a  single-cover  joint  (eq.  1),  so  that  rivets  of  a 
larger  diameter  may  be  employed  in  the  latter  than  is  possible  in  the  former 
for  corresponding  values  of  — . 

Chain-riveted  joints  (Fig.  709). 

/,(w-nd):  -S-f,Ndt; (5^ 

^"■'V^"!"/!  when  there  is  one  cover  only; (6) 

(S - jV-^-Zj  when  there  are  <i/o  covers (7^ 

This  class  of  joint  is  employed  for  the  flanges  of  bridge  girders,  the  plates    . 
being  piled  as  in  tigs.  714,  715,  716,  and  n  beine  usuallv  3.  4.  or  5 

In  Pig.  715  the  plates  are  grouped  so  as  to  break  joint,  and  opinions  differ 
H<  to  whether  this  arrangement  is  superior  to  the  full  buU  shown  in  Fig  716 


I  !  'H  I 


I 


772 


THEORY  OF  STRICTURES. 


Til"  iulvatitagos  of  the  latter  are  that  the  plates  may  1m;  etit  in  uniform  lengths, 
iui'l  the  flanges  built  up  with  a  decree  of  accuracy  which  cannot  lie  otherwise 
attained,  while  the  short  ami  awkward  pieces  accompanying  broken  joints  are 
dispeiisod  wi'h. 

A  good  practical  rule,  and  one  saving  much  labor  and  cxixnsc,  is  to  make 
the  lengths  of  the  plates,  bars,  etc.,  multiples  of  the  pitch,  and  to  design  the 


^    ^\    ^.    ^^    /'\    r~^   /~\ 


fb_ 


K^'     '•^"'J    '-^     \J     ^ 

Via.  714. 


^/-^^^^^/^/^.^^/-^^^^ 


V^ 


-r 


^i_ 


ni 


'  '^  \J  \^f  '^  KJ  \^  ^  ^  J  \.J  ^  ^ 

Fio.  71.1. 


-^y — O — C7 — '^     "^ — ^-r- 

FiG.  71C. 


covers,  connections,  etc.,  .so  as  to  interfere  with  the  j)itch  as  little  as  possible. 

The  distance  U'twcen  two  con.*cutivc  joints  of  a  group  (Fig.  713;  is  gener- 
ally made  e(]ual  to  liricc  the  jjitch. 

,\n  excellent  plan   for  hip  and  single-cover  joints  is  to  arrange  the  rivet 
ji-i  shown  in  Figs.  709  to  713. 

....  The  strength  of  the  plate  at  the  joint  is  only  weak- 

fli    '    I    I        cncd    bv   one   rivet-hole,  for  the   plate  cannot  tear  at 

I    I    I    j  •■  ' 

-|-- ] — I — I —    its  weakest  section,  i.e.,  along  the  central  row  of   rivets 

O    O  '    '       ("f)j  until  the  rivets  between  it  and  the  edge  are  shorn 


OXOTJD 


Oa9|    i       (Fif!-717). 

I  I  I  ! 


Let    ihere   lie    m    rows    of    rivets,  11,  2  2,  3  3,.., 


1    iP 
Fig.  717. 


The  total  numlx^r  of  rivets  is  evidently  m'. 

Ia'X  /,,  '/,,  </:„  Vi,  ...  l)e  the  unit  tensile  stresses  in 
'in-  i.  c  al(>iig  'he  lines  1  1,  2  2,  3  3,  .  .  .  ,  respectively. 
Then 


—  J2 

S={iv-  d)tU=='—nvji, 


for  the  line  1  1 ; 


=  (u-2<i)i</,^^(m'-l)/„"     "     "    2  2; 
=  («.-3r/)<7,=^(m'-3)/„"     "     "    3  3; 


=  (ir-4</)<7.=— (,m^-f))/„ 


It         tl  t( 


44: 


m 


Therefore 


5  =  (.f -(/)</.  =(.e-2rf) 


m'-l 


tq," 


EFFICIESCY  OF  RIVFTFA)  JOINTS. 


773 


Assume  that  /,  ^q,.     Then  w  =  (m-  + 1 )(/. 

IIi'iico,  by  substituting  this  value  of  w  in  the  first  of  the  above  relations, 
d      14/,      ^. 
y  = . .  J-     •Since  7,,  q,,  ...  are  each  less  than  /,,  the  assumption  is   u.stifiable. 

Coirrn. —  In  tcnt:i<>ii  joints  the  strenjith  of  the  covers  must  not  be  less  than 
that  of  the  |.lates  to  l)e  united.  Hence  a  xiiiyle  cover  should  \ic  at  least  aa 
tl'ick  as  a  oingle  plate;  and  if  there  are  two  covers,  each  should  be  at  least 
1.  ilf  as  thick. 

Wheti  I  wo  covers  are  used  in  a  tension  pile  it  often  happens  that  a  joint 
occurs  in  the  tc.'P  or  bottom  plate,  so  that  the  greater  portion  of  the  stress 
in  that  plate  may  have  to  be  borne  by  the  nearest  cover.  It  is,  therefore, 
considered  advisable  to  make  its  thickness  five  eighths  that  of  the  plate. 

The  numlx-r  of  the  joints  should  be  /educed  to  a  minimum,  r.s  the  intro- 
duction of  covers  adds  a  large  percentage  to  the  dead  weight  of  the  -lile. 

Covers  might  lx»  wholly  dis|)ensed  with  in  pcrjc'-t-jump  joints,  and  a  great 
economy  of  material  effected,  if  the  difficulty  of  forming  such  joints  and  the 
increased  cost  did  not  render  them  impracticable.  Hence  it  may  be  said 
that  covers  are  required  for  all  comprc><.sioti  joints,  and  that  they  must  l)e  as 
strong  as  the  ])lates;  for,  unless  ihe  jjlates  butt  closely,  the  whole  of  the  thrust 
will  l)e  transmitted  through  the  covers.  In  some  of  the  best  examples  of 
bridge  construction  the  tension  and  compression  joints  are  identical. 

22.  Efficiency  of  Riveted  Joints. — The  efficiency  of  a  riveted  joint  is  the 
rai  io  of  the  maximum  stress  which  can  be  transmitted  to  the  plates  through  the 
joint  to  the  strength  of  the  solid  plates. 

Denote  this  maximum  efficiency  by  jj. 

Let  p  be  the  pitch  of  the  rivets; 


d 
t 

h 

mj- 

h 
N 

c 


diameter  of  the  rivets; 
thickness  of  the  i)lates; 
tenacity  of  the  solid  plate; 

"    "    riveted  plate; 
shearing  strength  of  the  rivets; 
numlxT  of  rivets  in  a  pitch  length; 

uitlo  of  the  strength  of  a  rivet  in  double  shear  to  its  strength 
in  single  shear. 


Then 


T/i  =efficieni  y  as  regards  the  plate   = 


ij, -efficiency  as  regards  the  rivets  = 


(p-,r)tmj, 

ptf,  ' 

m(p-iT) 
P 


■N'^d'f. 
'^PlJT'' 


(1) 


(2) 


The  efficiency  of  the  joint  is,  of  course,  the  smaller  of  these  two  values; 
f.nd  the  joint  is  one  of  maximum  efliciency  when  ij,  =  ijj  ■=  ij ;  that  is,  when 


i     : 


y- 


774 


THEORY  OF  STRUCTURES. 


m 


p-d         4 


Ptft 


or  (p-d)tmJt^eN'j-dy, (3) 

In  this  expression  the  quantities  m,  ft,  N,  and  e  arc  constants  for  any 
given  joint,  being  of  necessity  known,  or  having  been  fixed  beforehand;  and 
the  equation  thus  expresses  one  condition  governing  the  relations  of  the  three 
variables,  p,  d,  and  t  to  each  other.  It  is  obvious,  however,  that,  in  order 
to  determine  the  values  of  any  two  of  these  variables  in  terms  of  the  third, 
another  relation  between  them  must  be  postulated.     In  short,  in  designing 

a  joint,  the  value  of  one  of  the  three  ratios  -->  ~  <i"d  —  must  be  fixed. 

a  t 

P 

C.KSfE  I. — Suppose  that  the  ratio  —  has  .1  certain  value.  This  is  very  fre- 
quently the  quantity  predetermined;  but  it  is  most  usually  done  by  fixing 
the  value  of  ij,  n  very  obviously  involving  — ;  in  fact  5  -m  (1 j  . 

Equation  (3)  may  be  wTitten 

\    t  mjt 

If  the  ratio  —  be  denoted  by  A:,  then 

|-fiV^A-4+l (S> 

a         4   m/i 

.              m(p-d) 
Hut  since  ij- — — , 

P       »« 

t-m-n («) 

Therefore,  substituting  in  (f)), 

-'--cA'jt4; (7) 

m-r,         A    m}t'        ••••••••    v>/ 

and,  ultimately, 

t      4    '^_1_.    .         (gj 


EFFICIEXCY  OF  RIVETED  JOINTS. 


775 


The  process  of  designing  a  joint  of  maximum  eflSciency  for  a  boiler  of 

given  diameter  and  pressure  of  steam,  when  ij  (or  the  ratio  j)  is  fixed,  is  then 

as  follows:   Settle  the  number  of  rivets  per  pitch  (i.e.,  JV);    the  value  to  be 
allowed  for  e  (depending  on  the  nature  of  the  shearing  stress  on  the  rivets); 
and  the  values  of  m,  jt,  and  /,.    Then  k  is  known  from  equation  (8). 
But  t  may  be  found  from  the  relation 


pressure  X  diameter  -  ij  X  2//,, 
pressure  X  diameter 


«■ 


2nh 


(9) 


H.-ncr  since  k~-  is  known,  d  may  be  found;  and  since ^ ^is  known. 

"    '  a     m-i)  ' 

;;  is  also  fixed. 

Case  II  —When  j,  the  ratio  of  rivet  pitch  to  plate  thickness,  is  given, 

equation  (5)  must  be  otherwise  manipulated. 

Multiplying  it  by  -,  and  substituting  for  d  its  value  kt,  we  have 


A   p     mji     p 


Putting  this  in  the  form  of  a  quadratic  equation  in  it, 

fAT^t   /  ^     fAr,T  /.   7""- 


(10) 


(11) 


For  brevity,  substituting  A  for  — ,  T  for  ^',  and  R  for  ^,  and  solving  the 
<|uudratic, 


47'     I 

*  -  -  -y  ±  -VA  T  +  4^  TR. 


(12) 


The  method  of  designing  the  joint  is,  then,  as  follows: 
.1,  T   and  R  being  known,  k  may  be  found  by  substituting  their  values 
ill  ( quation  (12),  the  positive  sign  of  the  serond  term  being  taken. 

,.,(,-i).„(,-t').„(,.|), 

and  since  both  k  and  R  are  now  known,  the  thickness  of  plate  (0  may  bo  found 
:!-<  ill  Case  I,  by  equation  (9).  The  values  of  the  diameter  and  pitch  of  rivets 
follow  at  once  from  the  known  values  of  k  and  R 

This  method  of  designing  a  joint  apfjears  to  In    the  most  ratiima!  of  th« 
three.    For  the  greatest  pitch  for  which  a  joint  will  remain  steam-tight  de- 


il 


i  jl 


77G 


IHEORY  OF  STRUCTURES. 


pciuls  mainly  on  tlie  rclatidii  of  pitch  of  rivets  to  thickneas  of  plates;  althou;;h 
It  is  also  atTectcil  l)y  the  rolativo  size  of  rivets  and  of  rivet -heads. 

Case  III. —  If  -  ,  or  /.-,  i>e  predetermined,  the  value  of  ^omust  first  bo  obtained 

ni  order  ll-.at  the  i)late  thickness  may  he  found  by  means  of  equation  (9). 

Now,  T.  =111  -     -  inav  l>e  put  into  the  form 
P 

ntd 

i>  = ; 

in  —  1, 
ind  if  this  value  is  substituted  for  /)  in  e<|Uation  (4), 

ni'l         iiNr:     /» 


III  - 


',      ^    \      in,  I       / 


I'lom  this  is  tinallv  deduce*! 


i\rl./,+4infi' 


(13) 


The  [ilate  thickness  may  now  1k>  found  by  equation  (9);  the  diameter  of 
rivet  friim  </-<7,  and  the  pit<-h  from  n  =■  .In  the  above  investiffiitiona 

w  —  )J 

no  account  has  l)een  taken  of  the  effect  of  the  l)earinn  pressure  on  the  rivets 
or  plate. 

If  /r  hi'  the  allowable  l)earinf;  pressure  jter  projected  square  inch  of  rivet 
surface,  the  following  relation  must  obtain: 

(p-<l)t}nf,'N,lt}c .     (14) 


This  may  Ih>  written 


h 


(/(  -ihmjt 


(15) 


Then  if  /r  b"  estimated  by  this  e(|uation,  and  if  it  should  1)0  (troater  than 
4;5  tons  [H^r  scjiiHre  inch  in  a  lap-joint,  or  4.">  to  .")()  tons  in  .a  butt-joint,  such 
joint  will  fail  by  the  rivets  sliearitij;  before  the  full  strength  of  the  plate  is 
exerteil,  as  Kenncily's  c\|K'riiiirMit-<  siiow  that  with  these  values  of  fr  the  rivets 
(|o  not  attain  their  nalm.il  uiiiniate  sliearinj;  stren(:th  (viz.,  /«\  but  fail  at 
•<he:irini;  stresses  much  below  tlii-*. 

Auain,  the  maxinnini  allowable  ratio  -    (i.e.,  A)  as  the  preliminary  dututn 

for  ihe  dcsijrn  of  a  joint,  may  l>e  (i\cd  by  using  the  expression 


(16) 


/. 


deduced  ftoin  the  obvious  r<'lation — simil.'ir  to  (14)— 


(rnwin  suppests  the  relation  d  ^ '^\  I.) 


!J 


RIVETED  JOINTS. 


Til 


In  designing  the  joint  by  any  of  the  methods  give-,  above,  any  value 
obtained  for  k  greater  than  that  supplied  by  (10)  should  be  rejected. 

Sotr  oil  Friction  of  liivrtal  Jo/((Av.— Elaborate  oxiM-rinicnts  on  the  small 
displarements  produced  by  loads  on  riveted  joints  of  all  kinds  have  recently 
been  liiade  by  Cotisidere  (Aniiaies  des  Fonts  et  Chaussees,  ISSO),  Bach  (Zeit. 
d.  Ver..  189-.',  1S91,  IS!).')),  l)u|)uy  (Antiales  des  Fonts  et  Chaussws,  189:)), 
and  by  Van  der  Kolk  (Zeit  d.  \'er.,  June  1S97). 

These  show  that  the  frictional  resistance  produced  by  the  great  pressure 
of  the  riveting  is,  in  a  well-made  joint,  sufficient  to  transmit  the  required 
amount  of  force  across  the  joint.  In  fa-t  the  stanchncss  and  durability 
<n  the  joint  depend  ui)on  the  plate  friction,  and  not  Ujnm  the  shearing  strength 
of  the  rivets  or  the  tearing  strength  of  the  plates.  When  the  rivet  cools  it 
contracts  lengthwise,  and  the  longitudinal  tension  thereby  produced  induces 
a  cross-contraction  which,  added  to  the  diametral  contraction  due  to  cooling, 
in:ikes  the  rivet  in  the  finisli(>d  joint  a  loose  fit  in  its  hole.  The  shearing  .strength 
of  the  rivet  does  not,  therefore,  come  ii  to  play  until  the  plates  have  moved 
sufficiently  to  cause  the  rivets  again  to  Ix^ir  against  tb"  sides  of  the  hole. 
i:ven  when  thi;  diK's  hapix-n,  it  is  evident  that,  at  first,  only  a  few  of  the  rivets 
in  a  given  joint  ^vill  Ix-ar,  and  these  must  l)e  deformed,  or  must  give  way  more 
or  less,  Ix'fore  the  rest  r'  the  rivets  can  come  into  action.  Hence  it  is  the 
irictioiial  grip  of  the  |>lates  upon  each  other  which  i)revenfs  this  slipping,  and 
which  is  the  true  criterion  of  the  strength  of  the  joint.  When  once  the  plates 
slip,  a  slackness  of  the  whole  joint  will  Ik?  produced  by  a  reversal  of  load. 
:ind  in  the  ca.se  of  a  joint  which  is  reciuired  to  retain  a  lluid  under  pressure 
leakage   will  take  pl.-ice. 

The  latest  ex|ieriinenfs  by  \')<n  der  k  were  made  on  the  joints  of  bridge 
struts  and  ties  with  double-butt  strap  The  breaking  l<iad  was  not  deter- 
mined in  these  exiM'riments,  as  it  wa,<  not  considered  of  sufficient  imimrtaree. 
.\s  a  rule  the  stresses  in  riveted  joiiils  of  l.ii.l<;es  are  much  less  than  the  break- 
iiiU  loads  of  plates  of  even  the  lowest  teneiity.  The  question  is  not  which 
kind  of  joint  has  the  greatest  .statical  .str  xth,  but  which  joint  is  least  hkely 
to  lu'come  slai-k  under  the  action  of  reverst'd  loading. 

The  dis|ilacemeiits  observed  in  the  joints  imler  load  were  of  two  kinds — 
clastic  (or  dis.ipiMviring)  and  [-Ttnanent.  The  former  were  considered  the 
iiKire  crucial  in  dclining  the  In-st  form  of  joint,  as  the  (lermi.nent  set,  once 
taken,  is  hardly  increased  by  repealed  loading. 

The  elastic  e\tensions  were  smalle.  •  (1)  in  the  ease  of  hand-riveting,  and 
vMtli  holes  somewhat  too  large  for  the  rivets,  anil  ('.>i  in  the  cast-  of  ma<iiine- 
riveting  when  the  pressure  on  the  dies  was  much  greater  than  is  usual  in  prae- 
lie. 

It  is  very  lemarkable  that  hand-riveted  joints,  with  rivets  a  good  fit  in 
iliiir  holes,  allowed  larg<>  elastic  displacements  under  comparatively  small 
loads,  liivefinj;  with  the  machine  causes  the  rivets  to  fill  the  holes,  and,  unless 
a  very  great  [ircssure  is  applied  and  maintained,  produces  the  same  bad  results 
as  to  elastic  movement. 


,ii 


iv 


778 


THEORY  OF  STRUCTURES. 


V.x.    17. — Dcaign  nj  l27-}t.  through  rivetal  span. 

Divide  the  span  into  five  panels  of  2o.4  ft.  each.  Take  depth  centre  to 
<('ntre  of  (•hords  =  2S  ft.  and  width  centre  to  centre  of  trus.«e8  =  17  ft.  The 
live  load  will  consist  of  two  cons()li<hition  eu!iiiie.s  followed  '  ■■  a  uniform  train 
loud  of  3400  lbs.  jkt  lineal  foot  of  track  as  shown  in  Fig.  7Ki.     In  this  figure, 


sl     51    8t     5J  i!     51 

n       M        ni  5       Si      « 


~Z7^.  --,t3^>4-»^-  -,-/i-o-»  o'^-S-  .v_frU_i^iM<-5y^«-'-J;-5'^4«-iH|e»-i 


I    i  i  I  i    I    I  I i  I    I  Si  I 

■      "    ■    ■  s      g     i    5  i    -■     "-•''• 


a    «    -    a       2    t 
S    ■»    U    n       s    ^ 


r-m' — uo.oooJ 

S3^ 141,0«H' 

-si^ lei.ooOf 

— S6^ — 181.000' 

1\- — SOliOOOJ 

i»i ai.ooo 


i    ^   t 


-M- 


-*».4)00>l 

rahooo 

.KB^OOO*' 


-3«1.000>l 

loa.oooJ 

(SSiOOfrV 


8  «  a 


^100-- 


-««.0OOil 
-OtOOAjl 


Fio    71S. 

to  facilitate  calculation,  there  is  written  under  e-ii  h  wheel  the  B.M.  of  all 
|irccc(iiiij;  wheels  up  to  tliat    point. 

Also,  on  tile  horizontal  lines  is  shown  the  distance  from  the  front  of  the 
ciiiriiie  up  to  each  wheel,  and  the  sum  of  the  loads  up  to  and  including  the 
wheel  in  (|uestioii   is  j;iven. 

('oo|H-r's  siK'cificatioiis  will  lie  used  and  all  iKMidiiiK  moments  and  siiears 
will  lie  coni]iutcd  from  the  actual  wheel  concentrations. 

li.M.  at  I'diiil-iKiiiit  1,  Fig.  7l!t.— Consider  the  load  as  com'Ug  upon  the 
sjMti  from  the  right,     .\ccording  to  the  cntirinn  deduceil  in  .\rt.  8,  Cliaptcr  II, 


-,^|«tn.■l»(•i».4-l«• 


Fio    71!) 


the  B.M.  at  |)anel-point  1  will  lie  a  maximum  when  the  average  load  on  the 
left  is  equal  to  the  averain"  load  on  the  whole  siMin.     IMace  the  3d  driver  of 


DESIGN  OF  1S7-F00T  RIVETED  TRUSS.  779 

the  first  engine  at  1.     Then  x,  the  length  of  span  covered  by  the  uniform  load 
-U.b  ft.,  so  that  total  load  on  span  =  442,000 +3400X15.6  =  495  040  lbs' 

The  average  load  {jor  panel  =495,040  4- 5  =  99,008  lbs. 

With  the  3d  driver  on  the  right  of  1,  load  in  panel  01  =  79,000  lbs.  < 99  008. 
\\  hen  3d  driver  ie  on  the  left  of  1,  load  in  panel  01  =110,000  lbs.>  99  008  lbs 
so  that  the  criterion  is  satisfied  with  the  3d  driver  at  point  1.     Hence 

y«=jf7(24,030,000-f  442,000X15.6+  3400x15^^  ^31338912  ^^  ^ 


and 


B.M.  =  25.4ft  -  771 ,000  =  5,496,782  ft  .-lbs. 


Panel-pmnl  (2).— Criterion  is  satisfied  with  3d  tender  wheel  of  Ist  engine 
at  2.    Then  J  =13.2  ft.    Therefore 


R 


■j^y  (24,010; 


,000 +442,000  X  13.2 +?^i^^)  =5«Mlbs., 


a"d  B.M.=.50.8/?-4,159,000  =  7,905,242  ft.-lbs. 

Panel-point  (3).— Try  2d  driver  of  2d  engine  at  3.     Then  x  =  13.8  ft 

30453348 


and  R  =  ~-  (24,030,000  +  442,000  X  13.8  +  5M^13.8'\ 


127 


lbs., 


•">J  B.M.-76.2ff-10,181,000  =  S,091,010  ft.-lbs. 

The  «.M.  at  3  is  larger  than  at  2.  We  should,  however,  obtain  the  .same 
moment  at  2  by  letting  the  train  advance  from  the  left,  and  hence  we  must 
iis<>  the  larger  moment,  8,091,010,  in  computing  the  stres-ses  for  the  members 
\vhos<'  centres  of  moments  are  at  [wnel-point  2. 

M>iximum  S/imrs  for  Tnisx  .UpmW.v.— Hy  .\rt.  S,  Chapter  II,  the  maximum 
-heir  in  any  jwnel  will  occur  when  the  load  in  that  panel  is  equal  to  the  aver- 
■nzv  load  jKT  panel  on  the  whole  sfMin.  For  panel-point  1  this  occurs  when 
tlie  3d  driver  is  at  the  i)anel-i)oint.     Then  x=15.6  ft.  and,  aa  above. 

,.    31338912 

H ,._,7 246,763  lbs. 

TK      f  CD     17000xlS+31000x(10  +  5) 

Therefore  S-R p,^- ^  -216,409  Iba. 

Panel-point  2.— 3tl  driver  at  panel-point  2. 

«- j27('*>l'*'-'0"0 +-*0'-.000 X4.2)  - 156,145  Iba., 
1  nd  therefor  S  - 125,791  Ibo. 


iJi 


^1 


li 


"80  THEORY  OF  STRUCTURES. 

Pdiul-piiint  3. — '2d  ilrivor  at  3. 

/!•  =  —-'  S,S3ti,000  +  L'fiO.mK)  X 1  .S)  =  73,3S7  lbs., 

iiiid  therofoiv  S  =  ."),S,.">S4  lbs. 

li.M.  jiir  .'itn'tKiirs,  the  Iniijth  hcim/  L'.").4  jf. 

The  jn'catcst  M.M.  will  occur  when  four  drivers  .-iro  on  the  stringer,  as  shown 
in  Fin.  720.     lArt.  S,  Chapter  II.)     Then 


I 


j.v.|:,-|V.^W. 


->>   -1.86' 

I 
I 

I  ■■■^    7'2v 


A'  =  .7^,(31,000x4.-).8)  =.-).-,,897  lbs., 
-•>.4  ' 

.111(1        H.M.  - .-).-).s<»7  X  1 1 .4.") -31,000 X5 
-4s.),()20ft.-lb.s. 


The  preatost  llonr-ljcam  concentration  wnl 
occur  with  the  loads  so  placed  as  to  jjive  the  max.  U.M.  at  the  centre  of  a 
span  of  two  panel  lengths.  Place  3d  driver  at  panel-point.  Then  load  on 
floor-beam 

=  . ,y^( 31, 000 X. SI. (5  + 17,000 X 7.4 +20,000 X 22.2)  =  122,023  lbs. 
For  (ireatest  end  shear  on  stringer  place  1st  driver  at  panel-point.    Then 

Ii  =  S  = ., .  ^  i  20,000  X 2.4  +  31 ,000(  10.4  -^  l.-).4  - 20.4  +  2.>.4)  i  =  80,270  lbs. 
Table  of  Momf.xts  Axn  .Sheahs  for  Live  Loads. 


I'anel- 
pninlH. 


Mompnts  for 
S|  itui . 


Slipars  for 
.S|tan. 


.1,  KtC.TS.'  ft.-ll..s         2ir),l()9  lbs. 
V.OlM.fMO       "         I     V2\7m     " 
SOltl.diO       "         1       .5S,.').S4    " 


Mftnieiit!!  for 
Trii!*9. 


•.'.71S,391  ft-.lbs. 

i.o).-,,.v).">     " 
4,0  ».-),.'«>.';     " 


Sheurt)  ,'or 
TruM. 


108,20.5  ft. -lbs. 

ii2,,sn.-i  " 

29.292    " 


Ma\.  n.M.  for  striiiicer  =  48-.,020  :-2 -242,.")10  I't.-lbs. 
'        ((iiicetitratioii  lor  lloor-bi-ani -=  122,023  lbs.,  or  (il,OI2  lbs.  at  each  end. 
"      end  shear  for  stringer-8<),27ti -:- 2  =  44,(5,38  lbs. 


Alliiiriililr  flriysis: 

TimlxT  (cMrcnic  fibre) 

Mediutn  steel,  teiisil<'  stresses: 

Lateral  sw.'iy-br.Mcing  for  wind  strains 

Hottoni  tiaiijrcs,  riveted  lloor-beams  and  strineers.  . 


l.(X)0  11)s.  |jersq.  in. 


.    18,000   "      "    "     " 
.   10,(KK)    "      "    "     " 


Live  L.iad.       Deal)  Loa.i. 

Bottom  chords,  main  dijigoiuiLs,  and  long  verticals.  .  lO.fKK)  20,000 


DESIGN  OF  127-FOOT  TRUSS. 


Live  Load. 


781 


Dead  Load. 


20,000 -9(W 


17,000-90 


/ 


13,000-60 


('omprcusife  xtirsses: 

Chord  .sopmeiits 10,000-45- 

Posts  of  throuf;h-l)ridges 8  500-45— 

'  '  r 

Latoral  struts  and  rigid  bracing  (wind  .strains) 

Rivets  9000  lb.s./sq  in.  in  shear,  L-,,000  lbs.  in  bearing;  80  per  cent  of 
Mhovo  for  floor  system;    reduce  one-third  for  field  rivets. 

Tics  and  Guard  rails.— Upacm^  the  stringers  7  ft.  centre  to  centre  we 
tnay  use  the  same  ties  and  guard-rails  as  f-r  the  plate-girder  span  E.xample  44. 
(  haptor  \  II.  Ill 

*VriHj,cr..-Assume  dead  load,  including  weight  of  stringer,  as  700  lbs. 
|>cr  foot  of  span  or  350  lbs,  per  foot  of  stringer. 

Then  B.M.  (dead  load)  =  J  X350 X25.4'  =  28  225  ft  -lbs 

"     (live      "   )=  242,.510     " 

Total    "  270,735     " 

Take  a  38"  X f "  web.     The  efTective  depth  -36"  or  3  ft. 
Then  flange  stress  =  270,735  +  3  -  90,245  lbs. 

•""^  "     area    =  90,245^10,000  =  9.02  sq.  in.  net. 

U.se  two  6"X4"Xr'."  angles  =  10.62  sq.  in.  gross  aiTa=9.5  sq.  in.  net 

:i.>.a  for  each  flange,  the  flanges  being  assumed,  under  the  specifications    to 

cirry  the  entire  iH-nding  action. 

/  Tii^^  "o^i!*"w,*  "  ""^'^'^  ^*""  ""^''■'nediate  stiffeners,  spaced  3  ft.  centres, 
'•!',"'  i\„  1  ^'^K  ""*^'"^  ^°^  ""'•  ^t'ffeners.  For  stringer  bracing  use 
■h    XJJ    Xf"  angles,  as  shown  in  Fig.  721. 


..L^L,, 


Flo.  721.  p,„  722 

/•7(wr.Amm«.— Dead-load  eoneentr.  per  strinpjr- 350x25.4-   8  890  lbs 
Liv*'     "  "  "        "  -61,012  lbs.' 

■^^t*'  "  "        "  -69^  lbs. 

.\ssumc  weight  of  floor-l)eam  -2600  lbs.    Then,  Fig.  722, 
H.M.  due  to  weight  of  floor-beam  -  j^  X2600  X 17      -     5,525  ft.-lb8. 
"         concentration  -69,902X5  -.'i49'5l0     " 


Tota'BM 35503- 


t 
ill  I 


i:;i 


i;!  . 


782  THEORY  OF  STRUCTURES. 

Assume  a  48"  X  f "  web.     The  effective  depth  =  3.84  ft .,  so  that 


and 


the  flange  stress  =  355,03.5 -f  3.84  =  92,4.")"  lbs. 
"        "     area   =   92,457-10,000  =  9.25  sq.  in. 


Use  two  6"x4"Xi*,"  angles  =  10.62  sq.  in.  gross  area  =  9.5  sq.  in.  net  area. 
End    stiffencrs    3i"x31"Xi";     intermediate    stiffeners    3i"x3i"xJ", 
spaced  about  4  ft.  centre  to  centre. 

Stringer  Details. — The  end  shear  in  the  stringer  must  be  transmitted  from 
the  web  of  the  stringer  through  the  end  stiff  ling  angles  to  the  floor-beam 
and  thence  in  a  similar  way  to  the  post.  The  riv  ts  connecting  the  end  stiffeners 
to  the  stringer  web  are  in  double  shear,  hence  their  bearing  value  on  the  J-inch 
web  plate  will  govern. 

The  bearing  value  of  a  J-in.  rivet  on  a  J-in.  plate  is  15,000  x}  X. 8  for 
floor  system  (see  specification)  =3938  lbs.  for  shop  riveting,  and  =  2619  lbs.  for 
field  riveting. 

End  shear  on  stringer  =  44,638  lbs. 

Hence  the  required  number  of  rivets 
=  44,6384-3938  =  11+,  say  12. 

These  may  be  placed  partly  in  the 
angles  A  and  partly  in  the  fillers  B, 
Fig.  723. 

Value  of  |-in.  rivet  in  single  shear 
=  .6013X9000X0.8  =  4329  lbs.  for  shop 
riveting  or  2886  lbs.  for  field  riveting. 

In  the  connection  of  the  .stringers  to 
the  floor-beams,  bearing  will  govern;  and 

is  only 

2619   lbs.     The   concentration   from   two   stringers  =  69,902   lbs.     Hence  the 
numlier  of  rivets  nviuin-d  =  69,902 -^ 2619  =  26  +  ,  say  28. 

The  rivet  s])acing  in  the  flanges  may  be  determined  as  in  the  ease  of  the 
platc-ginlcr  Ex.  44,  Chapter  VII. 

CoHiiertiiin  nj  End  Stiffcner  to  Weh. — 
The  load  =  69,902 +  1300  =  7 1,202  IKs.  and 
the  re(|iiircd  nunilK>r  of  rivcts=  71 ,202 -^ 
3938=  IS +  ,  say  19.  Put  10  in  row  C,  9 
in  row  I),  Fig.  724. 

End  Stiffdicrs  to  Pout. — Single  shi-ar 
will  }.';i)vcrn.  The  value  of  each  rivet  = 
28H(')  ll)s.  and  the  required  nunilx>r  of 
rivi'ts  =  71,202  :-'S.S6  =  24  +  ,  say  26. 

Dnidl  iiid  Sirrssrs  Fig.  719. — Assume 
a  dead  ioiid  of  !KK)  lbs.  jmt  lineal  foot 
of  truss.  Let  /,*  =  panel  dead  load.  Then 
i;  =  900  X  25.4  =  22  ,S(iO  lbs. 


Fio.  723. 
since  the  riveting  is  done  in  the  field  the   Ijearing  value  p«>r   rivet 


Also,     tan  0- 
DtunO- 


.907 
20,734 


sec  0  -  1 ,35 
/)  sec  0=- 30,860. 


Fig.  724. 


DESIGN  OF  ler-FOOT  TRUSS. 

Assume  J  of  the  load  concentrated  at  the  top  chord. 

Then,                      stress  in  01-12  =  2Z)  tan  0  =  41 ,468  lbs.  T. 

"  "  2-3   =3/)  tan  5  =  62,202  "  T. 

"  "  00,  fee  =  3D  tan  0  =  62,202  ''  C. 

"  "  Oa    =2/)  sec  0-61,720  "  C. 

"  "  a2    =  Z)  sec  0  =  30,860  "  T. 

"  "  al     =iD           =15,240  "  T. 

"  "  62    =JD           =   7,620  ••  C. 


783 


Live-load  Stresses. 

Stress  in  01-12=2,748,391  -4-28=   98,157  T, 
"      "     2,3   =4,045,505-;- 28  =  144,482  7, 
"      "  afe,  6c  =  4,045,505-^-28  =  144,482  C. 
"      "     Oa 
"      "      a2 
"      "      63 


al 
62 


Equivalent  Live-Load 
Method  (for  comparison). 

96,760  T. 
145,014  7. 
145  140  C. 
108,205  secO  =  146,076  C.  144^020  c] 

=      62,895  secO=  84,910  7.  86,412  7. 

=      29,292  secO=  39,544  7.  or  C.      43,206  7.  or  C. 
=  floor-beam  concentration  =  61 ,012  7.      61 ,295  7. 
=  (shear  at  panel-point  2)  =29,292  C.      29,028  C. 


-^ind-Umd  Stresses.— Sperificatim.-'To  provide  for  wind  strains  and 
vibrations  from  high-speed  trains,  the  bottom  lateral  bracing  in  through 
bridges  will  be  proportioned  to  resist  a  L  eral  force  of  600  lbs.  for  each  foot 
of  the  span;  450  lbs.  of  this  being  treated  as  a  moving  load,  and  as  acting 
cm  a  train  of  cars,  at  a  line  6  feet  above  the  base  of  rail. 

The  top  lateral  bracing  in  through-bridges  will  be  proportioned  to  resist 
u  lateral  force  of  150  lbs.  per  lineal  foot  for  spans  up  to  300  ft. 

The  stresses  in  truss  members  from  assumed  wind  forces  need  not  be 
(vnsidered  except  (a)  when  the  wind  stresses  exceed  30  per  cent  of  the 
maximum  stresses  due  to  the  dead  and  live  loads  upon  the  same  member 
Mild  the  section  is  then  to  be  increased  until  the  total  stress  per  square  inch 
.i.K.s  not  exceed  by  more  than  30  per  cent  the  maximum  fixed  for  dead  and 
live  loads  only,  and  ^/>)  when  the  wind  stress  can  neutralize  or  reverse  the 
stress  ill  any  member." 

Lower  Lateral  System.— \\c  shall  use  a  double  system  of  rigid  lower  lateral 
bracing  (Fig.  725)  capable  of  carrying  either  compression  or  tension. 

Dead  wind  load  =  1.50  lbs.  per  lineal  foot,  so  that  the  panel  load  =  150x25.4 
-3810  lbs.  We  shall  assume  one  half  of  this  load  or  1905  lbs.  (-tf)  lbs  to 
!-•  .'arried  by  each  of  the  systems  Figs  726  and  727,  and  superpose  the 
results  to  obtain  the  total  stresses  shown  in   Fig.   725.     We  have 

tc  tan  0  *  2846  «•  sec  6  =  3423 

tan  0  =  1.494  sec  0=1.797 

Then,  Fig.  726      Stress  in  01  =nfe        =2m'  tan  0  =  5(i!)2  lbs. 
"      "   1 2  =  23  =  fee  =  3  «•  tan  fl  -  RX^H.   ' ' 
"      "  Oa  -2ir  sec  0  =  6846   " 

"      "  16  ~w    sec  0-3423   '' 


r 


Hi: 


if' 
f    ll 

111  I 


I: 


784 


THEORY   OF  STRICTURES. 


Fi< 


Similarly  tlio  -'r"ss(\-i  in  Fij;.  Til  may  Ix"  fnu?ul  and  honre  combined  as 
Fip.  72').     To  iiiul  the  clioni  stresses  from  the  moving  wind  load  of  450 

llw.  |ier  foot  run  we  have  only 
|^'->-T=''-"-7^;:'~/Tx     /T\   y\       to  multiplv   the   dead  load   wind 

stresses  bv  3.     llius 


Fi.i.  720. 


Fi( 


the  panel  load  h' 

=  450  X2.'>.1=1 1,430  lbs. 

Supiio.'se  this  moving  load  to 
eome  u|>iiu  the  span  from  the 
ripht,  Fis.  72S.  The  shear  and 
hence  the  diagonal  stre.s.s  in  each 
panel  will  Ix-  greatest  when  the 
load  covers  the  span  to  the  right 
of  that  panel.  We  shall  assume 
that  the  shear  is  divided  e(iually  Ix'tween  the  two  diagonals  in  each  panel. 


Hence  in  panel  1  diagonal  stress^.j  ■ 


1 

■  ■> 

V 

lOic 
.5 

sec 

0  = 

•20.540  lbs 

1 

y 

Oir' 
5 

se( 

0  = 

12,3J4 

i  i 

1 

X 

3-/ 

Sl>C 

0  = 

6,162 

tt 

The  coml)ined  wind  loads  in  diagonals  of  the  lower   lateral    system    are 
shown  in  Fig.  720. 


I7<i:«  T    l.'i"''"  T     •AVm  T 


\,^. 


i;u7t  c 


"     iiO'M  C  filial   C 


^X^ 


Fin.  728.  F'«-  729. 

Vpprr  LnUml  .S'/.v/cm.— The  stres.ses  in  the  tipper  lateral  system,  Fig.  730, 
due  to  a  load  of  1.50  lbs.  jkt  lineal  f.-ot  may  U-  found  similarly.  The  chord 
stresses  need  not  l)e  computed  as  titcy  v.il!  1h'  much  less  than  30  jx-r  ceiu  of 
the  .oinbincd  live  and  dead  load  stresses  ai.tl  will  Ik>  further  reducwl  by  the 
elTect  of  the  overturning  n\i>ment  of  the  wind. 

(tnrUirninc;  Moment  Due  in  \Yi»d.—  (n)  On  Vpprr  Chord— 1\\<?  wind  load 
carried  bv  the  upix'r  lateral  system  to  each  hip-2  X.3Sl()  =  7f)20  lbs.  This 
will  iiicn'as.-  the  reaction  at   the  ends  of  the  leeward  truss  by  an  amount 

n     7,()20  >:  ~  -  12,.5.50  lbs.     (Fig.  731 .) 
1 1 


PTf 


Fifi.  730. 


Fio.  731. 


Fio.  732. 


DESIGN  OF  127-FOOT   TRUSS. 


785 


The  rorrespondini;  romprossivc  stress  in  the  inclined  end  post  =  12,5.50  aecO=- 
10,942  lbs.  (Fig.  732-.)  The  eorresiKinding  tensile  stress  in  the  bottom 
chnrd  =  r2,.").")0  tan  0=11,393  lbs.,  the  latter  being  constant  throughout  the 
length  of  the  bottom  chord.  We  sludl  have  a  compressive  stress  of  the  same 
in.ignitude  throughout  the  leeward  toj)  chord,  but  as  the  direct  efTect  of  the 
wiriil  is  to  cause  tension  in  that  chord,  the  two  will  tend  to  neutrahze  each 
other  and  need  not   Ix*  considered. 

(/))  On  I'mi/i.—Tbc  wind  load  on  the  train  (4.")()  lbs.  jier  ft.)  is  assumed 
to  act  G  ft.  al¥)ve  the  base  of  rail  or,  say,  11  ft.  alwve  the  plane  of  the  bottom 
Literals,  Fig.  733. 

Panel  load  P=  2.).4  X  4.-)0  =  1 1 ,430. 

Hence  A*,  the  reaction  at  leeward  end  of  floor-beam  =  — V= -=7396  lbs. 

It 

Thus  the  overturning  moment  of    the  wind  on  the  train  produces  a  load  of 

Toitt)  lbs.  ix-r  panel  on  leeward  truss. 


f- 

._i;i. 

-> 

P  -^  11130 

■ 

-h 

1      II 

J      I 

^ 

a          b           t 

A 

/ 

\ 

X 

/. 

\ 

13410  TS 

ISUST  g 

,    4    s 

Fig.  733. 


Fig.  734. 


Fig.  734  shows  the  corresponding  stresses  in  the  end  jxist  and  bottom 
chord.  For  reasons  already  mentioned  we  need  not  consider  the  effect  on 
the  top  chord,  while  the  resulting  stresses  in  the  web  members  are  sufficiently 
small  to  be  disregarded. 

Table  of  Stresses. 


Member. 

T)ea.l 
Load. 

1. 

Live 
Load. 

2. 

Fixed 
Wind 
Load. 

3. 

Moving 
Wind 
Load 

4. 

Over- 
turn in  (T 
Moment 
Due  to 
W.  L.  on 

Top 

Chord. 

5. 

Over- 
turning 
Moment 

Due  to 
W.  L.  on 

Train. 

6. 

Maximum 
Combined 

Stresses. 

\^n 

f(17i:()(' 

4ii(;.sT 

414»>.ST 
G2:.()2T 

4i4f>ST 
414(i8T 
r.2i^n:c 
3n8:()T 

0 
1.5240T 
7620C 

14rfl70C 
OS  I  -,7T 
981,^,7T 

144482T 

981  .->7T 
9S1,->7T 
MH8:7C 
8I910T 
.39.>t4T 
(ilOI2T 
29292(' 

U)942(' 
11393T 
11393T 
11393T 

11393r 
1I393(' 

19970C 
1341<iT 
134  UiT 
20124T 

(i34ir.(-) 

;!.341G("i 

244  70M" 
187202T 
2213.>1T 

3nr.,-,o.5T 

Minimum. 
24383T 
l.-.84,5T 

'II  locward 

12         "        .... 

■-•;!      "     .... 

01  windward.  .  . 

12 

t\U~\tc. 

.'>ri92T 
1423()T 
1707GT 

.';f.92t' 
1423()C 

1707C.T 
42G9()T 
.-1228T 

!707('.("i 
.1209()(! 

'i2 

(.:{   

Iv  ol)tainod  by  combining  columns  1,  3,  and  5. 
As  there  is  no  approach  to  reversal  of  stress,  no 
other  members  need  be  examined  for  reversal. 

,'l 

/,  > 

Main   Truss  Srrlinn.i. — Bottom  chord  01,   12. 
section  in  these  two  panels. 


We  shall  use  the  same 


7S6 


TIIEOUY  OF  STRUCTURES. 


Section  for  dead  load  stress     41,468      -^ 20,000      -  2.07  sq.  ins. 
"   live      "        "        98,157      +10,000      -9.82"     " 


Total  load 


139,625;  total  section -11.89  '* 


.Avoruirc    unit    stress    for   dead   and    live    load    139,625  +  11.89-11,743    lbs. 

Section    required    for   combined    load— 221,354 

HrZT— — 1  +  (11,743  X  1.3)  =  14.5  sq.  in.  net.  Hence  cosn- 
"'  billed  load  governs.  Use  four  6"X4"Xj' 
angles— 19  sq.  ins.  p-oss  urea.  Deducting 
8  rivet-holes,  each  i  sq.  in.,  we  have  19—4 
=  15  sq.  ins.  net  area,  Fig.  735. 

The    angles    will    be    connected    by    stay- 
plates  spaced  about  3  ft.  0  in.  centre  to  centre. 
Bottom  Chord  23. 


Fio.  735. 


Section  for  dead  load     62,202+20,000 
"        "   live      "      144,482  +  10,000 


>  3.11  sq.  ins. 
■14.45  "    " 


Total  load  =  206,684;     total  section- 17.50  "    " 

Average  unit  stress  for  dead  and  live  loads  206,684  +  17.56-11,770  lbs.  Sec- 
tion required  for  combined  loads  =  306,505 +  (11, 770  XI. 3)  =20.0  sq.  ins.  Use 
four  6"x4"Xfi"  angles -25.64  sq.  ins.  gross  area -20.14  sq.  ins.  net  area. 

Diagonals.     ;2.    Section  for  dead  load    30,860  -f  20,000-1.54  sq.  ins. 
"        "   Uve      "       84,910^10,000  =  8.49  "     " 
Total  section  =  10.03  "     " 

Use  two  12" — 25-lb.  channels  =15  sq.  ins   gross  area  — 12.75  sq.  ins.  net 
area. 

bm-2in.  Dead  load  =  0. 

Live      ' '     39,,544  + 10,000  =  3.95  sq.  ins. 


Use  four  3i"x2i"xi"  angles  =  8.44  sq.  ins.  gross  area  =  6.94 
sq.  ins.  net. 

This  section  is  much  larger  than  is  actually  required,  but 
cannot  well  be  reduced  without  using  sections  below  the  specified 
limits. 


Fig.  730 


al.     Section  for  dead  load  15,240  +  20,000  =  0.76  sq.  ias. 
"       "    live      "      61,012  +  10,000  =  0.10  "    '* 

Total  section -6.86  "   " 


:,1 


Use  four  5"x3"xi"  angles  =  11.44  sq.  ins.  gross  area -8.44  sq.  ins.  net 
area 


DESIGN  OF  ler-FOOT  TRUSS. 


787 


Top  Chord.— The  same  section  is  required  throughout 
«n  Tn^  T  ^"-^Oylb-   channeb,  Fig.   737,   laced   top  and    bottom-23.52 
sq.  ins.    r  about  a«s  AB-5.43  ins.   (Carnegie),    i-25  ft.-300  ins.,  and 


—  -  56.    Hence 


and 


20,000-90^-14,960  lbs. 


10,000 -45y-  7,480  " 


n 


Fio.  737. 


FiQ.  738. 


The  section  for  dead  load -62,202 +  14,960-  4.15  sq  ins. 
"       "        "     live     "    -14,482-i-  7,480-19.32  "'   " 

Total  section -23.47  "    «• 

Hence  the  assumed  section  is  satisfactory. 

Vertical  Post  62.— The  lightest  convenient  section  we  can  use  under  th* 
specifications  will  be  two  12"— 25-lb.  channels,  which 
will   be  found  to    exceed  the    requirements   for   mere 
strength. 

End  Post  oa.  Fig.  738.— In  addition  to  the  com- 
pressive  stresses  in  the  end  post,  there  is  a  bending 
moment  due  to  the  wind  pressure  on  the  top  chord. 
The  wind  load  carried  to  each  hip,  as  already  shown, 
is  7620  lbs.,  of  which  we  may  assume  that  one-half,  or,' 
3810  lbs.,  will  be  carried  to  the  pier  by  each  post.  It 
may  also  be  assum  d  that  the  post    is    fixed   at   A, 

Fig.  738.  by  the  portal  bracing,  and  at  Bhy  the  anchorage,  so 
that  there  is  a  pomt  of  inflection  at  C  and  the  B.M.  on  the 

post   is  3810  x|-=  3810X180 -685,800  in.-lbs..  as  ;  is  about 

Fio  739.        fj^'  °!,  l^  ''"\    '^^^  *^«'*™«  ^bre  stress  caused  by  this 

A.v.         h^^,^^  *°  ^^^  •=*"*^  ^y  the  compi«88ive  stress,  must 
not  exceed  the  specified  limits.  ' 

Try  the  following  sections.  Fig.  739: 

One  cover  plate  20"  Xi"  - 10.00  sq.  ins. 
Two  15"— 50-lb.  channels  =  29.42  "     " 
Four  flats  4"  X  J"  =10.00"    " 

Total  section -49.42  "    " 

The  rover  plate  is  used  to  make  the  section  capable  of  resisting  bendine 
::.  a  transverse  direction.  The  flats  are  riveted  to  the  bottom  of  the  channels 
to  balance  the  cover-plat^  .and  krop  the  centre  uf  gravity  of  the  sectfoH  t 
the  centre  of  the  channels,  so  that  the  centreK>f-gravity  line  of  the  end  posts 
w,U  intersect  that  of  the  other  members  at  connecting  points. 


iJi 


id 


H!! 


H'* 


f  -  I 


788  THEORY  OF  tiTRLCTURES. 

Moment  of  Inertin  nhont  AH. 

C<m-r-plate=T's>'-H)-<(i)'  +  10x:.7-j'  =  001.9 

Chamicls      =-.>x40l.'.7  =  «'>-'i-4 

Flats  =JX,'3X4X1.25'  +  10XS.1L'.V-  fifil.5 

Total  moment  of  inertia  =2068.2 

Moment  «/  hirrtin  about  CD. 

C.ver-plate  =  .'j  X  J  X20»  =  333.3 

Channels      =^2  X  11.22  ^29.42  X(ti.S-)=  ^1112.4 

Flats            =;;Xix2X4'  +  10xS.2.r  =  t>!«.9 

Total  moment  of  inertia --2439.6 
or  about  2200,  allowing  for  ri%et-holes. 

Alx.ut    AB,  '•  =  \"^Mr''=<i^"   '"»•;    ^   (unsupported   length  in  vertical 
plane)  =  45;j  ins.     Hence  —  =  70  so  tliat 

17,000-90 ^10,700  lbs. 


and 


8,r)(K)-45— =  o,3.')0 


tHTtion  for  dead  load  =   61,720-10,700  -   5.77  sq.  ina. 

■=27.30  "    " 


live 


146,076-   j,3.tO 


Total  load  =207,796    Total  section -33.87  "     " 

Allowable  stress  for  dead  and  live  loads 

=  207,796  ^  m.Ory  -  628.3  lbs. 
Allowable  stress  for  combined  loads  -628;j  X  1.30  =  8168  lbs. 

(W.-)800  y  1()._>.-,  ,, 

The  fibre  stress  due  to  iMMuhng  =  -jjoo  =.il .w 

The  stress  due  to  combined  compression  loads  -49.V2 


Total  iibn'  stress     8147   " 

And  na  this  is  slightly  less  than  the  allowable  intensity  (SKW  lbs.)  the  sec 
tion  is  sjilisfactory. 

ClionI  Splins.—  lu  the  top  chord,  for  convenience  in  erection,  only  one 
splice  will  1k>  used,  placed  in  the  middle  panel.  The  abutting  ends  should 
be  |ilaned  so  as  to  in.-  --•  jierfect  contact,  and  in  that  case  full  n'liance  may 
»x'  placed  on  llirir  Ix-ario);  against  each  other,  the  function  of  the  splice-plate^ 
U-ini:  to  hold  the  two  «>cticms  in  phor.  The  Ixittom  chord  will  U'  spliced 
in  the  second  p.inel  from  each  cn.l.  The  rivets  may  U'  nm»ngi-d  in  double 
shear,  so  that  their  U-aring  will  limit  their  value.  IVaring  value  of  a  I-in- 
r.vot  on  i-in.  angle  -  l.'i,(KK)  v  J  v  J  .,(i.-,6J  for  shop  rivets  or  4:J7.i  for  field 
.ivels.    Thirty  |)tr  cent  extra  whift-  wind  bad?  arr-  inilmied  gi\T»  tho  value 


DESIGN  OF  127  FOOT  TRUSS. 


789 


r>rKS,S  =  4(),  whioh  is  the  numlx'r 


of  a  field  rivet  =5f)SS  lbs.     Heiuc  2'21,3r)4 
of  rivets  required  in  the  s|)li<'e. 

[.tiurr  LnUral  lirtuiiuj.—Thv  lower  lateral  bracing  ^Fig.  740)  will  be  riveted 
to  the  bottom  Haiiges  of  tile  stringers  at  .1,  li,  V,  and  JJ,  the  jxjuits  of  inter- 
sv'ction;  so  that  tlie  greatest  unsup- 
|H)rte<l  length  in  a  vertical  plane  will 
U'  AC  or  III),  that  is,  about  l.'il  ins. 
The  gn-atest  stres.'^  in  a  diagonal  is 
'_'7,3St)    lbs.    coniijre.ssion    or    tension. 

llie  ratio 


e^l 


-must  not  exceed    120,  and       ^    ~~^^°~ 

\vc  sh.'ill  then-fore  use  two  4"x;j"x|" 
angles  (  =  4.!)  sc).  ins.)  with  the  4-in. 
legs  riveted  ba<'k  to  back  at  intervals 
of  1  ft.  (Fig.  741.)     r=l'.'(),  thenfore 

^   ^     B      13;()0()-(}0^-5800    lbs. 


F 

.1 
1 

i . 

! 

1 

; 

i 
1 

:, 

lioituui  Cbord 
FlO.  740. 

lol:- 1.26=120.     .\llowablc  stress 
that    the    section    required 


so 


IS 


Fia.  741         27,3.H()-:- .>SOO«4.7  sq.   ins.,  or  slightly  less  than  the  sj-ction 
used.     This  stctioii  will  of  course  lie  larger  than  is  neeessar- 
for  the  panels  nean-r  the  centre  of  the  s[mu;   bui  any  reduction  would  make 
/ 
-  too  great,  so  that  the  same  section  will  be 

used  for  all  bottom  laterals.    Transverse  angles 

MJ  and  liC  should  also  bi>  used,  to  provide  for 

the    longitudinal    thrusts    arising  fmni    braked  /r; 

trains.     ,\t  the  intersect  ion  O  one  of  the  diagonals  rK 

must    1h!  spiic(>d  as  shown  in   Fig.  742,  and  a 

suHicient  number  of  rivets  should  Ix-  us«>d  here 

as  well  as  in  the  end  cont'-ctions  not  only  to 

carry  the  maximum   stns;   liut  to  develoiie  the  Fig   742. 

full  stn-ngth  of  the  diagonal. 

T/j/xr  lAiUml  liriirimj.—'\'\n'  stresses  in  the  up|xr  laterals  an-  very  light, 

and  the  ratio       will  lie  the  governing  consideration. 

The  distance  lieiween  the  outer  rivets  in  the  connecting  plates  m.-iy  lie 
tjikcti  as  1,"».{  ins.    If  two  angles  4"  v;j"  y  J"  nre  us<-<l,  arningi'd  us  in  Fig.  743, 

r- 1.2  and        -121. 


Q     O      O    O  1  O 

1 

<i    o     y    o 

^^- 

_    «      J 

The  distance  back  to  back  of  the  angles  is 
l."i  ins.,  the  same  as  the  ilcplli  of  the  top  clionl. 
For  the  transverse   struts   use   four  unglcH 
:<"X2J">^I".  with  a  single  line  of  lacing. 

PoiIiiIm.  The  total  wind  load  at  c;:ch  iiip 
is  7<»2l)  lbs.  Assuiniii!:  that  the  end  jxmts  an- 
fixed  at  the  lower  end,  there  will  lie  a  |Miiiif  of  c(>ntr,'ille\un>  !>t  .1,  Fig.  744. 
Ijfh  jK^st  may  !«■  !i.--!inied  !-..  •••.•!!!--r:r  !  =  .  <h.-  pier  <>!!!•  !i;ilf  >.f  tin  wiiii!  hsitl 
ut  the  hip,  and  tlH-n-foiH'  //,  the  horizontal  reaction  at  .(,-asiO  IIhi. 


Kki    7I;{. 


S: 


I 


-^! 


790 


THEORT  OF  STRUCTURES. 


Also,  taking  moments  about  -4,   1'  = 


7620X23.1') 
17 


=  10,377  lbs.    Stress  in 


OY^V  spc  45°  =  14,673  lbs.    Or  the  stress  in  OY  may  be  obtained  by  taking 
moments  about  L\,  whence 


"l 

J^O      u 

jsno 

H-MIO 

V 

H-3«0 

it 

*'         1 

A    • 

Fio    744 


^  stress  in  OY  ^3810g3.15  ^^  ^^^  _  ^^^^^  ^^^  ^ 


Taking  moments  about  O  of  the  forces  to  the  left, 


stress  in  XX .3810x23  15 -10.377x8  5 _q 


Hence  XX  and,  conseciuently,  l/iA'  have  no  stress     Taking  moments  about  F, 


.     „„     3810(14.6.5+8.5)     ,-,„„,.      _ 
stress  m  l/iO- ^,- -10  377  Iba.  T 

0.0 


The  wimo  section  may  be  used  for  all  the  portal  members  as  for  the  upper 

iatmi's. 

I  P(«W(7«.— Chock  the  strength  of  the  jwrtal  members  and  design  the  con- 
ned ions.) 

The  stresses  in  the  sway-bracing  at  the  centre  panel-points  are  indeter- 
ininatc.  Thcs*'  braces  make  the  structure  more  rigid  and  dimini.sh  vibra- 
tion.    Hcyotid  securing  a  satisfactory  value  for  the  ratio  —,  their  design  is  a 

ni.itter  of  judgment.  At  the  point  of  their  connection  with  the  post  a  dia- 
lilirajtm  must  lie  inserted  U-tweeri  the  two  channels  of  the  post,  to  avoid  undue 
Ix'tiding  stn's>.  ■;. 

.S7i,.r.«.— The  total  weight  of  the  sjwn,  itwluding  the  track,  will  Ije  about 
'.'•.•.").(XM)  lbs.,  of  which  11'.»,.'>»M)  will  lie  carried  at  each  end.  The  gn-atest  reac- 
tion du<>  to  the  live  load  at  either  end  will  U'  l.'82,000  lbs.  Hence  the  total 
load  to  l)e  c.'irried  bv  each  sh<K'   will  be 


l(llJ,.-.0()  .2H-.>,(KK))  or  1<)7,0»K)  lbs.,  nearly. 


The  allowable  pressure  on  the  masonry  is  2.50  lbs.  fxT  mjuare  inch.  Hence 
hearing  area  n'<|uin'd  '  l<t7.(MK)     2."iO  -  7HS  m\.  ins. 

The   Iwise-piate   will   lie   niaiie  27"  "-'30"  "810  sq.   ins. 

Tlie  shoe  at  one  end  must  rest  ii|K)n  rollers,  so  as  to  provide  for  expansion. 
The  niller  shot-  is  shown  in  Figs.  74.'i,  74(1,  the  angles  and  vertical  pl«to»  being 


-'     a&h 


DESIGX  OF  1S7-F00T  TRUSS. 


791 


of  }-in.  metal,  and  the  horizontal  plates  J-in.  thick.  A  5|-in.  pin  may  be 
used,  the  method  of  desif^ing  this  pin  will  be  shown  in  a  subsequent  example. 
The  allowabic  bearing  pressure  for  this  pin  per  inch  of  length -5^X15,000 


Fio.  745. 


Fio.  746. 


=  7S,750  lbs.,  so  that  the  length  of  bearing  required  on  the  vertical  plates  is 
107,000  +  78,750-2.5  ins.,  while  the  four  }-in.  plates  provide  3  ins.  The 
strength  of  the  vertical  plates  should  also  be  checked  as  columns.  These 
plates  must  transfer  the  wind  pressure  to  the  pier.  The  wind  pressure  on 
the  siwn  is  1.50  + 600  =  7r>0  lbs.  per  lineal  foot,  so  th^t  the  horizontal  pressure 
at  each  pedestal -i(1J7  v7.10) -LM.KIO  lbs.,  half  of  which. or  say  12,000  Iba., 
must  be  carried  by  each  web.  The  weakest  section  will  be  at  B,  and  the 
li.M.  at  B-12,000"x7i- 90,000  in.-lbs. 

90000x6 
Hence  fibre  8tpp.s8-7,j^  ,..,,- 10,000  lbs.  per  square  inch. 

As  this  combined  with  the  direct  compression  would  be  somewhat  large, 
a  diaphragm  should  Ix?  provided.  The  smallest  diameter  permissible  for 
the  friction  rollers  for  a  127-ft.  span  is  3|  ins.  and  the  allowable  pressure  per 
lineal  inch  is  3i  X.'JOO- 937.5  lbs.  Hence  the  total  length  of  roller  required 
for  one  shoe  is  197,000+937.5-210  ins.  We  shall  use  eight  roUefB  261  «ns. 
long. 

Provision  should  Ix-  made  for  a  change  of  length  of  about  IJ  ins.  in  either 
direction,  or  3i  ins.  in  all. 

The  U'st  detail  for  a  roller  bearing  is  a  cast-steel  box  in  which  the  rollers 
may  be;  phurd.  The  1k)x  is  then  filled  with  oil  to  protect  the  rollers 
against  rust  and  accumulations  of  dirt.  For  the  fixed  shoe  a  cast-steel  grid- 
iron may  lie  u«>d,  the  siMices  lietwin-n  the  nbs  Ix'iiig  filled  with  cement  Each 
shoe  should  Ix?  fastened  to  the  masonry  by  means  of  anchor-lxilts  1^  ins.  in 
diameter. 

Cambtr. — Sufficient  camlx>r  may  Ix'  proviiled  by  lengthening  the  top 
chord  it  '"•  in  every  10  ft.  In  one  panel  length  this  will  amount  to  A  in., 
so  that  the  length  of  the  r»iiel  at  the  top  chord  will  lie  25'  4H"-t-A"-25'5|". 
The  diagonals  muit  of  course  be  lengthened  in  proportion.  The  make  up  of 
the  nicmbera  and  a  general  elevation  of  the  span  are  shown  in  Fig.  790,  pain* 

sr.». 


i 


792 


THEORY  OF  STRUCTURES. 


Ex.   18.  Df.vi';//!  «/  a  ,")■_'()-//.  Kifinij  Kpdii. 

The  span  will  bt;  dcsiiincil  to  carry  a,  double  track  railway  between  the 
tnis-scs,  while  on  cautiK-vcrs  outside  of  the  trusst's  provision  will  be  made 
for  coinbiiuil  hif;hw,iy  and  motorway  traffic,  and  for  pedestrians.  The  fol- 
lowing live  loads  will  )h'  used: 

For  railway  striti-jcers,  Waddell's  Class  "  U"; 

For  railw.'iy  lloor-ljeani.s,  hangers,  and  subdiagonals,  Waddell's  Clasa  "  S" 
combined  with  the  effects  of  the  loads  from  the  cantilevers; 


il! 


QO- 


oc^   00 


1 


-84 —  r6i^lu-r,i 


Jioiorwk;  Azk-  CoocfDtrmtioiM  . 

Fi(i.  747. 


For  main  truss  members,  Waddell's  Class  "  V  "  combined  with  the  affects 
of  cantilever  loads; 

For  the  roadway  jxirtion  of  the  cantilever  brackets,  Class  "  B"; 

For  each  motorway  track,  the  live  load  shown  in  Fig.  747. 

Figs.    748  and   749  give   the  curves  of  Equivalent  Uniform  load  for  thia 


111       ■Jt        3U        III         M        '■"         '•«        ">       Wl        iw 
f<j';in  il'  I'Vt't 

C.  U.  WOTUHWAV  LOAC^  FUR  PLATI  OWDIM 

I  ic    7  IS 


last.  The  ('(luivaleiit  iiiiifunn  loads  for  otluT  classes  may  Iw  found  in 
Waddell's  "  1)..  I'oiiiilHi."; 

For  the  -^iili'walks.  Ciiiss  "  C." 

The  span,  Ki :.  7".tl,  p.  SI:',,  «ill  eitnsi>^t  of  one  tower  panel  of  ,'}()  ft.  S  ins. 
fid  fourteen  panels  of  .!  I  ft.  \\,\  ins,,  or  altogi-ilier  .'»'.'()  ft.  )  in.  In  com- 
!'!i!!!ii_'  t!ie  loads  !!■•■  !>.!!!■,•!-.  '.v!!l  !k-  r!-i-.i!!»!-i)  :{.'  f!.  !iit!»,  the  di^tnnce  from  rentri? 
to  centre  of  trusses  iM'ing  :!l  It.  s  ins. 


DESIGN  OF  5S0-FOOT  SWING-SPAS. 


793 


The  details  of  the  Hoor  and  handrails  are  in  part  governed  by  the  .pecifica- 
fons.  The  ties  may  Ix.  compute.l  as  in  prevous  examples.  The^neril 
arrangement  of  the  Hour  is  as  showu  in  Fig.  750.  generai 


too      110       IW 


■sa     800     3UI     310     5a     5» 
Span  111  Feet 

E.  U.  MOTORWAr  loWM  FOR  TNUM  MAM 

Fm.  749. 


Stringcrs.-{a)  Sidewalk  Stringer.-lMening  to  Fig.  750  it  will  be  seen 
that  the  sidewalk  stringer  will  carry  one  half  of  the  sidewalk.    The  dead 


Tl»i«lo'«lo-isc.  toC. 


Fio.  7.')(1. 

1-a.U-ill  consist  of  the  handrail,  one  half  the  floor  and  the  weight  of  Ihe  girder 
ilsclf,  say  I.}!)  lbs.  ix-r  lineal  foot  of  ginler. 

w..  W. '""*''  *''"'  "  ^''"  '"'  ■'"'""•  '^"'-'^^■^  ^^-  f^'"-  «q"a"'  foot.     Hence 

J^'"'  '"•'"' -  13!)  lbs.  per  linear  foot  of  stringer 

Live      "     -7<i..")X2..5 -l<)l    "     "       "        <<    k        «< 

Inipai'f  (.'5.'>-ft.  sjKui) -.'»4'^[.  .     "UKi    "     "       "        "     "        «« 
Total  load —4li3    "     "       "        '«    <«        <« 

The  max.  R.M.  -  J  x4;W  X.r>'-flfi,400  ft.-lbs. 

J!f  ff,"'*^""  '"''*^'  '^'  •^'•■•'K"''''  '"^  '"   previous  examples.     We  ihall  use 
a  30    x,\     web;    four  angles  3"xii"xA"  for  flanges,  and  the  usual  end 


• 


r 


1 

in 

g- 

sIHI 

^: 

■IBJ 

i: 

aB 

ir 

'  -II 

'f 

1 

^ll 

794 


THEORY  OF  STRUCTURES. 


stiffoners  and  fillers,  the  total  weight  being  2130  lbs.,  or  about  60  Ibe.  per 
lineal  foot  of  stringer. 

(h)  Motorway  Stringers. — Outer  stringer.    The  dead  load  will  be  as  follows: 


Handrail 

Floor-timber 

Kail  and  spHces 

Weight  of  stringer  (say). 


60  lbs.  per  foot  of  stringer 

204    **      a      **     **        <  < 

on     it       It       it      tt  It 

J.)?      It        it         it       tt  n 


Total  dead  load =419 

Live  load  from  sidewalk 191 

Im|)a<'t  load  from  sidewalk  (54%). .. .  103 

Live  load  from  motorway 1020 

Impact  load  from  motorway  (74.89c).  "61 


Total  load. 


=2494 


The  max.  B.M.  =  ix2494x35'  =  382,000  ft.-lbs. 

We  shall  use  a  36"Xft"  web  and 
6"x3i"Xl"  angles  for  flanges,  the 
total  weight  of  the  stringer  with 
details  being  41.">7  lbs.,  or  119  lbs. 
per  lineal  foot  of  sixin.  The  inner 
stringer  will  carry  the  same  load, 
less  that  due  to  the  sidewalk.  The 
flange  angles  may  lie  reduced  to 
.')"x3i"Xj",  the  total  weight  being 
391.-)  lbs. 

lutrrmeiUntc  Cnntilrirr  BrnckHn. — 
Fic.  7.51.  Concentration  from  sidewalk  stringer 

at  .1,  Fig.  751. 


De.'id  load  -  l.jn  x35 

Cantilever  and  laterals  (say).  .  .  .  .' .  , 

Total  dead  load 

Live  loail  (7()-ft.  span)  73X2.5X35. 
Impact  (7()-ft.  span)  45.4''; 


4Sr>5  lbs. 

735  " 

.5600  " 

63S.S  " 

29(K)  " 


Total  load -  14,888 


Similarly  assuming  tlvit  the  iH)rtions  of  the  weights  of  the  cantilevers 
and  laterals  coni-cnt rated  iit  li  and  ('  an'  KKM)  lbs.  and  4.'>(X)  lbs.  respectively, 
we  h;ive  the  tittal  concentration  at   /i^(W,175  lbs.  and  at  r  =  61,975  Wnt. 

The  stresses  may  now  Im-  determined  gra])hicallv  and  the  s<>ction.s  and 
details  designed  in  the  usu.nl  way.  The  weight  of  each  intermediate  canti- 
lever will  tx'  nl>out  510  lbs.  The  general  dimensions  for  the  cantilevers  nt 
each  end  of  the  spfin  will  i)e  the  same,  but  as  they  have  to  carry  only  the  load 
U|)on  half  of  one  |>unel,  inst«>ad  of  half  of  two  [wnels,  they  iray  be  made  lighter. 


DESIGN  OF  A  StO-FOOT  SWISGSPAN.  795 

The  weight  of  each  will  be  about  4350  lbs.  Hence  the  total  wci  ht  of  metal 
in  the  cantiWers  wiU  be  5100x28+4350x4-160^00  lbs.,  or  210  lbs.  per 
lineal  foot  of  span. 

(c)  Railway  stringers: 

Equivalent  live  load,  Class  "  R  "  (35-ft.  span). . .  -     8450  lbs. 

Impact  74.77% _     6320    '* 

Dead  load  (including  floor) .     1100    " 

Total  load „  15,870  '• 

per  lineal  foot  of  track,  or  7935  lbs.  per  lineal  foot  of  stringer. 
B.M.  -iX7935x35'  =  1,215,000  ft  .-lbs. 
Assume  a  web  of  62"  xf";  the  effective  depth -4.87  ft. 

Flange  stress- 1,215,000^  4.87-250,000  lbs.  (about). 
"       area   -2.50,000-f  14,000-(i  of  web) -17.8-2.9 
=  14.9  sq.  ins. 

Use  two  6"X6"X}"  angles  =  16.88  sq.  ins.  gross  area -15.38  sq.  ins.  net 
area. 

The  weight  of  one  stringer  will  be  8637  lbs.  or  (say)  2;50  lbs.  per  lineal 
foot  of  stringer.  The  tower  span  is  only  30  ft.  8  ins.  in  length,  so  that  the 
stringers  will  weigh  about  6250  lbs.,  each. 

The  stringer  bracing  will  consist  of  4"x4"x|"  angles,  attached  to  the 
top  flanges  and  arranged  as  shown  in  Fig.  752,  with  a  cross-frame  at  the  centre 

Botlcm  Chonl 


Bolloai  ChdrJ' 
FlQ.  752. 


Fio.  753, 


of  each  panel,  Fig.  753.  Including  the  necessary  connecting  plates  and  rivets, 
it  will  weigh  80  lbs.  per  lineal  foot  of  sfian.  Hence  the  total  weight  of  stringers 
will  Im'  S7.')0x.-.« +62.^,0 x4.. -,15,000  lbs. -900  lbs.  per  lineal  foot  of  simn. 
Or,  total  weight  for  stringi^rs  and  bracing -990 +80 -1070  lbs.  per  lineal 
f(M>t  of  span. 

Intermetliate  f7(H»r-ifom.-(.— Concentration  from  railway  stringers: 

Kquivalcnt  live  load  (70-ft.  s,)an),  Class  "  S  ". .  .   -    ft320  lbs.  per  lineal  foot 

Impact  70.18% _     44.50    "     <<       <i       << 

Ik-iui  load  (not  including  tlooring) 


44.50 
-      900 


ToUl  load -11,670 


1 1 


It 


!    ! 


!  !  : 


It 


796 


THEORY  OF  STRICTURES. 


000 
Hence  the  dead-load  coiiceiitratioii  for  each  stringer  is  -:^  X35  =  15,750  lbs., 

and  the  corres|M)ndinj»  live-load  concentration  =     .,      x3.")  =  lS,S,r)(X)  Iba. 

The  (loor-lM'ain  and  its  cantilevers  will  then  Ik'  loaded  as  shown  in  Fig.  754, 
to  which  must  lie  added  the  weight  of  the  floor-beam  itself,  which  will  be 
assumed  as  IS,.")tX)  lbs.  uniformly  distributed. 


m 

aj2 


3i  i 

ii| 

i 

11 

i 

lis 

ill 

^-4^ 

^d5 

ir^ 

Hi 

Fio.  751. 


The  maximum  positive  H.M.  will  occur  in  the  fl(M)r-beam  when  the  latter 
carries  its  full  load  and  the  cantilevers  carry  only  the  dead  load.  The  mini- 
mum |)ositive,  or  maxinumi  negative  H.M.  (should  reversal  take  place)  will 
occur  when  the  cantilevers  are  fully  loaded,  and  the  floor-beam  carries  only 
the  dead  load. 

The  H.M.  at  lentre  of  floor-beam  due  to  full  load  on  same 
-  40,S,.-,(K)  X  12.;«  -•-»04,_'.->0  X  7  +  J  X  l.S,.->00  X30.67  =  +3,678,000  ft.-lbs. 
\ty.  H.M.  due  to  dead  load  on  cantilevers 

-1<»,37.-)X3  '-lt»,3(K)xrj+."j,()OOXlS._>5=  -    3.')f>,000      " 


Maximum  i«)sitive  H.M =  k3,.32-J,000 

Pos.  li.M.  due  to  dead  load  on  lloor-lteam 

=  3i,.VHi    r.'.;w-i.".,7.")Ox7  ^ixi>*.">oo>:w-fi7..  =  -  .340,000 

Xrg.  H.M.  due  to  full  load  on  cantilevers 

-  1 ,07.")  X3  +t>S,17.")  X  iJ  +  14,S7.J  X  18.25=  -1 .275 ,.500 


Maximum  negative  H.M.  .  .. 


-    02C.,.5(M) 


Tlic  liaiip'  stress  will  n'vcrse,  and  hence  (Waddell's  Specifications,  p.  17) 
we  nuist  add  to  the  sectional  an-a  nMjuired  for  the  greater  stress  three  fourths 
of  that  n'(|uired  for  the  less.  Or  we  may  add  three  fourths  of  the  smaller 
H.M.  to  I  lie  larpcr,  and  design  the  section  for  the  sum,  viz., 

3,.J22.0(K)^  J  X!)2f),.j(K)-  4,017,(KK)  ft.-lbs. 

.\sstmie  a  web  of  0(»"xj":  the  effective  depth  =  K  ft.;  then  flange  stress 
- 4.01 7,(HK):-S-.-)02, 125  U)s.,  and  flange  area -.502,125 -:- 14,000 -(J  of  web) 
-  35.S7 -•'..()     •.MI.S7  sq.  ins. 


DESIGN  OF  A  520-FOOT  SWIXOSPAN.  797 

Use  two  angles,  8"  x  8"  x  A" =17.52  sq.  ins. 

"    Olio  cover-plate,  16"  X  I'd" =9.00    "      "* 

"    one  plate,  16"  Xi" ". . . .    =   8.00   "     " 

Total  arra =Qd -.o    "      "    „-«„»        i- 

.■»4.oj  gross  section 

=  :M..V-'-4.:J7  =  30.1.">  sq.  ins.  net  section,  which  is  sufficient. 

It  will  be  fountl  that  the  outer  cover-plate  must  extend  to  about  the  outer 
striiijrcr  as  shown,  Fig.  7.5.5. 

The  floor-lx-am  must  be  cut 
;iHay  at  the  lower  corner  to  make 
room  for  the  bottom  chord  and 
l>in.  The  web  will  be  reinforced  by 
I  wo  plates  :>•)"  xyx  :,'h"  as  shown. 

The  floor-beam  will  weigh  17,500 
lbs.  (as.sumed  weight  18,500).  Total 
weight  of  fl(X)r-beams  =  17,i500xl4 
=  245,000,  or  470  lbs.  per  linear  foot 
of  span. 

(N.H.  The  weights  of  the  two  floor-beams  at  the  tower  will  not  be  con- 
.•^i.lcred  now,  since  they  will  be  carried  to  the  pier  without  inducing  stress  in 

the  trusses.) 

Lowfr  Laterals— Thne  cases  must  be  considered  for  the  wind  stresses 
in  the  lower  lateral  system. 

(n)  .\  ilead  load  of  30  lbs.  per  square  foot  of  exposed  surface  when  the 
span  IS  open.  This,  from  a  rough  computation,  will  amount  to  670  lbs. 
jier  lineal  foot  of  span. 

Hence  panel  load  =  ic  =  670x35 -23,450  lbs.;  sec  fl- 1.516; 

taa  »=  1.139;  —77—  =  17,800;  u-  tan  0  =  26,700. 


Fig.  755. 


The  stresses  will  be  as  shown  in  Fig.  756,  the  stres.ses  in  the  diagonals 
JH-ginning  at  the  left  being  successively  i,  IJ,  2i 


„,   ^.         tf  sec  0 

OJ  times  — ^ —    and 


I  he  chord  stresses  in  the  same  way  being  i,  U,  3J,  6J,  lOJ,  15i,  21},  28}  times 

ir  tan  0. 


\  \ 

1       \ 

§ 

8 

! 

8 

8 

tf 

X 

X 

X 

X 

X 

X 

X 

•.7WC 

M,««)C 

ae,(MC 

miwc 

«»,KIOC 

«ue,Moc 

M7,0WC 

«H,(IMC 

Fio.  756. 

(h)  The  span  closed,  and  a  movins;  l.uid  of  1 100  lbs.  r-er  Hnca!  foot  on 
one  arm  only;    u-panel  load  =  1100x35 "3S,.500.     The  atreaws,   Fig    757 
may  be  found  as  in  the  riveted  truss,  Ex.  17.  ' 


i  ! 

r! 
II 


I:  ■ 


798 


THEORY  OF  STRUCTURES. 


(r)  The  span  closed,  and  a  moving  load  of  1100  lbs.  covering  both  arms. 
In  this  case  we  have  a  girder  continuous  over  four  supports. 
Panel  load  =  1100x35  =  38,500  lbs 


»«">     "s.ooo    ioJMO    wSoo    iM;3oo     i»,ow     ii,m  ■ ' 
Fio.  757. 


The  following  table  gives  the  reactions  in  pounds  at  the  end  and  centre 
for  loads  at  each  panel-point  of  one  arm,  and  at  the  corresponding  panel-point 
of  the  other  arm,  Fig.  760. 


Pointn 

Reaction  at 

Reaction  at 

Loaded. 

End  in  Lbi. 

Centre  in  Lbs. 

1-1 

30,800 

7,700 

2-2 

23,500 

15,000 

3-3 

16,200 

22,300 

4-4 

10,400 

28,100 

5-5 

5,400 

33,100 

6-6 

1,900 

36,600. 

The  stresses  for  each  panel  load  may  readily  be  obtained  either  analytically 
or  graphically.  For  example,  consider  a  load  at  panel  point  3;  7?- 16,200  lbs. 
Hence  the  shear  in  the  first  three  panels  is  16,200  lbs.  and  the  corresponding 
stress  in  each  diagonal 


10200  sec  e 


=  12,290  lbs. 


The  shear  in  the    -niaining  panels  is  1 6,200 -38,.500-  -22,300  lbs. 

•)'>30Q 
Hence  diagonal  stress  =  -— ,-  sec  (?=  -17,570  lbs. 


Fig.  758. 

Fig.  758  shows  the  stresses  found  by  combining;  the  efT«cls  of  (he  several 
loads. 


DESIGN  OF  A  BgO-FOOT  SWING-SPAN. 


799 


The  greatest  stress  in  the  lower  laterab  is  thus  115,700  lbs.  T  or  C,  in  the 
panel  next  the  tower. 

We    shall    use    four    angles    con- 
nected by  angle  lacing,  Fig.  759. 

The  top  angles  will  be  riveted  to 

the  stringers  at  their  intersections,  but 

the  lower  angles  will  be  unsupported 

in    a  horizontal  plane  for  nearly  half 

their  length,  or  say  240  ins.    Try  four 

angles  each  5"x3V'xi",  arranged  as  shown,  Fig.  761:    area -12.2  sq.  ina. 

/     240 
r-2.4ins.;  — -— -loo.    Hence 


allowable  stress  per  sq.  in. -16,000-60—- 10,000  Iba., 

and  the  section  required -115,700 +  10,000 -11.57  sq.  ins.,  so  that  the  above 
section  is  ample.  For  the  sake  of  rigidity,  and  to  avoid  the  use  of  a  large 
number  of  different  sections,  we  shall  use  the  above  for  all  lower  laterals. 

The  total  weight  of  the  lower  lateral  system  will  be  about  84,000  lbs.  or, 
say,  160  lbs.  per  lineal  foot  of  span. 

The  loads  on  the  upper  laterals  will  be  much  lighter,  but  the  sections  will 

lie  determined  by  the  ratio  — ,  which  for  wind  bracing  must  not  exceed  140. 

The  section  may  be  similar  to  that  for  the  lower  laterals,  using,  however 
4"X3"XJ"  angles,  and  2i"xi"  bars  instead  of  angles  for  lacing,  in  the 
diagonal  members  [/,[;„  etc.,  and  5"X3i"  angles  for  the  transverse  strute 
l\l'„  etc.,  Fig.  762. 

Instead  of  making  the  upper  lateral  bracing  continuous  throughout  the  entire 
length  of  the  span,  it  is  omitted  in  the  panels  t/.C/,  and  {/,£/,.  Bracing  shown 
by  the  dotted  lines.  Fig.  760,  is  inserted  in  the  plane  of  l\M^  (Kg.  761) 


Lq         Ul  Ui        Ut         u«  u,        0,  u,       u, 


^XIXIXIXIX^  M 


■-«       .  ">  "•  U.  U4  U,  U,  U,        0, 

Fio.  760. 


nnd  a  portal  similar  to  "lat  in  the  plane  of  f/.L,  is  introduced  in  the  plane 
'/„/.,.  The  object  of  this  is  to  transfer  the  wind  load  on  the  top  chord  to 
tho  pier  in  the  most  direct  manner  possible,  which  tends  to  secure  both  economy 
ni.l  rigidity.  The  bracing  in  the  tower  panel  U,U,  consists  of  adjustable 
r.Mirs,  simply  intended  to  keep  the  tower  posts  in  position.  The  i^rtals  in 
p;itiils  f7,/.j  and  MJ^  may  be  designed  rs  in  the  riveted  span  Ex  17.  All 
tlic  portal  struts  consist  of  four  angles,  5"  X3i"  xf",  arranred  as  in  the  laterals. 


800 


THEORY  OF  STRUCTURES. 


The  total  weight  of  metal  in  the  portals  and  upper  laterala  will  be  about 
104,000  lbs.,  or  200  lbs.  jx-'r  lineal  foot  of  span. 


Ili! 


IK 


s 


U,*4_U»*4 


Fig.  7ti2. 


L 


Via.  764. 


The  stresses  in  the  vertical  sway  bracing  between  the  trusses  are  indeter- 
minate, and  the  ratio  —  will  be  the  governing  consideration  in  their  design. 

We  shall  use  struts  of  the  same  section  as 
U:       Ut  the  diagoiials  of  the  upix-r  lateral  system, 

|\/         Uj^ -i«  the  general  arrangement  being  as  shown  in 

l/\|        DXJ  Figs.  702-766.     The  horizontal  transverse 

strut    is   brought   to   within   23   ft.    (the 
specified  clearantv)  of  the  base  of  rail. 

The  total  weight  of  metal  in  sway- 
bracing  will  be  about  7.5,400  lbs.,  or 
145  lbs.  yicT  lineal  foot  of  span.  Hence 
the  total  weight  of  upper  and  lower 
laterals,  jjortals,  and  sway-bracing  will 
be  100 +  200 +  145  =  .50.5  lbs.  per  lineal  foot  of  sjian,  of  which  about  460  lbs: 
will  Ix!  carried  by  the  trusses. 

The  dead  load  carried  by  the  trusses  may  now  be  in  part  computed  and 
in  iMirt  assumed. 
Dfnii  L<mdx. 

Tracks  and  tiooring 1910  lbs.  per  lineal  foot 

.^idcwalk  stringers,  2  at  00  lbs 120    "     "       "       " 

Motorway       "         2  at  110  lbs 2.38    "     "       "       " 

2  at  112  lbs 224    "     "        "       " 

Cantilevers 100    "     "       "       " 

laterailmicing  for 100    "     "       "       " 

Riiilwav  stringers  ;ind  bracing 1070    "      "       "       " 

Floor-li^ams 470"     "       "       " 

Handrails 300    "     "       "       " 

Bohs  for  floor 48    "     "       "       " 

Lateral  system 400"     "       "       " 

Trusses  and  ojierating  machinery  (assumed) 3700    "     "       "       " 

Total  dead  load =8950"     "       "       " 

or  4475  lbs.  per  lineal  foot  of  each  truss. 


DESIGN  OF  A  etO-FOOT  f^iWING-SPA! 


801 


The  average  panel  load  would  be  4475x35=156,600  lbs.,  ut  je  panel 
load  will  evidently  increase  from  the  ends  toward  the  tower,  so  iuk.  we  shaU 
assume  the  loading  given  in  Fig.  765. 


1  »V>«iC|., 


.•>»S,uW)e  L,  i_  l.oj'j.gouo 


I  1      I      I 


Li 


I  I  ^  §•  ? 

Fig.  765. 


When  the  span  is  open  the  entire  weight  rests  on  the  rentre  pier,  and  we 
have  two  cantilever  arms  balancing  each  other.  When  J.e  span  is  closed 
the  same  is  true,  except  in  so  f.-ir  as  provision  is  made  for  lifting  the  ends 
ami  giving  tnem  a  firm  bearing  by  means  of  wedges  or  equivalent  devices. 
I  he  cnd-liftiiig  gear  will  be  designed  to  exert  an  uplift  of  120,000  lbs.  at 
each  end  of  each  truss. 

The  stresses  will  be  considered  under  four  conditions  of  loading. 

(1)  Span  open,  dead  load  only  acting.  The  resulting  stresses  are  readily 
obtained  graphically,  Fig.  766,  and  are  shown  in  Fig  765 


^^y^ 

xf 

X 

Fio.  766. 


f2)  All  uplift  of  120,000  lbs.  at  the  end  of  the  truss.  The  resulting  stresses 
■i.^^  found  graphically,  arc  indicated  in  Fig.  767.  As  this  uplift  may  not 
■i  I  ways  act,  the  resulting  stres.ses  are  considered  only  wben  the  v  increase  the 
-'re.s.ses  due  to  other  conditions  of  loading. 


802 


THEORY  OF  STRUCTURES. 


(3)  Live  load  on  one  arm  only,  which  is  then  assumed  to  act  as  a  single 
epac. 


L^imoioT  L,  i.»«.0WTC,      332,000  7      L»     388,000  f      L 

i 


Equivalent  live  load,  Class  "V," 
"  "      "      motorway 

"  "      "      sidewalks 


Fro.  767. 

'  245-ft.  span -4525  lbs.  lin.  ft.  of  truss 

*  '  <*        _      ""R       *  (         it       il      li  it 

it  **      ,m    'XM)     **       **      an       a 


Total  live  load -5590 


Hi. 


Rpferrinft  to  Fig.  754  it  will  be  seen  that  the  most  unfavorable  loading 
on  a  truss  will  occur  when  the  motorways  aii<l  sidewalks  outside  of  that  truss 
and  the  railway  tracks  Ix'tween  the  trusses  all  carry  their  maximum  load, 
while  the  motorway  aii:5  sidewalk  on  the  other  side  are  empty.  This  condi- 
tion of  loading  may  n-adily  pn'vail  over  one  or  two  i>anels,  but  is  (juife  unhkely 
to  <lo  so  over  the  whole  s|mn.  Its  effect  will  thi-refore  lie  coiisidere<l  in  com- 
puting the  stn-sses  in  the  floor-lx>am  hanja-rs  and  sub-diagonals,  but  wil; 
be  ignoreil  in  pro|K)rtioninK  the  main  truss  members. 

Panel  live  load  -  .'(.V.N)  v ;j.-,  =  lo:.,().j()  lbs. 

The  choril  St H'sses  will  lie  gn-atent  when  the  live  load  covers  the  whole  span. 
The  reliction  at  each  end  is  then  l!».'>,f».'i()  v;{  ,,v<ti,i).50  IIjs.,  and  a  single  stress 
diaKTiim  (Fig.  708)  gives  all  the  choni  strcAnes. 

To  find  the  stresses  in  I'J.t,  V,l'„  i',L„  aaaumc  a  reaction  of  100,000 


Fio.  7M 

lbs.  at  the  Ifft  end  o'  the  truss.     Fiir.  7rtO  gives  the  resiilting  straams,  vii.. 
99,000.  70.000.  and  70,400  lbs.,  n-s|H?i  tively.    Now  tl»e  strem  in  U,Lt  u  a 


DESIGN  OF  A   680-FOOT  SWISG-SPAS. 


803 


maximum  when  L.  and  all  the  panel-points  to  the  right  of  it  carry  the  fuU  live 
r^  f'r-    The  corresponding  reaction  at  L.  is  ^  X  195,650 X (1  +2  +3  +4  +5) 
-41V50  lbs.    Hence  the  stress  in 


l\L, 


4192.50x99000 

100000        "415,000  lbs.  (nearly). 


fror^tho  n'i/t^''/  '^"'  '""'^  *  '"""i^um  stress  when  the  live  load  extends 

Corresponding  reaction  at  /.,=-{  X  195,6.50 X (1+2  +3  +4)  -279,500  lbs. 
so  that 


and 


...«-.:..  77  7      279.500x70000 

stress  .n  U,L,^        ^^^^      -  =  ,95.000  lbs., 

.     ,.  ,      279.500x79400 
stress  in  I  ,/..» :^^^^^^^ —  -=222.000  lbs. 


100000 


Mm.larly  the  tensile  stress  in  U\l,  will  be  a  maximum  when  the  live  load 
extends  fron.  the  lejl  end  of  the  span  up  to  /.,.  l{.action  at  /.,.167,7r 
From  tig.  770  the  stn-ss  in  UM.  due  to  a  reaction  of  100,000  lbs.  at  A,  is 
2.35,000  lbs.  Hence  the  maximum  tensile  stiww  in  /. J/,  - 235,000  X  "'^'''" 
-304.000  lbs.  '  !"<>•"' 

When   the  live  load  advances  to  /.„  it  will  be  best,  on  account  of  f!u> 


Fio    770. 


Flo.  771. 


.  .ondary   menilxTs,  to    make  a   special    diairrHm     Fiff    771       P»o„.- 
/.,-279..5()OI»H.,whcn,vstn.s.si„.l/;f-,-4rM(K)li;     *       '"     ^"'"""   "^ 

The  stn.ss,.s  in  ,hc  hangers  T  A,.  .1/,/.,.  ,„„|  .,/./,.  ^j,,  ^ 
tlie  hvc  load  covers  (w.j  paiu'JM.  K«"iM8t  wm  n 


Equivalent  live  load  for  sick-walk  (70-ft.  simui,  CIh-s  "  ('  ")  -  n  v  -;    i«k  iu. 
Iier  lineal  f.Mif  of  Bi<k'walk.  '     '•»  x.i-.ififi  lbs. 

("on.-entration  on  fl<Ktr-»)cain  -36.5  x. 15-  12,77,5  ||„, 

K.|uivalci.t  live  I.Hid  f..:  inot.rway  -  N;iO  ll'm.  ,».r  foot  of  track 

ConciMilration  from  one  track  on  rt<K)r-ln.am  -  l4.Wx.35- -50  0.50  Iha 

F>,uivalent  liN-,.  |.M,d  for  railway  (Class  "  S  ")-«2.35  »«,.  ',..;  f.w,.  ..f  t„,k 

t^nrentration  rr..in  oIk-  tmck  on  fl«H.r-bi.«m-(«25x;W->2l  375  lbs 


I! 


^1  I 


804 


THEORY  OF  STRUCTURES. 


Assuming  one  sidewalk  and  motorway  unloaded,  we  have  the  floor-beam 
i.nd  cantilevers  loaded  as  in  Fig.  774,  where  li  is  the  reaction  on  the  hanger, 
the  concentration  from  each  track  being  placed  at  the  centre  line  of  the  track. 

Hence  /?x:iO.(57  =  221,37.") x.mf)7  +  ")0,0.'>0x38.17  + 1-',775X45.79,  or  ft  = 
:«)3,IK)0  lbs.  (nearly)  =  stress  in  t',A,,  MfL^,  and  MJ.^.  The  sub-diagonals 
MJ-i,  .U„/^5  will  e;ich  transfer  one  half  of  the  load  in  the  hanger.  Hence 
stress  in  each  is  lx;J03,(KX)  sec  0  =  2.33,000  lbs.  C. 

The  stress  in  the  hanger  VJ'i  will  l)e  greatest  when  the  live  load 
extends  from  />,  to  /-,.  Taking  the  equivalent  uniform  loads  for  a  span 
of  14(1  ft.  in  the  sjune  way  as  alxive,  we  find  the  stress  to  lx>  535,500  lbs.  T. 
Fig.  773  shows  the  stresses  in  all  members  for  a  live  load  covering  one  arm  only. 


Fio.  773. 


(4)   Live  load  covering  lx>th  arms,  which  then  form    a  girder  continuous 
over  four  sup|xirts. 

K<|uivaleiit  live  loads  .V_'()-ft.  sjian: 


Sidewalks,  4;{.."i  x.'i. 

.Motorwiiy 

lUilwav  (Class  "  \'  ' 


Total  loail. 


■   '.MS  Ills,  per  lineal  foot  of  trUM 

;    425    "      "       "        "     ■ '      " 
41',)0    "      "       "        "       '      " 

-  JHIt't     '  *       '  *         *  *  *  *      '  *       ' ' 


Fund  load  ^4S;Wy3.'>'- Kilt,  155  lbs.  ^'ic     It  will  iie  assumed  that  a  panel 
load  is  placed  sucivssively  at   |i!inel-|M>ints  1-1,  2-2,  etc.,   Fig.  771 


The  reactions  in  |Miunds  for  each  |KMilion  of  the  live  load  may  be  com* 
puted  by  the  Theorem  of  Three  MonH-nls.  as  follows; 


I  ivi»  l.iini 

r»iifi  !"• 

1  «i 

rit. 

«,  -  ».. 

«.  «•• 

Htrrw  IHacrwn. 

t  1 

1.'t.i,;«io 

.U.HJMl 

Fig,  7(i« 

'2  2 

urj.hiti) 

ii«'>.:<(Mi 

••     7(19 

3-3 

7\.r<i>0 

(tr.iido 

"    770 

4  4 

4.5. 1(N) 

l'JI.(MM) 

"     771 

^-^ 

24,(Kin 

II.'>,IIN) 

",   772 

6-« 

OJMM) 

I'.m.iuQ 

"    773 

r  4- 


j  k' 


DESIGN  OF  A  520-FOOT  SWING-SPAN. 


805 


A  separate  stress  diagram  is  required  to  give  the  stresses  for  each  position 
of  the  load,  Figs.  778-83.    The  stresses  in  lbs.  are  as  follows: 


Mem-  i 
ber 


Panel  Load  at 


'",'-'7 


Panel- 
puint  1-1, 


19000(K' 

13.5()(K)T 

840(X)T 


SrjOtKK' 
480n<K" 
29000T 
70000C 
49(X)0T 
otiOOOC 
flSOOOT 
fi.'iOOOT 
32000C 
3200(K' 
2.'>(KX>T 
13()0(K' 


Panel 
point  2-2. 


144000C 
103000T 
1G2000T 
4000T 
4000T 
17.'J(XM)C; 
102tMX)C 

.54000T 
KKKKX)! 

980()0T 
112000C 
13()O0OT 
13000()T 

O.'jOOOC 

tyjoooc 

4WXK)T 
24000C 


Panel 
point  3-3. 


lOOOOOC 

715  OT 

1190(X)T 

llOOOT 

llOOOT 

12I0(X)C 

1570(X)C 

75000T 

7fXXX)T 

49(XX)C 

S.VXXIT 

1940(X)T 

1940(X)T 

970OOC 

9"..  OC 

«)  (H    <T 

34iAJOC 


Panel 
point  4-4. 


040(XK; 
45000T 
77000T 

132000T 
3.'50{X)T 
77(KXK' 

1020(XK' 
78000T 
4.';(K)0T 
31{XX)(" 
3.5(XX)T 
410(X)T 
87000T 

1360(XK" 

13()000(.' 
70.500T 
3500(K' 


Panel- 
point  5-5. 


Panel- 
point  tt-6 


Total 
Tension. 


ZMoaac 

240(X)T 
410C0T 
71000T 
7I000T 
420(X)C 
.54(XX)C 
7(XXX)T 
24(KK)T 
17(XX)(' 
20(XK)T 
22(XX)(' 
2i:(XXK" 
174(XXK' 
1740(X)r 
(W«MX)T 
32(XXJC' 


13000(' 

9000T 

IfiOOOT 

2700()T 

1240OOT 

15(XX)C 

21000(" 

4.'5(XX)T 

9(XK)T 

fiOCXM" 

7(XX)T 

9(X«X' 

9(XXK' 

8fi(XXK' 

21.5(XK)(' 

39<XX)T 

20(KX)(' 


Total 

Com- 

prenion. 


387500T 
4990(X)T 
245(XK)T 
245000T 


.3.'>1000T 
•:48(XXiT 
147(XK)T 
I17000T 
38900(,T 
470000T 


3090(X)T 


5450000 


51.50000 
4840000 

70000C 
103(XX)0 
1(;8(XXKJ 
72000C 
3I000C 
S9flfXX)0 
7190000 

1580000 


To  allow  for  impact  a  per<-entage  /  is  to  Iw  addetl  to  the  live  loads.     For 

4(N)  1  on 

railway  and  motorway  I.Mtds  /  -  ^  -  ^,  and  for  sidewalk  loads  l~j ' y^,  /  being 

the  length  of  span  which  must  be  covered  to  produce  the  maximum  stnm 


Fio.  775, 


Fio.  777. 


Fiii.r78. 


Fni.  779. 


Fio.  7'X). 


in  the  m«»mlicr  under  conw«lentli<ni.     In  the  preient  caw  the  foil: 

age  V.^l^i^s  mjiy  lie  ii.k.'U: 


If 


lowing  aver- 


S06 


THEORY  OF  STRUCTURES. 


No.  of  Panels  Loaded.  /. 

1 73.0    per  cent. 

2 (iH.l      "       " 

3 G4.1      "       " 

4 60.25   "       " 

5 57.()     "       " 

6 .')4.S     "       " 

7 51.2     "       " 

Hoth  arms 3ti..S     " 

The  various  stresses  may  now  Ix-  combiiu'd  as  in  the  following  table: 
Combined  Sthessks. 


<;l«  1. 

i    Case  2, 

(  iwc  .1. 

Live 

Liiail. 

One  .\rni. 

in  Lbn, 

(use  4, 
Live 

No.of 

Im- 
littct 
Kac 

lor. 

.512 

Total. 

Mpin 

I)ea<l 
l.oa.1, 
in  I.Im. 

Iplifl 

I.<;.ul. 

in  Lbs. 

I.na.1. 

U«th 

Arni«, 

in  Lb». 

I.-.4.-..IXHK') 

I'anel» 
Liia<letl 
i     for 
1  Max. 

7 

I         .t. 

.t)S. 

425.oaoc 

Iwr. 

Tension, 
in  Lbs. 

127  .(NX) 

Com- 

prexsion, 
in  Lbs. 

/,„r, 

U*7.«K»T 

170.00()C 

s:w.oooc 

1,298.000 

/.„/.. 

BO.OIKM' 

120.000  r 

,587.000T 

<:W7..">00T)i       7 

512 

.■«ii  .wx)  r 

giN,o(xj 

90.(XX) 

/../.; 

iT.I.OIKM' 

LtM.iKwr 

»;to,(HM)r 

(4Wt,(K)0") 

1       7 

512 

425.(XX)r 

1.1N4.0(X) 

275,(XXI 

'■j-h 

1.IM,^.(HH>(' 

:t.'i.'.(KHH' 

as7.0(K)i' 

i24.->.(K)0T 

'       7 

512 

:r)2.ooor 

;M(i.(xh) 

1  .U45.0IX) 

1 .1 . 

LN,),0()()l' 

L'OH.OIKX" 

M."..0O(K' 

lol.l.OOCK' 

7 

.512 

4;<.'I,IXNK' 

2MO,0(X» 

1 .2(N).(XX) 

'\i\ 

.'hU.(KK)T 

-'71.IHI0C 

K.SO.(KHK' 

(4«4.0tKX-; 

7 

512 

4,i«.000«' 

5:u.ooo 

1 ,07U,U(X) 

i\i.s 

iio.oixir 

;«W.O(KIT 

2 

.ONI 

200,(KX)T 

019.(NX) 

i\i., 

2(>2.0(Kk  • 

1 19.000T 

41J  (K)OT 

(24N.IK)OTi 

15 

.570 

2:f9.000T 

511.U00 

70.I)(KM' 

15 

..ION 

20.(HXK- 

;<5s.iKin 

'J.J 

:«tL>.(MK)!" 

N4.000C 

10.).(KKM- 

(lOH.IXIDC) 

4 

.0025 

117.,5(KK' 

945.000 

1  47.OOOT 

15 

.:<0N 

54.(XX)T 

50.'<.(KK) 

',./., 

SSJ.IMMK' 

'.lo.(K)OT 

222,(KH)I 

(I17.(HKiri 

4 

.(M)25 

ria^-iooT 

0S.5OO 

lliX.IHXK.' 

15 

308 

02.000c 

012.IKX) 

','Hv 

:i.">.ooi)  • 

35,000 

>/,r., 

l.slt.lKKI  1 

I12,IM)(H' 

.TOI.IKKIT   I.WO.IHKrl'l 

:i 

041 

2.5a.(xx)r 

1 .3:Ui.(XX) 

HH.lMHir 

lia.lKKR- 

405.0001' 

(47t>.0<)OT) 

4 

0025  2!«0'|K)0T 

1. S.W.I'  •) 

',1/, 

:«').(X)UI' 

35.000 

>'./.. 

I.T.'HMir 

.MH.IKWT 

2 

•INl      iiWi.OOOT 

0:10.000 

."''••■■  i 

ll><i.()l)l)(' 

a.W.lMKM' 

2 

tWI       1.->H,0(KK' 

517,000 

f  :k ' 

;«)L>,oooT 

5:15  .(Xior 

4 

(ki25  .wn.oixir 

1,1«)..V)0 

:iTs.(XKtf 

.'il.tHHK- 

7.->4.0(KI»- 

(.'•(•O.OOIKi 

7 

.512      :l.S<i.(KKX' 

1 ,572.0IX) 

' ..  u„ 

.Ti.lHKIl' 

35.000 

U„/.„    1 

IW.CXHJT 

:io:t.()ixii' 

a 

.«S1      Jim.lMIOT : 

648,000 

."«/"  1 

.Mi.OtHM' 

,'>4.lin(H- 

NWi.lHHK'   (7tg.r)l)IM') 

7 

.512     4.'i(t.0(MM'  '. 

1.021. (NX) 

'7I,  j 

l.4i.-..()00l' 

J.in.oiKiT 

i,-> 

30H      114..''X»0T 

1 .850,.'>(K) 

f:/-r    ! 

Tl:t.(Kttt<' 

'   I.VI.IKKK* 

15 

M\n       .'i7..'i(KM' 

926.SO0 

I  ..".iHi.lKHVr 

.■»47.(iihk; 

.          :ti7.iKKir 

15 

MVi      12K.(HI01' 

2.071. (XX) 

N  H       Kiciires  in  pnrenlhe*"<  (    )  are  no»  itirlmletl  in  lotnl». 

Tlie  win<l  ..*( res.Mv-  in  the  rhitnU  will  not  affei't  the  section  reijuiretl.  and  are  ll:erefor«  n<)t  coo- 
•itli-K'  I  in  this  table 

Where  ft  meml)er  carries  .streH.ses  of  op|x)sile  kinds  the  ana  of  .section 
reijiiinMl  must  U-  rompuied  for  tension  and  eonipn>ssion  separately,  and 
thni'  fnurthn  of  the  sinallir  area  added  to  the  (tn-i»ter.  Thus,  for  the  top 
chord,  try  the  following  si-riiori: 

(hie  cover-plate  Sti'Vi'ii" 20.25  »»i    ins. 

Four  to(i  anRles  .■Jl"x31"x|",  .      !».!)2    "      " 
Two  wcl.s  30"  .  !<'■ 4S.75   ''      " 

Two  l)oMom  aiinles  fl"x(i"xj''-     1''»>W    "      " 


w 


J 


=^ 


I 


in.   78i 


1 

L 


Total  Miction aj.M)    "      " 

Uimupiiorted  JenRth,  /-35'-r.J5",  ftay  427  ins. 
Moment  of  inertia  alxuut  axia  ('/)- 18.117, 


DESIGXmo  A  BtO-FOOT  SWINQSPAN. 


807 


and    rbout    axis    .4fi=  13,1'!X).     Therefore  r  =  ^^-ii.8    and    f-36.2. 

Hence  the  allowable  stress  per  sq.  in.  =  18,000-70^  =  15,470  lbs. 
Thus  the  section  required  for  conipn-ssion  {;',{/, 

=  1  ,-'0(i,000  ^  1 0,470  =  7S.0  sq.  ins.; 
the  si'ction  retjuired  for  tension  l\L'.^ 

=  2S0,000-Hl6,(X)0=17.5sq.  ins., 
and  the  total  section  required 

-=78.0  +  1x17.5  =  01.1  sq.  ins. 
The  section  required  for  compres.sion  L\L\ 

=  1 ,070,0(H)  :- 1  -,,470 -  69..3  sq.  ins.; 
the  ^section  required  for  ten.sjon  l\l\ 

---yiX  000-^  10,000 -.-JS.li  sq.  ins., 
and  the  total  section 

=  09.3 +Jx33.2  =  94.2sq.  ins. 

The  above  section  may  bo  used  for  the  entin>  top  chord. 

Top  chord  section  l\l\  carries  tension  only,  and  eve-bare  may  be  used 

S.ct.on  re(|ui red  =  2,071,000  :  18,000=  1 1.1.1  so.  i„s.  LV  four  bare  1'"  x  1 8'"' 
an.l  twolwrs  12"xl,',"-ll.->..iHq.  ins.  "       " 

.Ml  other  memU-rs  may  lie  desipned  in  a  similar  manner.  Takine  f<.r 
<\:.inpl,-.  the  m.-nilx.rs  which  meet  at  l\,  wc  may  use  for  M.U.  four 'bars 
12"xHr'-S7.0  sq.  ins.;  for  l\L,  four  l,ars  10"xlf|"-72.5  sq.  ins.;  and 
for  L,.i/„  the  following  section,  Fig.  782; 

One  cover-plate  30"  x  ,'," 20.2.1  sq. 

Four  top  angles  3J"x3J"x A" 11.48   " 

Four  webs  (in  ixiirs)  30"  X  J" fio.oo   ' ' 

Two  IxHtom  angles  (»"x()"x  j" 10.88    " 


ms. 


Total  section «,  los.fii    "     << 

The  stresses  from  all  these  m.-mliere  are  transmitted  through  the  pin 


The  allowable  liearing  on  the  pin  is  22,000  lbs.  p««r 
sfpiare  inch  of  the  surface  obtained  by  multiplying 
together  the  diameter  of  the  i>in  and  the  length  over 
which  the  n)emlM>r  liears  upon  it.  .\ssume  10  ins.  as 
tiM'  diameter  of  pin  (',.  Then  allowable  liparing  per 
inch  .  •  length- 22,000 X  10 -220,0(K)  lbs.  The  largest 
eyc-li  are  12  ins.  wide,  so  that  they  can  safely  carry 
12X1 8.000 -216.000  ll)s.  p..r  inch  of  thickness".  The 
2200()0  »».  in  }x.ari??g  afTonlt-ti  by  it  iO-:n.  pin  is 
accordingly  anipk  for  the  eye-bara. 


at  U 


Fio.  782. 


\t 


808 


THEORY  OF  STRUCTURES. 


The  length  of  bearing  on  the  pin  roquin-d  for  (7,.V,- 1,572,000-^220,000 
=  7.1')  ins.,  of  which  the  webs  provide  '2  ins.,  so  that  5.15  ins.  in  the  whole 
nH>mb«'r  or  2.57  (say  2|;  in.  on  each  side  must  lie  made  up  by  means  of  pin- 
plntrs.  We  shall  reinforce  each  web  with  six  plates  i'»"thiclv  arranged  as  in 
Figs.  783-4.     These   i)lates  should  Ix'  of  sufficient  lcn':;th  and  width  to  di» 


Fio.  784. 

tribute  the  pressure  from  the  pin  uniformly  over  the  entire  section  of  the 
member.  The  two  in.side  plates  in  this  member  are  extended  so  as  to 
form  a  hinge.    Plates  of  the  following  dimensions  may  be  used: 

Two  plates  2f."   xA"x6'0"    inside 
29"   xA"xO'  li"     " 
25i"Xrt"x:r()"       " 
20J"  X  A"  X2'  Si"  hinge-plates 
20i"xA"x;{'r)"    outside 
23i"x,',"x3'7r     " 

The  pressure  on  each  plate  will  U-  A  x 220,000  =  96,000  lbs.     Hence  the 
number  of  rivets  required  to  transfc-  the  stress  in  plate  A  across  plane  t'B 


Fio.  785. 


-%,n00  :  fiOn,  -i-iy  Ifi,  ,in;i  .it  irnst  -ixtef-n  addifiohnl  rivets  will  he  required 
at  each  successive  plate  until  the   web  is  n-ached. 


ni.SIGX  OF  .1   Seo-FOOT  SWING-SPAN. 
Figs.  785-6  show  the  joint  at  C/,  in  plan  and  elevation. 


80» 


Fig.  786. 


The  greatest  stress  in  the  pin  l\  will  evidently  occur  when  both  arm« 
of  the  s,«n  ar.  fully  loaded.  The  stresses  in  the  various  members  wUlthTn 
be  as  .shown  in  Fig.  787.  *" 

Following  are  the  horizontal  and  vertical  components  H  and 


V: 


l\l\ 

l\l\ 

r,-w, .„__ 

YA'* -    899'.400 

'-'t^i 


It 
+     128,900 
+  I,8.51,.'i00 
1.080,000 


V 
+  23,100 
+925,700 
-926,400 
+  771,400 
-764,000 


Positive  stresses  are  supposed  to  act  to  the  right  and  upward-  newtivA 
to  the  left  and  downwa«l.  Distributing  the  .XrL  in  eachTem^r  ^on^ 
Its  c-omponents  .n  proportion  to  their  sectional  ar^-as,  Figs.  788  and  789row 


Fio   787. 


Fio.  788. 


Fio.  789. 


the  Wdmg  of  the  pm  in  honiontal  and  vertical  planes  respectivelv  Onlv 
.me  h,.  f  „  ,hc  p,„  ,H  .hown  as  the  arrangement  of  the  memLrs  i^  .^-nXl 
r>.al  ulK,ut  he  .viitrf  line  of  th..  truss.  The  load  from  each  pie.*  is X 
assume,  t..  l«  .-oncentratcd  at  the  cntri.  of  its  !».aring  upon  the  nin  ^ 
a.sum.l  .lead  load  .,f  .3.5,000 11..  „t  l\  may  be  distributed' veM he  ^n'l^hi.I^ 
each  upward  component  and  increasing  each  downwaiti  one  by  about  2000 


I 
if 

\i 

ji 


810 


THEORY  OF  STRUCTURES. 


lbs.     The  bending  moments  in  a  horizontal  and  vertical  plane  may  now  be 
computed  for  each  ix)int  in  the  pin. 


Heni)ix<!  Moments 

IN  Horizontal  Plane  (Mb). 

Point.                Load. 

Shear.         j    Lever-arm. 

Increment  of 
B..M. 

Total  B.M. 

1 

2 
3 
4 
5 
6 
7 

+   04, .'■.00 
-»49.7(K» 
+  »K»..V)0 
-270,(HKt 
+  312,.'>(H) 
+  312,.T(K) 
-270,000 

0 
+   04,.V)0 
-.38.->,2(H) 

-  84.7(K) 
-.3.54,7(K) 

-  42.2(M) 
+  270,300 

1.08  ins. 
2.82  " 
1.7.-)  " 
3.28  " 
2..-)     •' 
1.78  " 

+     108,300 
-1,0.S0.204 

-  148.22.'-. 
-1,103,410 

-  10.5,.i(H) 
+    481,134 

+    10S.300  in.-lbs 
-    977.904       " 
-1,12(5,129       " 
-2.289,.54.5       " 
-2,.39.5.04.5       " 
-1,913,911       " 

Bendixo  Moment.s  in  Vehtical  Plane  (i\/i-). 


a 

-191,0(K) 

1 

+     9,0(K) 

•> 

+  .384,(HM) 

3 

+  148,(MM) 

4 

-234.(HK) 

h 

-I91,(M)<I 

.5 

+  I.->4.(HH) 

6 

+  1.-.4  .(MH) 

t 

-234  .(KMt 

-191, 
-181 
+  202 
+  3.-.(), 
+  110, 
-  71 
+  79, 
+  233, 


,000 
,4(K) 
lUM) 
,0(K) 
IMK) 
4(H) 
r>(H) 
(■>(M) 


44  i 

08 

82 

7."> 

87 

21 

-» 

78 


6.57,040 
.304.7.->2 
.571, .332 
(il3,').-0 
218.042 
90,024 
199.(i(M) 
415,808 


0.57.040 
901,792 
390,400 
223.040 
441.1.32 
3.51,108 
.5.50,108 
90.5,910 


in.-lbs. 


HoiKv  2,4iw  ,000  = /  = ', — 

'  r  4 


The  Kreatest  n-sultant  B.M.  will  evident  ly  lie  at  (>,  where  H.M.  -  VSln'  +  M  v' 
=  \  (2.3it.-.,04.5'  t-.-).-|(),i;).8-)  ==2,4.->7,(HK)   in.-ll).>'. 
.Mlowahle  fibre  •■'tre.ss  in  jnn  is  27,(KK)  lbs.  jK-r  squan-  inch. 

whea'  r  is    (he    radius  of  the  pin, 

and  therefore  r  =  4.S7  ins.  and  diameter  of  pin  =  9.74  ins.,  so  that  the  a.s^umed 
diameter,    10  Ins.,  is  on   the  safe   side. 

Other  joints  and  niemlxTs  may  lie  desiRncd  in  the  same  way. 

It  will  lie  necessary  to  determine  the  deflection  of  the  ends  of  the  span 
due  to  its  own  weight  ami  other  causes,  to  a.s<-ertain  how  much  the  eyc-liars 
in  [KUiels  l'J\,  rj',  must  Ije  shortened  in  order  to  brinjs  the  (vntre  of  the 
pins  at  the  i-nds  of  the  s|wn  to  the  sjuiie  elevation  as  those  near  the  centre. 

The  deflection  J  of  u  given  point  due  to  the  distortion  of  any  member 
is  ifiven  by  the  fonnula 

J     -  A'  • 

wIh-H'  p  is  the  stress  per  square  inch  in  the  meml)cr; 
/     "      ^cnjtth  of  tlie  meml)er; 
M     "      »ilr!-ss  in  the  mctnU-r  due  to  a  unit  load  placed  at  the  point 

wh<>s«'  d<>flecli<in  is  sought ; 
J?=-thc  c<x'fficient  of  elasticity  (Young's  miKlulus). 
I)owii\vard    deiteclioiis    and    (eiisde    stresnes    will    tx'    coiisiderud    posiiive; 
Upward  detlectiuna  and  compreiuiive  stresses,  neyalive. 


DESIGN  OF  A  StO-FOOT  SWING-SPAN.  811 

Deflection  at  End  of  Span  Due  to  Dead  Load.~p  is  obtained  by  dividine 

A     -.(,000,000  lbs.     The  following  Table  can  now  be  prepared: 


Mera- 

Iter. 

/./-. 

I.  J.. 
{..'.: 
/..;.; 

r.c. 
r,L. 

'if'!- 

r.i.: 


Strem. 


-  9(),(M«) 

-  27."),()<MI 
- 1 ,04.5.(K«» 
-l,0.39,.-i<;0 
-1,42.5,(KH) 

+  280,(HK) 

-*-  .')31,(MI0 

+  I,.')9fl,n0() 

+  1,42.';,(KK) 

f  127,(KH) 

-  378,(MK; 

-  .'■)J2,(K)0 

-  2(e,0()0 

-  3S2.()(K) 
+  (>89,(KK) 
+  .S14.(1(K) 
+  :«)2,()()0 

-  713,(KK) 


Area. 

71   0».5 
KM  095 
99  97 
09,97 
470.40 

9.-).  8 
9.J.8 

11.5  5 

103.. "5 
99. 5.-. 

108,01 

142. 38 
('8.43 
(WIS 
7.")  (K) 
87.00 
4(i.98 
70.44 


-  l,22»i 

-  2,(142 
- 10.4.'-.3 

-  io,.ms 

-  3,030 

+  2,921 
+  5,."3!» 
+  13.8KS 
+  13,780 
+    1,27(1 

-  3,483 

-  3,.59(i 

-  3,83.') 

-  .5,8,30 
+  9,187 
+  9,3.5r. 
+  6.440 
-10,1.50 


838.4 
419.4 

8:18.4 
838.4 
184.0 

420. 0(i 
1278. 1,S 
9.38.13 
1S4.(M) 
.593. '-.S 
.5,52  7.5 
.5.52.7.5 
.593  .53 
048  81 
.5.52.7.5 
.5.52  7.5 
49.5  0 
1140  0 


K 

.0.307 
.0382 

..3o:;o 

.3000 
.0193 

.0430 

2442 

.4480 

.087.5 

.02()2 

0(.«M 

.0080 

.0784 

.131.5 

1749 

1784 

no 

.399 


-1.000 
-1.695 
-2.912 
-2.912 
-2.579 

+  1.723 

+  2.2.38 
+  2.879 
+  2.5735 
+  1.414 
+  .445 
+    .445 

-  .984 

-  .790 
+  .928 
+  .928 
+  .695 
-1.228 


B 


+  .o;«i7 

+  .0049 

+  .8800 

+  .8730 

+  .049«) 

+  .0742 

+  .540.5 
+  1.29(K( 


+ 
+ 


.22.52 
.0.300 
.0;{(M) 
.0.'tf).5 
.07W4 
.  10:!i 

mvi) 

.1000 
.07(4 
51.50 


Hi-nce  total  deflection  due  to  dead  load -.5. 11 99  ins 


■•     (-ri')- 


Sinnlarly  the  deflection  due  to  the  uplift  at  the  ends  may  bi-  obtaine.l 
The  amount  18  -l.fi«i(i2  ins.  ""laimo. 

The  pin-holes  will  Ix-  about  A  in.  larger  than  the  pins,  and  hence  then- 
W.I1  Ik.  a  play  at  each  end  of  a  pin-connected  memlx^r^A  or   01.w7n 
^^^   (onsidenng  this  play  as  a  distortion  of  the  member,  corresponding  to 
^    \n  the  alx>ve  table,  it  may  be  shown  that  the  deflection  at  the  end  due 
to  this  cause  is  +0..54I9  in. 

The  temperature  of  the  top  chords  and  web  members  will  often  evw-d 
hat  of  the  lK,ttom  chords.     A.s.suming  n  mean  difference  of  15°  F    for  the 
top  chonis  an,l  7^  F  for  the  posts  an.l  .liagonals. , he  elongation  of  th^  former 
,.r  un.t  of  length  will  be  1.5  XO.00000<J«7  =  O.O.K)l  in.  and  of  ,he  latter  OoSS). 
in.     This  will  cause  a  deflection  of  the  ends=  +0(iK31   in 

Similarly  the  lengthening  of  the  lop  ,.hor,ls  and  diagonals  on   ac-ou-.t 
..f  camlxT  w,ll  cause  a  deflection  at  the  end8-3.1904  ins      Hence  the  to 
»lellection  at  the  ends  will  1»  ^  '  "'" 

.5.1199-1.6662 +0..'i4l9+0.fiS31  +3.1964-7.8751  ina. 
This  will  he  taken  up  by  shortening  the  eye-bars  in  U.V.  and  If  V.     Tho 

"l';«;.M  %^:^S:.  .hT.  "°  •"°""'  ""'■  •»"' ""  "  •^•«" 


812 


THEORY  OF  UTRLCrURES. 


.;  ffi 


DESIGN  FOR  SWING^PAN. 


813 


814 


THEORY  OF  STRUCTURES. 


C 

•V 
C 

S    si 


3  i 

c  = 

03   I 


s 


•3 


-  * 

o 


Is 


S   •" 


E 

-a 


'fi 


1111 


Q       Q       q       Q 


ii* 

ill 

111 


ii     I 


«    lO 

lO 

ilS 

c 

1   ) 

i 

P   E 

e 

J  8 


££ 


I 

e 

I 

3 


Six    cS^'^      S"*    2:-'    ^    •-S-" 
•8«    .Si^xs-* -s:'«.2x'«s:.S«8-2.-=3J: 


si 


a;  a. 


^3  ::3  ? 


1^ 


(S3 


.SeJ 


II 


S         e     so 
..         »     it  it 


=       I    II 

■i     -S  1.5 

2      >j  ^»j 


<5 

03  = 
O 


:::i 
> 


23 

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THEORY  OF  STRUCTURES. 


T.M1I,K    OF    f  OADS    FOR    IIlOHWAV    HrIIXJES. 


City  am 

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hriilitr 

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EXAMPLES. 

1.  .\  bridge  of  .V  equal  sptins  crosses  a  s|i!in  of  /,  fppt :  the  wpijjhts  in  tons 
per  lini'al  foot  of  tlic  nritiin  (tinlcrs  of  the  plutfnmi,  |x'nnancnt  way,  etc.,  and 
of  the  live  load  are  w,,  u'„  w„  re8|)ectivply.     Show  that 

LA 


whore 


A  - Vj(pk  +  r]  +  u;(pk  +  (/)      ili'l     /{  =  pk  f  r, 


k  lH>in(t  the  ratio  of  span  to  depth,  and  />.  </,  r  numerical  coefficients.     Hence 
ril.-Mi  di'lermine  the  limiting  span  of  a  girder. 

If  -V  i.s  till-  cost  of  a  pier  and  if  1'  is  the  cost  |H>r  ton  of  the  sufierstructure, 
liiid  tile  vahic  of  ,V  wliich  will  in.ikc  tiie  tiilul  cost  |ht  lineal  foot  a  minimum, 
and  prove  that  this  is  approximately  tlie  case  when  the  spans  are  so  arranged 
that  the  cost  of  one  span  of  the  bridge  structure  is  e(|Ual  to  the  cost  of  a  pier. 


A  IIS.  .\  span 


Cost   is  a  minimum  when  .Y -/./< -/.*'--;—,  and 


till'  iiiinimiim  cos 


pl.+r ' A  A^ 

(lit, 
1  -   - .  I  ».Y,  ai>pr:ixinintely. 


2.  The  platform  of  a  single-track  britlgc  is  sup|>orted  u|)on  the  top  chord* 
of  two  Warren  girders;  caih  ginler  is  KM)  ft.  long,  and  its  br.ieing  is  formed 
of  till  etpiilatcral  triangles  ibase  10  f  I .  I ;  the  dead  weiclll  of  the  bridge  is 
'.KX)  lbs.  |M'r  lineal  foot ;  the  greatest  total  stress  in  the  seventh  sloping  memU-r 
from  one  end  when  a  train  cl•os^es  the  bridge  is  ll..'{!H.*<  lb".  Di-termine 
the  weight  of  the  live  load  tier  lineal  fool.  Prepare  a  t.ibie  showing  the  greatest 
Htress  in  each  bar  ..t.^l  bay  when  a  single  load  of  l.'),(MH)  lbs.  cronsi's  the  girder. 


Alts. 


^KgSygfevvvv 


2771?  Ibn.  [Mr  lin.  ft. 


h     h    H    U 

Fio    7TO 


Stresitcs  in  diiignn»b: 


i/.,    (/,.,    •.'J^  ;{  Ions. 


J 


EXA  \fPLES. 


821 


Sfrpsses  in  comftressioii  chord: 
SircsM's  in  ti'iisioii  clionl: 


c.  "=11  J\  3  tons. 


^~loj\  ;ij 

.    ...  '.-1-'   N  .<  tons. 

3.  .\  Uam-n  p,r,!..r  .•omp<w,l  of  eight  .-.luiiatcTal  trianpl.s  has  ,t.s  u,.per 
.  ...nl  .M  ..o.„pr..ssKm  un.l  has  ..very  joint   loaded  with  a   weight  of  2  to^. 

t.cal  HtrutH.      rhenimn^Wft.     Find  the  stre.«..s  in  all  the  meml,  ,< 

Anx.  Hays  in  eompresMion  ehord:  lot-   ov;};         i>d-i;;\  .■^ 
•■W=-1S§V3;     4th-L'ls  f'toni,. 
MayH  in  tension  ehord ;  1st-   'A\\.\:      lM-1(»s.{ 

^.  ,     .  3d  = -ON  3;       4th-L'liv'5"tons. 

•stresses  in  verticals;   In  each  vertical  ->  tons. 
J^tresfK-s  in  diagonals :  i  sf  -  1 0\' :7;      i>,I  - .s§  \  :T; 

;jd-7jN:i:    4th-r.\37 
5th-4§N;f:    «th-31\J; 
.   „.  ""'='  -^;<:    Hth-i\3ton9. 

4.  A  Warren  girder,  «-iih  a  platform  on  the  lower  Ihh,«..  carries  a  loa.f 
"f  iO  tons  at  the  centr,-.     F.n.l  the  stress  in  each  .nendH-r.  an.!  also  find  the 

rent^i  hi"    ;h '''"  "'  '""7  '•""•"  "*'"'  "'"  *-""'  "'-  -"     ^'^'--^  i"  the 
rentre  bays.     There  .ire  su  l.ay^  ji,  ih  ■  l.,xv<-r  .-hord. 

Ann.  Stress  in  each  diaKoii.,!  -  •;■%  ,;  i,,,,.. 

Tens,  chord:     stress  in  1st  Lay  V'v  3';     L'd-IO\'if:     3d-V\3tons 
Con,,.,  .hor.! :  stn-ss  in  1  st  hay  -  'fK%  .M  -i«  v  Jf.     3J  .  .^^  3  ^„„^ 
Weight  at  each  joint  -"Hi  tons. 
5.  A  Warren   gir<ler  for  a  single-track   railway  bridge  .insists  of  eiriit 

m  .>... rf)  Ibs^due  to  engine,  l.-ilH)  lbs.  due  ,..  train,  and  m  lbs.  d-.e  to  hridiT 
me  "II^'T  rr'  r  ",""""'.'*•""«';:'  "«"h  tensile  and  compressive,  in  the  meml-ers 
m  t  .>  >ertieal  p  aties  m.m.Hhately  on  the  right  of  the  .se.t.nd.  third,  and 
i.'urth  apices  .11  the  compression  .honi.  Also,  find  how  many  :.„,  Hvets 
..re  n,,uired  to  eoni.eC  the  .liagonals  me.  by  thes,.  pla.ies  ^iih  thechonU 
-.1     0  pn-ve,,,   any  tendency  to  longitudinal  slip  l.tw. the  sup,H>rt   and 

.,"""■  '"•'--'.•I"-  "r^'  »".l  -cond  api,..s  in  the  tension  chord. 

-hear  strength    of  rivets  U-ing  lO.tXK)  lbs.  |».r  s«pinn-  inch  ) 

i'M.  Chd.  (C.)  „i,M.I()    MU-.M:  -1,350;  (T.)  130.785;   m.o>4:  .'15.245 
i'"'^.  .'.'«/   < ,  ,<ir.-:  -,4  Slit;  :{|.2!)4;  Rivets,  0,  9.  15. 

iVtermine  the  maM.nun,  .treses  .„  each  ...cr.U.r  when  a  live  !...,  i,       Ti, 
.-.   l-r  lineal  f.H.t.  pre.vdcd  by  a  con,,.n,rated   load  of  IK,hh.  Ib-      ,,oss,.s 

Z  rh^'f.  r""'7       'T'"^  ^"'"*  "  '"■"''"'      ■'■'"•  ""-•^""'^  ..n.1  verticals 
an'  riveted  t..  angle-iron.-  forming  part  ..f  the  H«ng..v 

How  many  J-in.  rivets  ar,-  rcpiind  for  the  connection  of  the  si-veral  mem- 
U-rs  n.eet..g  „,  ,he  third  «,x. .  in  the  up^r  chord'     (23,  6,  and  i:[,     H^i 


'»    M 


822 


TUBORT  OF  BTRUCTURE8. 


manynre  required  in  the  first  bay  of  each  chord  to  prevent  longitudinal  8lip?(15 
in  tension  chord  and  26  in  compression  chord.)     Rivet  stress  -  8000  lbs.  /sq.  in. 


Ans. 


<, -/,=.')!  ,r.-.><Hbs.; 
«, -/, -L'()<)..-..S()  lbs. 
0,  =   !M  ,•.'•_'•.'  lbs. ; 
f.-2l7..'.{()lbs.; 
</,  =1(».{,(MM)  llw.; 
d,-    71,147  lbs.; 

d,-  4;v7:r.  liw.; 

rf,„=    I'.t.Uli)  lbs.; 


<  -^-12.3,2S0lh,., 
;   /„-/„.  = -'I  !»,")4<»  lbs. 
f, -I.'>t,.V.K)  lbs.; 

(/.-(V.'.OSSlbs.; 

(/,  =  (il  ,777  lbs.; 

<y,-. •}.•),;{<>:}  lbs.; 

</..-lL',.S4(ilbs.: 


^ -/,- 178,260  lbs. ■; 

r,-l!KvVJOlbs.; 

rf,-Sl,.-)r)l  lbs.; 
(/„-.-»2,.-W{9lb8.; 
'/,- 27,424  lbs.; 
</„-    «>,207lb8. 


The  strcs-ic-  ,/„,  ,/,,,  and  </,j  are  of  an  opimsitp  kind  to  those  due  to  the 
dead  Imul.     Tlic  iri.i\itiiimi  load  on  each  vertical -•_'()..">()«)  lbs. 

7.  If  .1  font-  of  .'.(XK)  lbs.  striki-  the  biiifom  chnnl  of  the  (jirder  in  the  pre- 
oediiiR  <iiiestioii  ;.t  20  ft.  from  om.  end  and  in  a  dirertinn  inclined  at  30°  to 
the  horuoiiiul,  delerinine  its  elTecf  u|M)ri  the  several  members. 


I 


Ant. 


Xn 


Yh- 


Yd^  Yl 
2      ;}  ' 

Ya—ah-tH--.  . 
Yit  -  .vT  -  rif  -  (//) 

„     sri()(x)N  .\ 


Ik.    801. 

•Y..    .'lOOON  :r 

■■■"  l.<  27 

_lj»_i(iono\'3 

■  ■■"   7    '   ^    27 

1()OOOV3 

27 

.{."lOOOV  u"  „ 

lbs.; 


•no-np 


Ibit.; 
Iba.; 
iba.; 


llw.;    /(/-         ..       Ib«.; 

8  .\  Iiitfice  ifir.ler  iCJO  ft.  lorifi  Hiid  20  ft.  de»'p,  with  two  syntemn  of  right- 
anglc  in.iiKl.s.  .arries  a  tliiid  load  of  H0()  llw..  fur  lineal  fo<i|.  Determine 
the  itreatest  stre-w-^  in  the  diaK<iimli«  and  chonl^  of  the  fourth  buy  from  one 
•nd  when  a  live  load  of  liK)  llw.  \wr  lineal  f<K»t  pa-  «•  over  the  girder. 


EXAMPLES. 


823 


Ana.  If  riveted:  Diagonal  stress  -  37,200V2"lbB.; 

Chord  stress      -450,000  lbs. 
If  pi/i-connec/ed;  Diagonal  stress  -  44,800^^  and   29,600 V2"lb8.  • 
Chord  stress      -460,000  lbs.  in  compression  and 
-•*'*0,000  lbs.  in  tension. 

I  ?■  r  . .  .""^  T"^"  ^  ^*-  '"""^  "'"*  ^  ^*-  'l^P  ^'"•'•'*«  »  uniformly  distributed 
load  of  144,(KX)  lbs.  Find  the  flange  inch-stres-ses  at  the  eentre,  the  sectional 
area  of  the  toj)  flange  being  561  sq-  ins.  gross,  and  of  the  bottom  flange  45 
sq.  ma.  net.  ^ 

What  .should  be  the  camber  of  the  girder,  and  what  extra  length  should 
he  given  t..  the  top  flang,^  m  that  the  Ix.ttom  flange  of  the  loaded  girder  may 
be  truly  honzontal?     (A' -2<M)00,(XK)  lbs.) 
Ana.  3185.8  lbs.;    4000  lbs. 

X,  - .29735  in. ;    x, - .2987  in. ;    a,-*,- ,VM  ft. 

10.  \  lattice  girder  80  ft.  long  and  10  ft.  deep,  with  four  systems  of  right- 
angle  triangles,  carries  a  dea.l  load  of  1000  lbs.  ,xt  lineal  fm.t.  Determine 
the  gn-atest  stresses  in  the  diagonals  met  by  a  vertical  plane  in  the  acrenth 
bay  from  one  end  when  a  live  load  of  L'.JOO  ll>s.  ,*r  lineal  foot  passes  over 
the  girder.  I ),.sign  the  flanges,  which  are  to  consist  of  plates  riveted  together. 
The  hitti.-.-  bars  are  riveted  to  angle-irons.  Find  the  number  of  J-in 
rivets  n-qiiired  to  .onnect  the  angle-irons  xxith  the  flanges  in  the  first  bay 
lO.OtX)  lbs.  i^r  .square  inch  being  the  safe  shearing  strength  of  the  rivets.  "  ' 
Ana.   If  riifteti:  Diagonal  stress  -  10,«kM A  v^2" lbs. 

If />i/i-ronnfr<«/.-       "  "     -9062i>/2;    OioOViT;    15,468iV^; 

1 1,875V  2  11,8. 
II.  A  lattn-e  girder  of  40  ft.  span,  5  ft.  depth,  and  with  horiiontal  chords 
has  a  web  com|mH,..l  of  two  systems  of  -ight-angle  triangles  and  is  designed 
to  support  a  deml  and  a  live  load,  each  of  J  ton  |»r  lineal  foot,  upon  the  bottom 
chonl.  Determine  the  maximum  stresses  in  the  members  of  the  third  bay 
from  one  end  met  by  a  vertical  plane. 

Ana.  If  riveted:  Diagimal  stress  -  V/ v'iFtons; 

Chord  stress       -;}3f  tons. 
If  piH-ffj«ncr(rt/;  Diagonal  stress -fjvi>    and     flv^tona* 
Chonl  stress       -.TiJ  tons  in  tension, 

•'W  tons  in  compression. 
i».  A  lattice  truss  of  100  ft.  s,Min  and  10  ft.  depth  has  a  web  composed 
of  four  systems  ..f  right-angle  triangles.  The  maximum  stress  in  the  diago- 
nal  joining  the  sixth  a|>ex  in  tlie  upjH-r  ch.ml  to  the  fourth  ajx'x  in  the  lower 
IS  16  tons.  Find  the  dead  Um\,  the  live  load  being  1  ton  per  Uneal  foot,  assum- 
ing tlie  truss  to  be  (a)  riveted,  (6)  pin-connected. 

^   ,  Ana.  (<i)  ..VMton;     (M  1.062  tons. 

ij.  A  lattice  ginler  of  40  ft.  sfmn  has  a  web  comf¥»se<l  of  two  RVBtema 

of  triangles  (Iwse-lO  ft.)  and  is  designe<l  to  curry  a  live  load  of  IflOO  lbs. 

per  lineal  f<H)t  and  a  dead  load  of  1 200  lbs.  [mt  lineal  f<M)t.     Defining  the  stream 

length  of  a  member  to  he  the  firoduct  of  it«  length  into  the  atrmia  to  which 


i>-4  THEORY  OF  STRVCTVRE8. 

it  is  suhjectpcJ,  fiiul  the  depth  of  flip  truss  so  that  its  total  stress-length  may 
U'  a  ininimum.  .l„.s.  Hivptod,  1 1 M  ft. ;    pin-<'onnectod,  1 1.9  ft. 

14.  Trpparp  a  tahip  pviiiR  thp  .strps.-<ps  in  the  wvpral  members  of  a  single- 
interspctioii  dpck-truss  for  a  doul)lp-ira(k  bridge  of  34'.'  ft.  .span,  33  ft.  depth 
and    with  pi};htp<'n   paiipls.     The  jmnpl    pnginp,    live,    and   dead    loads  are 
121,000,  (i.>,(XM),  and  40,000  lbs.,  nspectively,  |)er  truss. 

15.  Prepare  a  table  giving  the  stresses  in  the  several  members  of  a  double- 
interseption  through-truss  of  342  ft.  span,  .{3  ft.  depth,  with  eightifn  panels 
and  a  double  track.  The  panel  piiginp,  train,  and  dead  load  are  121,000, 
05,000,  an<l  40,000  lbs.,  resiH-etivply,  yn'X  tru.ss. 

16.  Pr(>parp  a  tablo  giving  thp  stresses  of  the  several  members  of  a  double- 
intprs«'etion  through-truss  of  I.'>4  ft.  »\m\,  20  ft.  depth,  and  with  elpvpn  panels. 
Thp  paiipl  enginp.  live,  and  bridge  loatls  are  91,000,  48,000,  and  23,000  lbs., 
resijeptively,  fx-r  truss. 

17.  Prpjmre  a  table  giving  the  stresses  in  the  several  members  of  a  through- 
truss  for  a  double-intprscption  doublp-traek  bridge  of  .342  ft.  span,  40  ft. 
dppth,  and  with  ninptwn  panels.  The  panel  engine,  live,  and  dead  loads 
^re  90,000,  .')3,000,  and  43,2(K)  lbs.,  resin-etively. 

18.  A  horizotital  eye-bar  of  length  /  in.  and  weighing  it-  |K>und8  per  lineal 
inch,  carries  a  force  of  f  |Miunds.     Taking 


-Wo -moment  of  resistanei'  of  section  —  .U, 


-U,  -  b««nding  moment  at  (•«-iit  re  of  eye-bar  due  t<»  its  own  weight  -  \wV; 
iW, -Ix'tiding  moment  from  direct  stress  /'  into  the  lever-arm  <i-Pd; 

/,  -unit  stn'ss  on  pxtn-me  fibre  at  cpntrp  of  s|>an  due  to  .W,  and  .tf, 
acting  together; 

♦"-•listancv  from  neutral  axis  to  extn>me  fibre; 

d  -  maximum  deHection  due  to  all  forpes  acting  together; 

/-nionipnt  of  inprtia  of  bur; 

£-m(Mlulus  of  plasticity; 

/,-unit  streas  uniformly  distributed  due  to  P, 

(Tt  Fl\ 
I  ^r^'fr)  -Mye. 

19.  A  horizontal  eye-bar  2"  ?;  1 2"  is  .W  ft.  long  centre  to  centra  of  pin- 
holes. Tne  direct  tpiisiie  stress  /,  on  the  bar  is  IM.OOO  llw.,  or  /^-4.32.000  lbs. 
The  nuHeriul  is  of  stei-l  and  the  weight  of  the  bar  is  Hl.ti  llw.  jier  lineal  ffxit. 

(a)  Kind  /,  by  the  formula  of  the  prece<ling  problem,  and  obtain  the 
maximum  fibre  stn'.sses. 

('()  Assuming  that  the  direct  lipnding  and  the  ten.sion  stresses  act  inde- 
pendently, find  the  extreme  fibre  stn-sses  in  ten.sion  and  compn-ssion. 

Ann.  (/i)20,168andl.';,83'2lb.s./sq.in.  (T.);  ih)  24,375  and  H,ti2.-.lbs./sq.  in. 

ao.  A  round  lateral  md  1  in.  in  diaim-ter  is  64  ft.  long  between  end  sup- 


EXAMPLES.  j^or, 

ports,  and  is  subjected  to  a  tensile  stress  of  8000  lbs.     Find  /,  bv  the  method 
of  Problem  18.  and  al«,  find  the  total  maximum  and  minimum  fiUst^ts 

,,    „, -»««.    4701bs./sq.in.;    10.658  lbs.  and  9706  Ibs./«q.  in. " 

at.  Hy  compannR  the  two  pre.-eding  j.roblems  it  will  be  seen  that  the 
bendmg  stress  for  the  2"  X  IJ"  eye-bar  is  much  greater  than  thaTof    he  ,-in 
round  bur    although  the  unsup,K,rted  length  of  the  nnl  is  ^' ^'aU-r       t 
7     1    t^!  ^"n  ""^  '^'■•'"  ^""^  '•♦""^^  /'  '^'^  •«  *  «'rt'»i"  rati^  of  depth 

two'!:.„^.lT."^'  "^  *  '"^  "*  '"'"•  '°"'^'  '"-""'«'  °"  ''>«  ohidjoininrthe 
t^o  panel-points    ,s  curved  to  a  ra.lius  of  20  ft.,  and  .arrie.  a  direct  com- 

dTt'o-thrrir:'^  '^-  ^^^-^  -  ^^^  ^--^in. --. ».  inchiz, 

23.  The  end  p.n  o  a  riveted  span  is  pjareil  21  ins.  U-low  the  eentrt-  of  the 

mZ:^  JT'  '*'%^'"^'"«  ^"T-t  P^x'— '  by  a  traction "o^dS 
100,000  lbs.  on  the  truss.     Is  this  a  faulty  detail?     Why? 

j^    T»,»  .„.  1         .  ..  ^"*-  2,286,000  in.-lbs. 

,.  »*•  The  otal  reaction  on  a  roller  (cylindrical)  shoe  of  a  span  is  l.-iOO-OOO 
lbs  The  aJlowablc  Ix-anng  on  the  limestone  masonrv  is  .JOO  lbs  pe;  .nuare 
mch  and  the  lH..-jr.nK  intensity  on  steel  rollers  is  ;,-fiOO,/  whe«.  p-jmLiZ 
per  meal  inch  of  roller  and  ./-the  diameter.  Find  the  si,.e  of  base  an^  h^ 
number  and  dimensions  of  rollers  „.,,uir..,l.  (.V.«.  The  roller,  should  be^ 
pro,K.rt.oned  that  the  distance  from  the  «lge  of  the  spaces  thev  oc^^upv  to  tS 

Taking  the  same  loading,  but,  instead  of  limestone  'using  granite  for  the 
coping,  the  ^anitc  having  a  safe  l>earing  strength  of  .^W  Ibs.'^p^  squl.1   nch 
ch.tcnnine  the  si^e  of  base  an.l  the  numl..r  and  .limensions  of  "y  JI/  n  lie™ 
requm-d,  the  rollers  to  occupy  a  space  lOJ  ins.  les.s  in  each  direction  thin  th" 
area  required  for  liearing  <in  the  masonry.  * 

An.    .^MK)  s<i.  ins  :  elei'cn  rollers  4  ill's,  diam  x57  ins.  long-2727  so   ins 
eight  rollers  7i  iiis.  diam.  x  42  ins.  long.  ^'        ' 

Ihs  "„?''•;"•»''"'''*/•';  '^"  "•"•'"  "^  "  "'"-bearing  draw.,,«,n  is  3.000,000 

1    The  diameU-r  o    drum  .s  IS  ft.      Allowing  ,he  same  i  itei.sifyT  the 

rollers  as  ,n  the  premling  example,  determine  the  size  and  numi;eT  requi«d! 

-     .    ,  ,       ,  •^"*-  •*«;  l«i"«.  diam.  X 7.81  ins.  faee, 

16.  A  drum  of  a  draw-s,«,n  is  20  ft.  in  .liameter.  A  load  of  4  000  000 
it™  Tl!r,r"  '^  -''-"y '»'^\"but<Hl  over  eight  ,K,in.s  of  sup^r;^^ 
drum.     Fffld  the  moment  on  the  drum  by  the  formula  .1/  -  ,',  Wl 

17.  A  bent  for  .  viaduct  is  30  ft.  high  from  top  of  maiilnrTlo^'w  of'T.il 
Th.  nJumns  an.  spare<l  a.  1.1  ft.  n-ntn-s.  ami  an,  subje.t  to  transvenn,  S 

«  SO.00O  lbs.     The  base  of  column  is  2  ft.  »,,u«n>,  and  the  anchor-b<,ltsTr^ 

to  hx  the  bot  oms  of  ...iu.nns.  supposing  the  t^olumna  to  U-  supp„r^«i  tn^T 
ven«ly  by  a  lloor-beam  5  ft.  dc^.p.  ^„,  "^^^^  «»*- 


826 


THEORY  OF  STRICTURES. 


a8.  The  train  load  on  an  elevated  steel  structure  is  5000  lbs.  per  lineal 
foot;  the  structure  is  divided  into  bays,  each  consisting  of  six  4r)-ft.  spans 
and  one  30-ft.  tower-span.  .\ll  of  the  traction  load  is  carried  by  the  longi- 
tudinal diagonal  braces  of  the  towers,  the  diagonals  licing  placed  on  an  angle 
of  4,")  degrees.  .V.ssuniitig  the  ctH-fTicicnt  of  friction  between  the  wheels  and 
rails  to  Ih' O.'JO,  find  the  greatest  stress  on  the  tower  diagonals.  There  is  an 
expansion  joint  Ix-twecn  each  bay.  The  diagonals  are  to  be  rigid  sections 
capable  of  carrying  either  tension  or  conipres.sion. 

If,  in.stead  of  using  to'wcr  bra<'inR  to  ])rovide  for  the  longitudinal  thrust 
from  traction  load,  the  coluniiis  are  made  rigitl  enough  to  resist  the  effect  of 
same,  ilctcrininc  the  Ix-nding  moment  on  each  colinnti,  assuming  the  same  load- 
ing and  that  tiic  structure  is  divided  into  ."f()-ft.  panels.  The  expansion  joints 
are  3U0  ft.  apart,  the  unsup|«)rte(l  length  of  columns  is  30  ft.,  and  the  mlumns 
are  fi.xed  top  and  l)ottom.  .l«.v.  l()ti,0(>7  lbs.  (T.  or  C);  4,.tO(),(K)0  in.-lbs. 
39'  Prepare   a   table   giving   the   stresses    in    the   several    memlH'rs   of  a 

singlc-inierseition  through-truss  of  \'A  ft. 

tnels. 
(or 
Fig.  802.  bridge)  loads  are  L'7,.tOO,  17,(»00,  and  8470 

lbs.,  resjK'ctively. 


ti   (a  ^    ^    '.   '       M   N   M-  N — ■    The    panel    engine,    live,    and    dead 


Ans. 


nia«. 


p 

t 
If 


\fult.i  2.'M)U  I  Mult. 


lU 
14 

N 
7 
A 
5 

4 


'  2.'>.(NK)  4.5 

2  J  .5(10  .3fi 

20.(K)0  2>* 

.  17,5(XI  21 

I.I.IKW  15 

I2..V)0  1(1 

n).unu  6 


IflOO 

nyym 


Sum 

«7,0()0 


57.l'i(M){M).niO 
44,H(XI  (M.SOO 
:<.'l.(i(M)  5I.|(X) 
24.(MI(I  .'<<t.(X)l) 
lli.(K!Oi  2H.5(HI 
9.fH)0i  19.(i00 


Mull. 


.55 
44 
.1.1 
22 
M 
0 
-11 


770 
42  ..1.50 

rw.Hso 

25.41(1 
1(>.»40 
S.470 

-8.470 


Sum. 


i:j9.:i.50 

I  I.-1.9S0 
0U.2I0 
«S,040 
47.470 
28.500 
11,130 


1.22  1 

1.22 ; 

1.22 
1.22 
1.22 
1.22 
1.22 


Total  Mu. 

I     Htm*. 

170.007 

i.ig.a5a 

1 10.0.57 
H.3.aOB 
57.914 
.14,770 
13,570 


Panel. 


Is 
Is 


Mult. 

10 
-1 
-1 
-1 
-1 


3270 

.12.700 
-3.270 
-3.270 
-3.270 
-  3.270 


Mult. 

2370 

Sum. 

tan. 

Panel 

St  rex*. 

Total  PaMl 
Strma. 

■l.-i 

lOA.n.'M) 

139.3.50 

!'f 

97  ..545 

97  ..545 

V, 

I0«i.r>50 

lUl..3K0 

72  ..300 

109.91 1 

34 

KO.-lMd 

77.310 

*' 

54.117 

224 .02H 

23 

.'.4.511) 

51.240 

•  * 

:{5.NA)( 

289,H96 

12 

28.440 

25.170 

17.619 

277.515 

30.  .\  seven-panel  si!igle-inters(>ction  tru.s,s  for  a  .single-track  bridge  has 
a  length  of  lO.i  ft.  and  a  depth  of  20  ft.,  the  lo.-id  being  on  the  lower  chord. 
Find  the  stres.scs  in  tlie  several  meml)ers  {a)  when  the  aix-x  live  load  is  12 
tons,  (/.)  when  the  live  load  is  ])r(Mluce(l  by  ( .mcentrated  loads  of  S,  8,  20, 
20,  s,  H,  S,  S,  and  a  uniformly  distributed  load  of  1..')  tons  |)er  litieal  foot, 
following  each  other  in  order  over  the  bridge  at  the  distances  of  5.5,9,8, 
H,  '.».."),  .'i,  .">..'>,  5,  and  U  ft.  a|>art. 


31.  The  two  trussc^s  for  a  IC-ft.  roadway 
17  ft.  .'}  ins.  dc<'i>,  and  of  the  tyjie  repre- 
sented in  the  ficure;  under  n  live  load  of 
II JO  lbs.  [XT  lineal  foot  the  greatest  total 
stress  in  ,1  li  is  3.'i,400  lbs.  IVterinine  tho 
permanent  load. 


are 


each  100  ft. 

A      C 


in  the  clear, 


Kio.  S<»3. 


EXAMPLES. 


827 


The  diagonals  and  ^•crt.cal8  are  riveted  to  angle-irona  forming  part  of  the 

Nr^7  «""'".,""' Y  *••"•  "''''"  *'■•'  ''''^''''^  ^"'  »»>«  connection  ^.4«  and 
t....  f  :1     '  *'7.'"""y  ""^  ""K'ired  between  .4  and  Cto  re«i.t  the 

/,^wl"i7  «"K'«-'r»»«  to  Hlip  longitudinally?    Working-shear  stress - 

m,(m  1  «  rn^r  square  ..u-h.  .4,,,.  708.6  lbs. ;  8.  4. 15. 

32.  .\  1  ratt  tru.w  unth  sloping  end  po«ts  has  a  length  of  l.W  ft.  centre 

i    '7\     \  '"'i.^n'^^'  "'  "^  ^*-  •^"'"'  »"  '■'•"♦'*'  '^'h  Pan^l"  «-^  ft-  long-  the 
doad  load  ..s  3000  lbs.  per  lineal  f.n.t.  and  th.-  live  load  12,000  lbs.     Determ  ne 

he  muld  e  of    he  b<,ttom  .hord.  and  in  the  memlK-rs  of  the  third  panel  met 
by  a  vertical  plane         ..l„.s.   mAH;  Hl.Ar.,  imim;  12.i;  iHyieJATtonT 

33.  A  ten-panel  smgL-intersectioM  through-bridge  of  170  ft.  «.«n  is  -'.^  ft 
f.  ms.  m  height,  and  has  .loor-U-ams  13  ft.  in  length.     How  manv  1-in.  rivets 
are  required  .«  the  th.rd  panel  from  one  end  to  connect  the  web  with  the 
ltT!:n~;^b;V*"'"''  "^■"  '•"-•  »-  •-  •■^•«««  »-  -^  »»•«  panel  dead 

.1/M.   17  9  and  12,  f  he  shearing  strength. .f  the  rivets  iK-ing  10  OOOIbs./sq.  in 

Timber  floor-lx-ams  are  lai.l  upon  the  upp,.r  chonl.  -J  ft.  cvntre  to  centre- 

beams  for  the   oa,l.„g  k.vcu  „.  F,g.  (V.'2,  p.  (VS.!.      Vis,,  determine  the  maxi- 

.In.-.  10  in.  V  10.4  in  ;  max.  chord  stresses -30.08  and  32.41  tons-  max 
diagonal  stress -S.99  tons.  i"ita,  max, 

,.„n^f'  ■'T':"'  "^  ■"**•  '•'•  "•  "•  ""•'  ^'  *"""  ^"""^^  '^'*'-'>  "t»"-^  in  order  over  a  ten- 
panel  single-mtecsection  truss  at  .listances  of  S.  .-ij,  41  and  4*  ft  anart  T)nil 
mine  the  position  of  the  loa.ls  which  will  give  h  Tavi^um  dTagonaU^d 
chord  stress  in  the  thini  and  fourth  pane...  Span  -  .20  ft.  «"l d.'Jh  -1  "t 
Ans.  The  max.  shears  in  the  3d  and  4th  panels  are  lf,.2S7.-,  and  13  5r'5 
tons  respectively,  and  .H-cur  when  the  3i-ton  wheel  is  fir.t  at  the  third  pane'l- 
pomt  and  second  at  the  fourth  panel-point;  the  maximum  chord  t^si„ 
the  3d  and  4th  ,»anels  are  .33.77.'5  tons  and  43..587.'5  tons  respectivelTand 

nrsi  b-ton  wheel  is  at  a  |)anel-j»oint. 

Ste'Llvt'T.'"'"  1^"  "TTu  "'  "'"'"tance  of  a  floor-beam  for  the  Sault 
and  nu  Z'^^  '"I^'  t'""."''"^  '••""=  ^''""^-»'<-'"ns.  16  ft.  6  ins.  long 
and     3  ft.  10    ins   apart;   the  .lead  weight  of  the  H.K,ring.  stringers,  etc  - 

n    a^'.ri    T'    ""'  "^"•T^«'^=    'he  live  I.,„.l  as  p'ven  in   Fig     ,22 
p.  ftS3.   the  loa.l  IS  transmitte.l  t.,  the  fl.«,r-lH.am  bv  four  lines  of  strin«'rs 

rft^iri-'er"^' ''-" '''-''  -^ '''-  "-♦  "^"  ^\ '-- pair.  Vhir:;- 

>i.    i„       .         ,    . ,        ,  'ln».  290.294  ft  .-tons. 

37.  In  a  truss-bridge  the  panels  arc  17  ft.  and  the  floor-l,eams  13  ft    in 
length.     Ixiads  of  S,  1L>,  12,  12    r>    10    10    in  «...!  in  ♦        t  1.  ■       ". 

in  order  over  the  bridge'at  thl  .ii;t-an,.e;  UV     4  '    J  4  "I   ^oVr^S  tu 

duo  to  the  pUtform,  etc.,  to  be  500  lbs.  per  lineal  foot.    An,.  82.2-225  ft!.ton,! 


MICROCOPY    RESOLUTION   TEST   CHART 

(ANSI  and  ISO  TFST  CHART  No    2) 


1.0 


I.I 


1.25 


1.4 


;-ia  ilia 

fm  i^ 
^  ^  IIIIIM 

1.8 
1.6 


^     APPLIED  IIVMGE     Ir 


'*"     •".■  1500        Wiiin, 


828 


TIIEOUY  OF  STRICTURES. 


38.  Witli  the  loading  given  by  Fig.  SO*  design  a  floor-beam  for  a  single 
track  bridge  with  panels  11  ft.  long,  the  weight  of  the  platform  being  450  lbs. 
per  square  yard,  and  of  each  longitudinal  200  lbs.  per  lin.- .;  /ara. 


/0„.Q»(?Mf;><f:)  ,.^,.^,.^^,^  „,y^      / 


FiQ.  804. 


Ans.  The  R.M.  in  cross-tie  is  greatest  when  the  3d  22,000-lb.  load  is  at  a 
panel-point,  and  total  max.  H.M.  =  i;W.SS3  ft.-tons. 

39.  Design  a  (ro.ss-girder  for  a  five-])anel  through-truss  bridge,  120  ft. 
long,  for  Cooiht's  standard  loading  E  50.  The  span  of  the  girder  is  ITft.'o  ins! 
back  to  back,  of  lonncction  angles,  and  there  are  two  lines  of  stringers  8  ft. 
apart.  The  dead  load  of  the  stringers  and  floor  system  may  be  taken  as 
7(H)  ll)s.  |ier  foot  run  of  bridge,  and  the  estimated  weight  of  the  cross-eirder 
i-i  :{(H)0  lbs. 

40.  Design  the  central  cross-section  of  a  plate  girder  having  an  effective 
span  of  50  ft.  and  a  depth  of  5  ft.  (5  ins.  centre  to  centre  of  the  flanges,  for 
('ooiicrs  standard  loading  E  .jO;  the  dead  load  may  be  assumed  to  be  0.50  lbs. 
[XT  foot  run  of  girder. 

41.  The  stringers  for  a  railroad  span  are  spaced  Sft.  centres.  The  tics  are  to 
!.(>  -paced  5  ins.  apart  in  the  clear.  Assuming  that  the  rails  are  strong  enough 
10  (l>slril)ute  each  wheel  concentration  ciiually  <.ver  three  ties,  and  that  the 
tinilMT  will  carry  safely  an  extn-me  fibre  stress  of  2000  lbs.  jx-r  .square  inch 
including  impact ;  determine  the  size  of  tie  required  for  a  whivl  load  of  25,000 
ll)s.  Distance  centre  to  centre  of  rails  =  4  ft.  11  ins.;  impact  SO  jH-r  cent  of 
•'uliveKmd.  ,,,,    Si"xl0". 

42.  Design  the  central  section  of  a  plate  girder  of  45  ft.  span  and  5  ft. 
lecp  to  carry  a  dea<l  load  of  ."iOO  lbs.  [kt  fo(.t  run,  a  live  load  of  3200  lbs.  per 

foot  run,  and  an  inipa.l  load  of  2400  lbs.  ^tL-r  foot  run;  also  determine  the 
lengths  of  the  flange  plates. 

43.  Design  a  cross-tie  for  a  double-track  op»-n-web  bridge,  the  ties  being 
18  ft.  5  ins.  centre  to  centre  and  the  live  load  for  the  floor  system  beinff 
8000  lbs.  |)er  lineal  foot.  "  * 

44.  D<'sign  a  stringer  for  a  Pratt-truss  bri<lg.<  20  ft.  long  and  3  ft.  0  ins 
back  to  back  of  angles.  The  live  load  is  :}2(K)  lbs.  jht  foot  run  of  stringer' 
and  the  estin.ate<l  weight  of  the  dead  load,  iticluding  the  weight  of  the  striniter' 
is  400  lbs.  i)er  foot  run.  ' 

45.  r>esign  the  s».cf  ion  of  the  top  chord  memlxT  of  a  bridge  f<.r  a  total  dead 
live,  and  impact  load  of  (i2().(KK)  lbs.     The  length  of  the  member  is  25  ft! 

and  the  inside  wi.lth  is  15  ins.     Use  tliB  column  formula  p /l  +      '' ^m 

^  V       llOOOr'/ 
17,000  lbs.  |)er  s(|uare  inch. 


EXAMPLES. 


8_'y 


46.  A  panel  of  the  top  chord  of  a  deck-span  is  20  ft   in  U,.^h .  »k    j-      . 
comp^ssive  stress  is  320.000  lbs.;   the  topThordls  subjS   '  '"'  '""' 
to  a  bending  moment  due  to  a  uniform  load   of  4000  lbs    per 
Imeal  foot.    The  allowable  extreme  fibre  stress  in  eiZer  tensbn      j, 
or  compression  for  the  combined  loading  is  16.000  lbs.  per  square  J 
inch.   Determine  the  bending  moment  by  the  formula  Af  =  A«,Z>     Fic 
Find  the  section  required  by  the  form  shown  in  the  diagram 

Ans.  One  18" Xf" cover;  ^f^o top 3"x3"xrx7.2  lb.  angles-  two20"Xl'' 
webs :  two  bottom  6"  x  4"  X  ft"  x  18. 1  lb.  angles.  * 

47.  The  figure  represents  a  counterbalanced  swing-bridge,   16  ft.  deep 
A  A  AAA   KK--4-      "'":  ^''°?^  supported   upon    the    turntable    at   A 
CjytVA    r^^?-       r  u    iL    \'*^*'^  ^'"^'^^^  '«  ^'^  'bs.  per  lineal  foot 


Fio.  80.5. 


MH- 


\:\ 


13^ 


ot  bridge;    the  counterpoise  is  hung  from  C  and  D 
Fio.  806.  *'"a  'ts  weight,  assuming  (a)  that  the  whole  of  it 

to  mLke  fh*.  r.„nti„n      '';'^"!""'"«d  t«  5;    «>)  that  a  portion  of  it  sufficient 
b^r  B£     Al^  H  '\'"*^  ^  "'^""'  '^  transmitted  to  .1  through  a  mem- 

ber BE.     Also,  determine  th.  stres.,es  in  the  several  members  of  the  truss. 
Arts.  Counterpoise  in  case  (a)  -26,162 J  lbs.; 
in  case  (6)  -22,lS6il  lbs'. 
A9.   T»,»  fi  ^^^  transmitted  through  BE  in  ca.se  (6)  -6962  lbs. 

the  counterpoise  is  suspended  from  CD.     Find  — -c 7P~^ 

Its  value,  the  joint  ,a  E  being  so  designed  that     - :X?:Vr^Z?:\A>^ 
the  whole  of  the  load  upon  the  bridge  is  always  '*■.:  'S.^?   V^^o 

transmitted  through  the  main  posts  EA,  EB  p,„  an, 

and  IS  evenly  distributed  between  the  points  of  support  at  A  and  B     PinH 

has\%;^n:rZdll^d  ?  ?tlf  ^^;?  r  '^'-^--^"^'^^^^^^^^ 

stress  diagram  when  the  bri£  is  ^^.  ICT^T^^'  "^ .^""^  ^'"-^  *''« 
the  stress  diagram  when  the  SJp  i^*^  ^^«"  ""<»  the  reactions  and  draw 
live  load  of  4  t^ns.  '^  "  ''"*'^-     »«termine  the  effect  of  a  panel 

50.  A  draw-.pa„  having  equal  arms  of  ,00  ft.  length.  div^l^Jd  inloLr 


I'K!     SOS. 


Fio.  HOI). 


S:«) 


THEORY  OF  STRUCTURES. 


1.«<1  is  concentrated  at  the  first,  second,  and  third  panel-points.  Also  find  the 
stress  in  A  when  the  bridge  is  open. 

51.  The  wedges  beneath  the  ends  of  a  draw-span  are  each  required  to 
exert  an  uplift  of  1(K),0(K)  lbs.  The  slop*,  of  the  wedges  is  1  in.  vertical  to 
o  MIS.  horizontal.  The  coefficient  of  friction  on  the  two  surfaces  of  the  we.lges 
IS  OM).  Required  the  horizontal  for.c  neces.sary  to  drive  and  draw  each 
wedge.  " 

-W/.v.  .50,000  lbs.:  10,000  lbs.  In  practice,  however,  it  usuallv  takes  as 
much  to  draw  as  to  drive  the  wedge  if  ojx-rated  onlv  at  long  interval's. 

52.  \  draw-span  -.iJO  ft.  in  length  is  divided  into  one  centre  panel  of  20  ft 
an.l  twelve  ordinary  panels  of  L'.l  ft.  ca.^h.  One  arm  of  the  span  is  subjected 
to  an  unbalanced  wind  load  of  T,  lbs.  jK-r  stp.are  foot  of  exposed  area,  or  100 
n.s  ,H.r  lineal  foot  of  arm.  The  diameter  <.f  the  pitch-circle  of  the  rack  is 
^ott  1  here  are  two  main  dnving-i)inions  gearing  into  the  rack.  Find  (a)  the 
tjjoth  pressure  on  each  necessary  to  overcome  the  unbalanced  wind  load- 
also  (,)  the  ],ower  required  to  turn  the  soan  against  this  load  through  one' 
tourth  revolution  in  two  minutes. 

The  total  weight  of  the  .Iraw-span  is  1,000,000  lbs.  Find  (c)  the  power 
re.,uired  to  turn  the  span  through  one  fourth  revolution  in  one  half  min- 
ute UJuler  ordinary  conditions;  also  (,/)  against   the  unbalanced  wind  load 


in  two  minutes. 


Use  the  formula  II.I'.  -~^^  "  for  determining  the  power 


required  to   overcome  friction,  accelerate  motion,  etc.,  where  IF -weight  of 
span  and  i-  -velocity  of  pitch-circle  on  rack  in  feet  per  second 

(c)  Determine  the  size  of  main  pinions  for  the  power  required  in  Case 
(r).  assuming  that  the  gears  are  to  he  made  of  east  steel  capable  of  a  safe 
extreme  fibre  stress  of  Ki.OOO  lbs.  ,x>r  square  inch.  Make  the  face  of  pinions 
two  and  one  half  times  the  jntch  and  use  the  formula  />-0.025VfP  where 
/)  -  pitch  re.piired  and  P  the  tooth  pressure  on  each  pinion. 

The  shafts  supporting  the  main  driving-pinions  have  boxes  located  so 
that  the  Ix-iiding  and  torsional  moments  are  equal;  determine  (/)  the  size  of 
shaft  rcpiired,  assuming  f.r  such  conditions  that  steel  shafting  will  carry 
safely  an  (>xtreme  fibre  stress  of  21,000  lbs.  jn^r  .scjuare  inch. 

Talxe  the  draw-span  to  Ix-  moving  at  the  rale  of  one  revolution  in  four 
minutes.  Required  (q)  the  force  at  each  of  the  two  ends  neces.sary  to  bring  the 
span  t..  rest  in  a  sp,.ee  of  dins,  .\ssume  (he  1,(KK).(X)0  lbs.  of  weight  to  be 
equally  distributed  Ihrough.ut  the  span.  Find  (h)  the  size  ..f  latch  re,,uired 
for  stopping  the  span  under  th<'  conditions  naiiiMl,  assuming  the  latch  is  of 
a  rectangular  s«-ction  of  steel,  an.l  that  the  resistance  is  obtained  by  bend- 
ing on  same,  the  unsupportc.l  length  or  lever-arm  l)eing  20  ins.  Working 
Htr«ss  on  extreme  fibre  -  l(i,(KK)  lbs.  jx-r  sipiare  inch. 

.W/.«.  .-il.JtK)  lbs.:  ;«).,-,,  14.S,  .-{.T,  34.2  H.P.:  .3J  ins.  pitchxSi  ins.  facex 
l-.S  ins  diam.xKl  teeth;  diam.-C.l  ins.;  (;,S,2(K)  lbs.,  assuming  the  weight 
of  the  draw-span  to  1h<  uniformly  distributed;  the  latch  to  1k>  10  ins.  wide 
X.).I1.-|  ins.  thick,  whi.h  presu|)po.s,.s  that  the  latch  is  caught  in  such  a  way 
.■w  to  '\,r<  \hv  jrftrMiis-  uf  fi,S,2t)0  lbs.  throughout  the  distance  of  (i  ins. 


EXAMPLES. 


831 


53.  A  bent  for  an  elevated  structure  consists  of  two  columns  spaced  ''0  ft 
apart  Between  the  two  columns  is  a  cross-girder  AB  which  must  carry  a 
dead  load  of  4000  lbs.  fjcr  foot  ,>f  its  length  and  a  live  load  of  6000  lbs  jjcr  foot 
Outside  of  each  column  are  cantilever  brackets  AC  and  BD,  which  support 
the  same  live  and  dead  loads  ,)er  foot.  Find  the  greatest  upward  and  down- 
ward moments  on  the  cross-girder;  the  moment  in  each  cantilever  at  the 
columns;  the  total  moment  for  which  the  cross-girder  should  be  proportioned 
and  the  maximum  and  minimum  loads  which  each  column  will  receive  The' 
cross-girder  should  be  designed  to  carry  the  greater  moment  plus  three  fourths 
of  the  lesser. 

Ans.  (a)  L.L.  on  AC  and  BD,  net  upward  moment  =300,000  ft.-lbs  ■  (6) 
L.L.  on  AB,  net  downward  mom™/ =.300,000  ft.-lbs.;  (c)  L.L.  on  CB  'max 
hKul  on  column  -21.1,(M)0  lbs.;  (d)  max.  B.M.  on  cross^irder- 52^,000  ft.-lbs  • 
(c)  L.L.  on  BD  min.  load  on  column  =  05,000  lbs.  '  "' 

54.  Find  the  maximum  stresses  in  the  several  members 
of  the  compound  \  bridge  truss  of  1(50  ft.  span  and  40  ft. 
depth  shown  by  diagram,  the  panel  dead  and  live  loads 
being  2  and  4  tons  respectively. 


-•«— — 
Kio.  810. 


55-  In  a  seven-panel  Pegram  truss  for  a  span  of  200  ft.,  the  upper  panel- 
/O^MXRVTVx    P"'"**  "''  '"  »  '•'"•ular  arc,  with  a  chord  of  160  ft.  and  a 

p,„    o,  '  ''^^^^  »'"e  of  15  ft.     Taking  the   dead  load  at  900  lbs. 

'"•  ""■  ami  the  live  load  at  1800  lbs.  per  lineal  foot  per  truss,  deter- 

mme  the  stresses  in  every  member  of  the  truss.     (See  Ex.  6,  p.  706.) 

56.  The  figure  represents  the  half  of  one  of  the  trusses  for  a  bridge  of 
0  ft.  span,  the  nanpl  (lr>n<l  mwl  \\\ra  inn^c  k»:..~  a i 


It'     It'      16'     It' 


^^^ 


--r<."^  .^y\.a^in\.sy  iiic  umi  ui    one  oi    me  in 

'20  ft.  span,  the  panel  dead  and  live  loads  being  6  and 
4  tons  rcs|K'(tivoly.  Determine  the  lengths  of  the  ver- 
ticals and  the  stresses  in  the  several  members  (a)  so 
that  the  stress  in  each  meml)er  of  the  lower  chord  may 
be  80  tons;  {b)  so  that  the  minimum  stress  in  each 
diagonal  may  be  zero. 

Am.  (a)  7.3,  11.73,  14.27,  and  15.02  ft.;  (6)  7.3,  11.62,  13..55,  and  1.3.55  ft. 

57.  Determine  the  live-load  stresses  in  the  members  of  the  cantilever 


Fio.  812. 


Fio.  813. 
truss  shown  by  Fig.  813  when  subjected  to  a  panel  live  load  of  55  ton«. 

58.  The  accompanying  truss  of  240  ft.  span  and  30  ft.  deep  is  to  be  d*. 


«], 


If 


S.i2 


THEORY  OF  STRUCTURES 


sigued  for  a  panel  engine  load  of  24,()00  lbs.,  u  panel  train  load  of  IS.OOO  lbs., 
!m  and  a  |)anol  bridge  load  of  12,000  lbs.     Deter- 

'   l\i.-^r\iii\i^T\i^i\i^  -■*   '""'"'    S'"-'i'l>''"'lly    the    maxln-um     stresses    in 

^Cl^Cl>fCL>^T>^^^  *   the  members  met   by  the  vertieal  MX.     Also, 

j  draw    a    stre.ss    diagram    for    the    whole    truss 

1'^  when  it  is  covered  with  a  uniformly  distributed 

'"'"    ^^*  live  load  of  I,SO,0()<)  lbs. 

59.  The  compression  chonl  of  a  bowstring  truss  is  a  circular  arc  of  80  ft. 
span  and  10  ft.  rise;  the  bracing  is  of  the  isosceles  ty{)e,  the  ba.ses  of  the 
isosceles  triangles  dividing  the  tension  chord  into  eight  equal  lengths.  Deter- 
mine the  maximum  stresses  in  the  members  met  by  a  vertical  plane  128  ft. 
from  one  end.     The  live  and  dead  loads  are  each  i  ton  [x^r  lineal  foot. 

Ans.  Chord  tension    =77.30  tons;  chord  compression  =80.3  tons; 
Diagonal  stress  =   7.9.")  tons. 

60.  Design  a  parabolic  bowstring  truss  of  ,S0  ft.  span  and  10  ft.  rise  for 
a  dead  load  of  J  ton  and  a  live  load  of  1  ton  jmt  lineal  foot.  The  joints  l)et  ween 
the  web  ami  the  tension  chord  are  to  divide  the  latter  into  eight  equal  divisions. 

6i.  The  compression  chord  of  a  bowstring  truss  is  a  circular  arc.  The 
depth  of  the  truss  is  14  ft.  at  the  centreand  4  ft.  at  each  end;  the  span  =100  ft.; 
the  load  upon  the  truss  =S40  lbs.  \wr  lineal  foot.  Find  the  stresses  in  all 
the  members.  Di-termine  also  the  maximum  stres.st>s  in  the  members  met 
by  a  vertical  J")  ft.  from  one  end  when  a  live  loa,d  of  1000  lbs.  |x;r  lineal  foot 
crosses  the  girder.     What  counterbraces  are  required? 

62.  .\  bowstring  truss  of  120  ft.  span  and  1.5  ft.  rise  is  of  the  isosceles 
braced  ty[)e,  the  bases  of  the  isosceles  triangles  dividing  the  tension  chord 
into  twelve  equal  divisions;  the  dead  and  live  loads  are  J  ton  and  1  ton  per 
lineal  foot  res]K>i'tively.  Find  the  maximum  stresses  in  the  memliers  met 
by  vertical  i>lanes  immediately  on  the  right  of  the  second  and  fourth  joints 
in  the  tension  chord. 

63.  Fig.  Sl.").l  represents  the  riveted  truss  for  a  span  of  126  ft.  centre-to- 
centre  eml  pins.  The  trusses  are  17  ft.  <-entrc  to  centre,  28  ft.  deep,  and  are 
each  divifled  into  six  panels.  Figs.  Hliili 
and  ('  show  the  lower  and  tip(M'r  lateral 
systems,  the  wind  load  on  the  former  being 
4.")0  lbs.  |M'r  liiieal  foot  wheti  the  span  is 
loaded  and  2(K)  lbs.  jxt  lineal  foot  when 
the  span  is  empty.  The  wind  load  on  the 
upiMT  system  is  l.V)  Ih-i.  |)er  lineal  foot. 
The  live  and  dead  loads  are  2  tons  and  J 
ton  rcs[KMtivcly  [K-r  lineal  fool  of  span.  (a)  Prepare  tables  of  maximum 
stresses,  (/,)  Mild  maximum  wind  stresses  in  lM)tt(mi  chords  and  diagonals 
of  lateral  systems,  (r)  Will  there  l)e  reversion  in  the  end  panels  of  the 
bottom  chonl,  .Mssuming  the  diagonals  to  In-  in  tension?  (tl)  Find  the  effect 
of  the  transferre(l  wind  lo.id  on  the  incline<l  end  posts  and  liottom  chords. 
(e)  Discuss  the  wind  effect  on  the  jwrtal  bracing,  Fig.  H15D, 


Fio.  SI- 


EXAMPLES. 


S33 


Fig.  810. 

/=  18.000-70^.) 
r 


64.  Determine  the  direct  and  transferred  wind 
load.s  on  the  lower  chords  of  the  pin-connected 
irus.s-hridge  shown  by  Fig.  ,810,  when  the  wind 
I'.-kI  per  foot  run  on  the  upfK-r  chonl  i.s  200 
ii).s.  and  on  tlie  lower  chord  is  oOO  lbs  per 
ln..t  run.  The  .lea.l-load  stres.se.s  in  the  panels 
of  the  lower  chord  are  84,000,  84,000,  114,000 
atul  125,000  lbs.,  as  indi.-ated.  Is  there  any 
tendency  to  reversion  in  the  lower-chord  mem- 
lx;rs?     (Max.  width  of  batter  post=23  ins.;  r  =  7in.s.,  , 

The  end  post  of  the  truss  is  subjected  to  a  total  direct  stress  of  55o'oOO  Ib8 
due  to  the  dead,  live,  and  impact  loads,  and  the  wind  load  on  the  upper-chord 
members  ,s  estimated  at  200  lbs.  per  foot  run.  Determine  the  section  of  the 
member  on,the  assumption  that  there  is.a  point  of  inflexion  at  N.  Am  AQ  so  (ins 

-^»  "'"^  "^'^  '•""'■'  ^'""  ^^"^  t"""^  «»»"^v»  by  Fig.  817  are  eVjOO  and 

...>00  lbs.  per  foot  run  respectively,  and  the  wind  load'^per^ot  run  is^  bs' 
for  the  uprx^r  and  .WO  lbs.  for  the  lower  chord.  Will  there  be  any  rever^on  ol 
stress  m  the  lower  chord?    Determine  the  wind-bracing  stn^ases 

fifi    t"tl^.  "'version;  diag.  stresses,  22  500;  16.070;  10,715;  6430  lbs. 
o-  f,         u^il:  f"«'''-t'-a('k  railway  span  is  divided  into  eight  panels  of 

'  ^'-^^h  ^^'  *""•"'"  "'''  ^^  ^'-  ''"-P'  '■""»'•''  to  centr,.  of  chords  and  are 
spaced  18  t.  centre  to  centre.  The  wind  load  on  the  upper  la;;t'l  s^tl 
.>  -'00  ll,.s.  iK-r  hneal  foot  of  span.  Fin.l  („)  the  n,axim.:m  bending  mome,^ 
m  .nch-pounds  on  each  inclined  end  post  or  batter-hrace  .lue  to  ttfra"  verse 


Fio.  817. 


Fio.  818. 


v-ind,  with  portals  as  shou-n  on  J-ig.  818.     Also  find  (6)  the  transferred  wind- 

:::i';i.r;;;:;;;;nri.r"^-''^"'-^  -"^'^^•'^  ^^—  -  *»>-'*^-^ 

C£-10,3501b8.(T.);FG-0,flF-0,a„dtG'-0  27,8.50  lbs.    (C); 


I  nmo  41IJ.1U)  r,,,,u,  4, 

Fto.  810. 


II 


! 


834 


THEORY  OF  STRUCTURES. 


centres  =  171  ft.    The  bridge  is  designed  to  bear  the  loading  given  by  Fig. 
(122,  p.  683.     Show  that 

(a)  The  stresses  in  every  panel  length  of  each  chord  are  greatest  when 
the  third  driver  is  at  a  panel-point;  and  find  the  value  of  the  si'veral  stresses. 
(6)  The  stres;sos  in  the  verticals  a  and  the  diagonals  b  are  greatest  when 
the  third  driver  is  at  a  panel-point ;  and  find  their  values. 

(f)  The  stresses  in  the  remaining  members  of  the  truss  are  greatest  when 
the  second  driver  is  at  a  panel-point ;  and  find  their  values. 

(rf)  The  maximum  stresses  in  the  verticals  d  vary  from  a  tension  of  64,000 
lbs.  to  a  compression  of  11,000  lbs. 

(f)  The  stress  in  the  counterbrace  c  is  nil. 

A  lis.  The  values  of  the  stresses  in  the  several  members  are  marked  on 
the  diagram.  They  are  deduced  from  the  distributions  given  in  the  table 
on  p.  685,  and  are  correct  wthin  a  very  small  jxjrcentage. 

68.  An  eight-panel  bridge,  Fig.  820, 
of  200  ft.  span  and  30}  ft.  rise  at  the 
centre  has  two  main  trusses  18  ft.  apart. 
The  dead  load  is  estimated  at  2000  lbs. 
and  the  live  load  at  4700  lbs.  per  foot 
run  of  bridge. 

Determine  the  stresses  in  one  of  the 

trusses  due  to  thi>  dead  load  and  the 

reversion  stres.ses  in  the  web  members 

due  to  the  live  load. 

Calculate  the  stresses  in  the  lower  lateral  bracing  and  the  reversion  stresses 

in  the  lower  chord  for  a  wind  load  on  the  upper  chord  of  200  lbs.  and  on 

the  lower  chord  of  480  lbs.  per  foot  run  of  bridge. 

69.  Determine  all  the  stresses  in  the  Hsdtimore  truss  shown  by  Fig.  821  for 


irxryiyiyixixb^ 


Fig.  820. 


Fio.  821. 


a  dead  load  of  000  lbs.  and  a  live  load  of  800  lbs.  per  linear  foot  of  truss.  In 
determining  the  stresses  in  the  verticals  it  may  bo  assumed  that  6000  lbs. 
of  the  panel  dead  load  is  concentrated  at  each  of  the  bottom-chord  panel- 
points. 

70.  The  figure  represents  one  of  the  trusses  for  the  Jefferson  City  440-ft. 
draw-span,  the  dimensions  and  dead  loads  in 
tonslM'iiigasiiidicatiHl.  In  determining  the  stresses 
in  the  viTti<'als  it  may  he  assumed  that  "-'j  tons 
of  the  (lend  load  is  ajipiicd  at  the  top-chord  panel- 
points. 

in)   Find  the  de:i(i-ioad  stresses  in  the  several 
memliers  win  n  the  span  is  o|(en.     (h)  Find  the  stresses  in  the  several  members 


EXAAfPLES. 


«% 


due  to  a  live  load  of  0.6  ton  per  lineal  foot  per  truss,  assuming  that  one  arm 
only  ,s  loaded  and  that  the  cantilever  arts  like  a  simple  girder  on  two  sup- 
ports (c)  Find  the  stresses  in  the  several  members  due  to  an  assumed  uplift 
of  1.}  tons  at  the  end  of  each  truss,  assuming  each  arm  to  act  as  a  cantilever 
resistmg  said  uplift.  (,/)  IJetermine  the  maximum  stres.ses  in  the  several 
members  for  a  live  load  of  0.6  ton  per  lineal  foot  of  span,  assuming  the  span 
continuous  over  the  four  points  of  support. 

««  ?'  fi'^'-'l  spans  for  the  bridge  are  of  the  Petit  type  and  of  dimensions 
as  shown.  The  live  and  dead  loads  are 
600  and  800  lbs.  per  lineal  foot  per  truss. 
In  determining  the  stresses  in  the  verti- 
cals it  may  be  assumed  that  6000  lbs.  of 
the  panel  dead  load  are  concentrated  at 
each  of  the  top-chord  panel-points.  Find 
(e)  the  maximum  dead-  and  live-load  stres.ses. 

71.  Figure  824  represents  one  half  of  a  truss  of  a  cantilever  bridge  span- 
ning  the  river  Agarno.  The  dead  load  for  suspended  span  is  940  lbs.  and 
the  hve  load  is  1678  lbs.  per  lineal  foot  per  truss,  (a)  Find  the  dead-  and 
live-load  stresses  in  the  members  (jf  the  suspended  span.      (6)  Find  the  erec- 


Fio.  823. 


-    -•J 
Fio.  824. 

tion  stresses  produced  by  cantilevering  out  the  suspended  span,  the  erection 
loads  being  indicated  in  tons  on  the  figure,  (c)  Determine  the  dead-load 
stresses  in  members  of  cantilever  proper  due  to  total  weight  of  suspended 
span,  namely,  108  tons.  "sptiiuea 

♦  K   ''"J"."  ♦J>'""K.»'-^P''"  "^lO  ft-  in  length  the  stringers  are  riveted  together 

aZf  Mh  '  "?'r  "T!"  "^  '"""•  '^^""''"^  '^'''  '^'^  ^♦""'^'•^^  «re  manu- 
factured the  cxa.t  l<-r,gth  for  span  with  no  I.K.d  thereon,  that  the  dea<l  load 

on  span  strains  the  metal  in  the  bottom  .nords  8000  lbs.  per  square  inch 

and  the  live  load  lO.tKK)  lbs.  ,K>r  square  inch,  find  the  effect^.n  the  stringer 

connections  when  the  span  is  swung,  an.l  when  the  live  load  is  on  the  strL 

lure.      the  span  is  divnlcd  into  seventeen  panels  of  .30  ft   each 

.1^...  Each  additional  application  of  the  live  loa.l  produ^vs  an  additional 
distortion  of  i  in.  f„r  each  ,,anel.     How  can  this  Ik-  remedied"' 

73.  Inn  riveted  joint  the  rivets  are  spaced  eccentricallv  to  the  line  of  stress 

t;^ir      f  't     """"  T"  ^''■'••*"  "^^-     "T^^"  -'-  "f  --h  riv'e" 
IS  0000  lbs.,  and  eight  are  used  m  the  connection.     Find  the  stress  in  uarh 

An,.  Total  stress  in   1  =  -UOOO  lbs.,  in  2=2000  lbs.,  in  3-2000  lbs.,  ,n 


i 


i:i 


836 


THEORY  OF  STRUCTURES. 


4-6000  lbs.,  in  5  =6000  lbs. 


in  0  =  10,000  lbs.,  in  7  =■  10,000  lbs.,  in  S  =  1 4,000 
lbs.  ix>r  inch,  showing  an  unequal  distribution 
of  stress  and  that  such  cci-cntric  connections 
should  \>e  avoided. 

74-  (iiven  an  elevated  structure  the  columns 


•i^OO  o,>.,|^P'«-''°°   of  which  are  unsujiported  for  a  distance  of  20 


ft.,  and  are  fixed  at  each  end,  the  dimen.sions 
and  arrangement  of  longitudinal  girders  and 
towers  being  as  shown  in  Fig.  826.     Find  the 


Fi(i.  S2."). 
maximum   fibre   stress   due   to   bending  caused  by    a   ma.\imum    variation 


Kx|mns(oii  I'CK'kot 


^^    J?    1    i    1    M 

k  25   ><     -10  -  X  --10--  >f,     -«0— >^  --40'-  M  -  40--  >K  iS-J 


n 


Fk;.  ,S2(>. 


Fui.  827. 


i!!: 


in  temporatun  of  100°,  assuming  the  coefficient  of  expansion  for  100°F  to  be 

0.(K)06,  and  taking  the  distance  from  centre  of  gravity  of  column  to  extreme 

fibre  in  a  longitudinal  direction  as  "J  ins.  Ans.  7."),tO  Ibs./sq.  in. 

75-  At  a  toi)-chord  panel-])oint  of  a  riveted  truss-siian  the  gravity  lines 

of  the  tiiree  memlxTs  assembling  at  this  point  do  not 

have    a   common    i-oitit    of    intersection.     Find    the 

Ix'nding  moment  about  the  jianel-point  for  the  con-  -(i- aI- 

(".iiiiMis  shown  in  Fig.  ><27. 

Alls.    1,(')(H),0(K)  in.-lbs.      Siirh   iiilerxrctimii'  ahould 

lie  (ivoiihd,  (IS  lh(  islrt  nil  fihri-  stnss  fruiii   ihr  hiiiiiiiig 

iiidiiiriits  is  frii/iiiNlhi  i/riiitir  lliitii    tliosi    pruiliiccil  hi/ 

the  (irtiiiil  li'iiils.     The  urnvit;/  liiiis  shmilil  all  intersect 

in  a  common  point. 

76.  Determine  the  stresses  in  the  several  members  of  the  three-hinged 

truss    represented     by    Fig.    828    and 
.  K,' ..  ,f/„  K,'>,  I,-.'  .  if,'^  ,f,'^  „•/.        liaving  a  panel  dead  load  of  2  tons.    If 

a  weight  of  4  tons  is  concentrated  at  each 
of  the  points  A  and  //,  find  the  stresses 
dcveloried  in  the  menilwrs  j.  .\lso  find 
the  positions  of  the  load  which  will  pro- 
duce no  stress  in  the  members  ij  and  z. 

77-  A  three-pin  arch  of  the  form  and  dimensions  shown  by  Fig.  829,  and 
having  a  span  of  ISO  ft.  and  a  rise  of 
40  ft.,  is  loaded  over  one  half  its  span 
with  a  uniformly  distributed  load  of 
1200  lbs.  jMT  linear  foot  of  truss.  l)e- 
♦ermine  the  stresses  in  the  members 
cut  by  the  planes  .\fi  and  Cl). 

78.   A  itirc..-i>in  arch  of  ]m  ft.  span  "'  '^^• 

has  a  rise  of  .12  ft.  and  is  divided  into  eight  panels.     The  live  load  per  foot 


EXAMPLES. 


8.37 


run  of  truss  is  3000  lbs.  Determin  the  stress  in  each  member  when  the 
span  IS  fully  loaded.  Also  determine, 
graphically,  the  line  of  resistance  of  the 
arch  span  for  a  live  load  covering  three 
fourths  of  the  span,  and  find  the  hori- 
zontal thrust  of  the  arch  for  the  given 
system  of  loading. 


Fig.  830. 


79.  Figure  831  is  a  skeleton  diagram  for  one  of  the  main  trusses  of  a  bridge 
of  80  ft.  span,  15  ft.  deep  at  ends,  and  I J  ft.  deep  at  centre,  pivoted  at  the 
ends  and  centre,  and  of  the  dimensions  shown.    Where  must  the  load  be 


A  ic 


Fig.  831. 

placed  between  A  and  B  so  that  it  may  produce  no  stress  in  X"!  Find  the 
ma.\imum  stresses  in  the  several  members,  the  panel  dead  and  Jive  loads 
being  2  and  4  tons  respectively. 

Show  how  the  stresses  are  modified  if  the  lower  boom,  instead  of  being 
straight,  is  a  parabola  of  the  same  rise  as  the  arch. 

8o.  The  figure  represents  a  half  bridge  of  128  ft.  span,  suspended  from 
the  point  P  and  hinged  at  O.  The  depth  at  P  is 
20  ft.  and  at  0  4  ft.  The  upper  ends  of  the  verticals 
0  lie  in  an  arc  of  parabola,  the  point  0  being  the  vertex. 
The  bridge  and  train  panel  loads  are  6  and  4  tons 
respectively.     Find  the  stresses  in  all  the  members. 

8i.  The  accompanying  figure  represents  a  portion  of  a  cantilever  truss 
the  horizontal  distances  of  the  points  A,  B,  C  from    ^A  ' 

the  free  end  being  /„  /,,  /„  resi)e.'tively.  The  boom 
ABC  is  inclined  at  an  angle  «,  and  the  boom  XYZ 
at  an  angle  ,?,  to  the  horizon.  Find  the  deflections 
at  the  end  of  the  cantilever  due  to  {n)  an  increase 
ku\B  in  the  length  of  AB;  (2)  an  increase  kJiV  in 
the  length  of  BY;  (3)  a  decrease  k.XY  in  the  length 
of  A'l';   (4)  a  decrease  k,BX  in  the  length  of  BX. 


Ans.  (1) 


/iA'sin.4/iA"' 

)«)'  COS.? 


^^^  ^'i^^TTTrX  -''^'=*^*  I'fiC-cot  BYX)\; 
k.XYL 


(3) 


BXsmBXY 


s;j8 


THEORY  OF  STRUCTURES. 


, .     ,    ( BX  pos  a     .  1 


If  k\-k^-k^'k,=k,  and  if  .!»•  is  parallel  to  HX,  and  ^LV  to  BY,  show 
that  the  angle  between  ir.Y  and  .Yl'  after  deformation 

=  lU(cot  .4«A'+(()t  BYX). 

Hence,  also,  if  the  truss  is  of  uniform  depth  d,  show  that  the  "deviation" 


of  the  boom  |)er  unit  of  length  is  constant  and  equal  to 


2k 


82    poternnne  the  d.ameter  of  a  steel  bridge  pin  subjected  to  the  stresses 

.00-     l      ^"'■"'""'"'  ""•'  '■•''■'''•'''  ''"mpononts  are  shown  by  Figs.  834. 
and  83a,  the  working  coefTicient  of  strength  being  25,000  lbs.  per  square  inch. 

Ans.  tjj  ins. 


-  V  6'2  ToBj 


.j>C2ToM 


BoruooUl  Reaotutim 

Fio.  834. 


MToni- 


Centre' 


XOTou 


Fig.  835. 


eS 

Fig.  830. 


r 

^>■D 


A<- 


Half 


Fig.  837. 


83.  Calculate  the  size  of  the  pin  subjected  to  the  stres.ses  shown  by  Figs 
8.J6  and  S37,  the  extreme  fibre  stress  is  not  to  exceed  23,000  lbs.  per  square 

Am.  5i  ins. 

84.  The  figures  show  the  magnitude  and  directions  of   the  stresses  con- 
.0  ic      <"e"tratedat  a  pin-connected  joint.       The 

8trcs.ses  are  distributed  along  the  pin  in 

vertical   and    horizontal    planes   as    indi- 

FbRCHiN     ''"'^'•-     ^'"^^  the  maximum  bendin?  mo- 

What 


FORCES  IN  FORCES  IN  ""■     "-^""^ 

HOR.  PLANE      VERT.  PL.     mcnt  to  which  the  pin  is  subjected 


1 10.  838.      F,G.  839.    Fig.  840.    should  be  its  diameter  with  a  fibre  stress 
ofl3J  tons  per  square  inch?  4  n..  5  J  ins. 

85.  \Ve  have  a  bottom  chord  point  L,  at  which  the  stresses  and  sections 
are  as  shown  m  tables  an.l  sketches  below.     Find  the  maximum  bending 


k                \ 

.     ^ 

ts.*."4-.  Av 

^ 
^ 

H^= 

3 

■\rt 

'*  *-♦.  \ 

x^\ 

0 
§ 

<^         \ 

■»  M 

e 

L  11,1.1190,(100  T 

\ 

l.LW.DOO  T       L.Ls 

8-a.n:,  l;,;- 

L 

4         CBiin;!  IJ^' 

i 


L4L5 


•TO.OOO   T  L4      1,03S.00«  T 

^^'"   ^-»'  Ra.S-12. 

moments  in  the  pin  for  these  conditions:   (a>  when  bottom  chord  strcs.ses  are 
greatest;  [h)   when  diagotial  stress  is  a  maximum. 


hi! 
...    ,j 


£:A^y4.VPLf:S. 

sume  the  packing  as 

follows: 

Memlier. 

Out  to  Out. 

In  to  In. 

C.  to  C. 

C.  toC.of 
Adjacent 

lA"  bar,     /,3/., 

32J" 

29t" 

31  A" 

Hembers. 

11"     "        /.,/.,, 
U"     ■'        I.  J., 

29}" 
2.H" 

2fii" 
22" 

28i" 
23»" 

IH" 
2i" 

li."    •'         hsl-t 
11"    plates,  r,/,4 

2U" 

IHI" 

19" 
15t" 

20A" 

nv 

IH" 
IH" 

IH"      •■       l■il.^ 
li"    bar,      i,/.. 

k:" 

lit" 
71" 

12H" 
9t" 

2A" 
IH" 

1|"     ■•        LiAj 

7»" 

4i" 

5»" 

1»" 

Ans.  345,000  in.-lba.;  464,000  in.-lbs. 

86.  A  seven-panel  pin-connected  truss  of  200  ft.  span  has  a  depth  of 
37i  ft.  at  the  centre  panel  and  of  28  ft.  at  the  hips,  the  upper-chord  points 
lying  in  the  arc  of  a  parabola.  The  dead  and  live  loads  are  945  and  2360 
lbs.  i)er  hneal  foot  of  truss  respectively.  The  percentage  by  which  the  live- 
load  stresses  are  increased  to  allow  for  impact  is  given  by  the  formula  per- 

,  400       ,  ,    .        , 

centage  =^^j^^,  L  being  the  length  of  bridge  covered  by  the  live  load  when 

the  stress  in  any  given  member  is  greatest.     Verify  the  total  maximum  stresses 
indicated  on  the  several  members  of  Fig.  843.     If  the  stresses  in  the  members 


?.^Fio.  843. 


Fio.  844. 


Fio.  846. 


0«,;33 


Louer  Lati'ra)  Systvin 


lu.sn     i;o,a)3 


of  the  upper  lateral  system  are  as  indicated  show  that  the  wind  load  is  taken 
at  200  lbs.  per  lineal  foot  of  span,  and  find  (a)  the  stress  nlong  the  bottom 
chord  due  to  the  transferred  wind  load.  Assuming  a  wind  load  on  the  lower 
Literal  system  of  490  lbs.  per  lineal  foot  of  span  when  a  train  i.s  crossing  the 
hndfjo,  verify  the  total  maximum  stresses  indicated  in  thfe  diagram.  Assum- 
ing  that  the  wind  load,  when  the  bridge  is  empty,  is  one  half  of  that  when  the 
bridge  is  loaded,  discuss  (6)  the  tendency  to  reversion  of  stress  in  the  lower 
chord. 

Arts,  (a)  34,243  lbs.;    {b)  in  second  panel  total  stress  due  to  direct  and 
transferred  wind  loads  is  81,563  lbs.,  while  the  dead-load  stress  is  S2,«64  lbs.. 


Mil 


Ml: 


M 


III 


S40 


THEORY'  OF  STRUCTURES. 


so  that  reversion  almost  takes  place,  and  it  is  therefore  advisable  to  make 
the  two  end-chord  panel  Icnjjths  rigid  members. 

87.  In  the  preceding  cxainplo,  the  unsuprwrted  lengths  of  batter-braces 
IS  30  ft.,  and  tiic  l)attcr-hraccs  may  be  considered  as  fixed  at  the  lower  ends. 
I'irid  (a)  the  moment  of  resistance  at  the  foot  of  such  a  br.ace.  If  the  work- 
\n<r  stress  is  estimated  by  the  formula,  working  .s/rcss  m  pounds  per  square  inch 

-lS,(H)«-7()--.  determine  (h)  whether  it  is  safe  to  use  a  section  made  up 

of  o„r  ■2l"x,V'  cover,  two  3J"x;?i"X->4-lb.  angles,  two  bottom  5"X3i''x54- 
II).  angles,  and  tu-o  IS'xf  side  plates. 

-bi.s-.  Ui)  l..-,4L',.S.-,8  in.-lbs.:  (h)  gross  /  =  2.'>.')7.7,  r  =  7,  r/f/ 7=2300,  max 
lH'.ding  stress  in  brace  =78.57  Ibs./sq.  in.;  required  area  =.50.6  sq.  ins.,  and 
ilierefore  section  is  am|)le. 

88.  The  dead  and  live  loads  of  a  five-pr-nel  riveted  tru.ss  28  ft.  deep  and 
of  127  ft.  span  are  800  and  2100  lbs.  per  lintal  foot  of  truss  resi)ectively.    The 

impact  effect  increas<>s  the  live-load  .stresses  by  the  factor     ^^ -  ,  L  being 

t!K>  length  of  bridge  in  feet  which  is  loaded  when  anv  given  member  is  sub- 
eeted  to  a  maximum  live-load  stress.  Verify  the  'toUil  maTimum  stresses 
indicated  on  the  several  mem»)ers  in  Fig.  84<-     The  wind  loads  per  lineal 

foot  are  l.W  and  450  lbs.  for  the 
ui)per  and  lower  lateral  sy.stems 
resjK'ctively.  Verify  the  .stresses 
in  the  several  memlxrs  of  these 
s\  'ems  as  indicated  in  Figs.  847 
and  848,  and  (0)  rem.irk  u})on 
the  method  of  designing  the  wind 
diagonals.  Also  find  (ft)  the  chord 
stress  line  to  the  transferred 
wind  load.  Will  there  (r)  be  a 
reversion  of  stress  in  the  chord 
panels? 

A)iK.  The  chord  and  diagonal 
wind  stresses  are  too  small  to 
have  any  appreciable  effect  upon 
the  sections  of  the  upjier  lateral 
menilRTs,  which  are  governed  by 


ZDIM* 


Fio.  84C. 


Fio,  847. 


848. 


/ 


/ 


for  the  transverse  struts  should   \yp    ''120- 


tlie    limiting  value  of  -  ;   the 

r  r 

'!•)  1I,3S1  llw.;  (,)  i„  ti„.  ,„„st  unfavorable  case,  i.e.,  when  the  bri.lge  is 
empty,  the  conil.med  transfetrrd  and  direct  wind-load  stresses  are  less  than 
th..  .•..rrespoM.ling  dead-load  stres.si-s,  and  there  is  therefore  no  ten.len'cv 
to  reversion  in  the  lomr  chords. 

8q.  .\n  A  trus.,  lie  Slv,  i,  i,k)  f,.  ,,.„,r,.  ,„  ,,p,„r,.  „f  „n,|  ,,i,„^  „,„,  ^^8 
a  d,-|.th  of  1(1  ft.  a.  the  centre,  the  trusses  are  spaced  17  ft.  .rntre  to  cnlre- 
the  live  and  de,-,d  I..  ,ds  are   l(MH)  an<l   !  KM)  Mw.  j-.r  lii.e.^!  f<»,t   „f  «pan  ^e- 


EXAMPLES. 


S41 


speotively;    each  truss  has  four  panels  of  25  ft.  length.    Verify  the  m»xi 
mum  stresses  mdicated  on  the  members  of  the  truss. 


Fio.  849. 


Fio.  850. 


90.  A  four-panel  A  truss,  Fig.  850,  has  a  central  depth  of  40  ft  and  its  Ipn<rf  h 

SZS       .,l.T"'°''"^P'"^'^  '^''■'  the  live  load^in  lbs.  pJri^tSr 

n  the  hve-load  stresses  for  impact  may  be  taken  at  iOO^Cim^T^  ^-h      T 
live  load  covers  a  length  of  /.  foot      VVrif,,  ♦!;    ,!,  i      ^""  ^^*^  +  ^)  when  the 

on  the  several  memSrs  of  the  t'"««     ^h      '^ '?"'^'""'" '--'''^^^ 
bracing  between  fl  and  C  U  l'^  IV.  y     Tl     '"'''^  °"  ^^^  "PP*^""  '*t«^*l 

wind  iL.    pTg  850  shows  thf«I        ^'  T^^  ^T'  '''"'^  ^"^  ^^e  transferred 

Assuming  that^hett  a  Utt  of'' •"^r'^  "T'T"  "'"'  '^'  "*»«»'  ^--"-"K- 
sup,K,rtedlen.rth«f^finH%MlK         '"/^'•'""'  «*  the  middle  point  of  the  un- 

»(.   workin,  .,™,   ,.  (,8,000-80 i),b..  per  ,^„„  i„eh,  de-™™  (., 
the  greatest  area  required  ia  23.8  sq.  ins.  '  ^  ^  "^-^  ^'^ 


CHAPTER  XI. 


SUSPENSION  BRIDGES. 


I.  Cables. — The  modern  saspension  bridge  consists  of  two  or 
more  cables  from  wliich  the  platform  is  susjx^nded  by  iron  or  steel 
rotls.  The  cables  pjiss  over  lofty  supix)rts  (piers),  and  are  secured 
to  iiiichorages  uiwn  which  they  exert  a  ilirect  pull. 

Chain  or  link  cables  arc  the  most  common  in  England  and  Europe, 
and  consist  of  iron  or  steel  links  set  on  edge  and  pinned  together. 
Formerly  the  links  were  made  by  welding  the  heads  to  a  flat  bar, 
l)»it  they  are  now  invariably  rolled  in  one  piece,  and  the  proiM)rtional 
dimensions  of  the  head,  which  in  the  old  bridges  are  very  imjK'rfect, 
have  been  much  improved. 

Hoofhiron  cables  have  Ix'en  used  in  a  few  Ciuses,  but  the  jiractice 
is  now  abandoned,  on  account  of  the  difficulty  attending  the  manu- 
facture of  endless  hoop  iron. 

M'ire-ro/^  cables  are  the  most  common  in  America,  and  form 
the  strongest  ties  in  proportion  to  their  weight.  They  consist  of  a 
numlKT  of  parallel  wire  rop<'s  or  .strands  compactly  bouml  together 
in  a  cylindrical  bundle  by  a  wire  wound  round  tlie  outside.  There 
are  usually  seven  stran<ls,  one  forming  a  core  round  which  are  i)lace<l 
tlie  rcMiaiiiing  six.  It  was  found  impossil)le  to  employ  a  seven- 
strand  cable  in  the  construction  of  the  Hrooklyn  Hridge,  as  the 
individual  strands  would  have  In-en  far  too  bulky  to  maiii|)ulate. 
TIk'  same  objcctiiin  licid  against  a  thirteen-strand  cable  (thirteen 
is  th«'  next  numlH-r  giving  an  api)roxitnately  cylindrical  sIia|H'), 
and  it  was  finally  decided  to  make  the  cable  with  nineteen  strands. 
.S'\cn  of  these  .'ire  presseil  togetlier  .so  as  to  form  a  centre  coie, 
aiouml  which  are  placed  the  remaining  twelve,  the  whole  being  con- 
tinuously wrap|K'd  with  wire. 

in  laymg  up  a  cable  great  care  is  reciuired  to  di-stribute  the  ten- 

842 


ANCHORAGE. 


843 


sion  uniformly  amongst  the  wires.  This  may  be  effected  either  by 
giving  each  wire  the  same  deflection  or  by  using  straight  wire  i  e 
^vl^e  which  when  unrolled  upon  the  floor  from  a  coil  remains  straight 
uiul  shows  no  tendency  to  spring  back.  The  distribution  of  stress 
IS  practically  uniform  in  unt listed- u-irc  ropes.  Such  ropes  are 
spun  from  the  u-ires  and  strands  without  giving  any  tmst  to  individ- 
ual  wires. 

The  back-stay  is  the  portion  of  the  cable  extending  from  an 
ancliorage  to  the  nearest  ]nor. 

The  elevation  of  the  cables  should  be  sufficient  to  allow  for  settling 
which  chiefly  arises  from  the  deflection  due  to  the  load  and  from 
changes  of  temperature. 

The  cables  may  be  protected  from  atmospheric' influence  by 
giving  them  a  thorough  coating  of  paint,  oil,  or  varnish,  but  wher- 
ever they  are  subject  to  saline  influence,  zinc  seems  to  be  the  only 
certain  safeguarii. 

2.  Anchorage,  Anchorage-chains,  Saddles.— The  anchorage  or 
abutment,  is  a  heavy  mass  of  masonry  or  natural  rock  to  which  the 
'■H«  <»f  a  cable  is  made  fast,  and  which  resists  by  its  dead  weight  the 
pull  ujK)!!  the  cable.  * 

The   cable    traverses   the    anchorage   as   in   Figs.   851    to  85'' 
passes  through  a  strong,  heavy  cast-iron  anchor-plate,  and,  if  made 


Fio.  851.  Fio.  8.52. 


Pio.  853. 


of  Wire  rope  has  ,ts  end  effectively  .secured  bv  turning  it  romid  a 
<l«'=.d-«<y,-  and  sphcinR  it  to  itself.  Much  care.  howev.T,  is  re.uiired 
I"  prevent  a  wire-ro,K.  cable  from  rasting  on  account  „f  the  gr..«t 
.-xtent  of  Its  surfme,  and  if  is  consider^  advi.sable  that  the  wire 
!H.rt.on  of  the  cabl,.  should  always  terminate  at  the  entrance  to 
'l'<;  anchorage  and  there  Ik-  attached  to  a  massive  chain  of  bars, 
"Inch  IS  cont.ruuH!  to  the  anchor-plate  or  pluiea  and  secured  by 
lj«»ifs,  wedges,  or  kej-s. 


il 


til 


i;:] 


iil 


S44 


THEORY  OF  STRUCTURES. 


In  order  to  reduce  as  much  as  possible  the  depth  to  which  it  is 
necessary  to  sink  the  anchor- plates,  the  anchor-chains  are  frequently 
curved  as  in  Fij;.  852.  This  gives  rise  to  an  oblique  force,  and  the 
masonry  in  the  jiart  of  the  abutment  subjected  to  such  force  should 
be  laid  with  its  l)eds  {XTixuidicular  to  the  line  of  thrust. 

The  anchor-chains  are  made  of  compound  links  consisting  alter- 
nately of  an  odd  and  an  even  number  of  bars.  The  friction  of  the 
link-heads  on  the  knuckle-plates  considerably  lessens  the  stress 
in  a  c-iiain,  and  it  is  therefore  usual  to  diminish  its  sectional  a.cii 
gradually  from  the  entrance  E  to  the  anchor.  This  is  effected  ii 
the  Niagara  Suspension  Bridge  by  varying  the  section  of  the  bars, 
and  in  the  Brooklyn  Bridge  by  varying  both  the  section  and  the 
number  of  the  bars. 

The  necessity  for  preserving  the  anchor-chains  from  rust  is  of 
such  imjx)rtance  that  many  engineers  consider  it  most  essential 
tiiat  the  pa.ssages  and  channels  containing  the  chains  and  fastenings 
stiould  \ye  accessible  for  periodical  examination,  painting,  and  repairs. 
This  is  unnecessary  if  the  chains  are  first  chemically  cleaned  and 
then  emlx>ddetl  in  good  hydraulic  cement,  as  they  will  thus  be  per- 
fectly protected  from  all  atmospheric  influence. 

The  direction  of  an  anchor-chain  is  changed  by  means  of  a  saddle 
or  knuckle-plate,  which  should  be  capable  of  sliding  to  an  extent 
sufficient  to  allow  for  the  expansion  and  contraction  of  the  chain. 
This  may  Ik-  accomplished  without  the  aid  of  rollers  by  Ix'dding 
the  saddle  ujwn  a  four-  or  five-inch  thickness  of  asphalted  felt. 

The  chain,  where  it  passes  over  the  jjiers,  rests  on  saddles,  the 

^-^^      C")      >>,-^^  object    of   which   is   to   furnish    hearings 

^1^  '  h-nc  .^  "'i^'i  '''•»•''>■  vertical  curves.  Either  the 
■  saddle  may  Ik-  constructed  as  in  Fig.  854, 
so  as  to  allow  the  cable  to  slip  over  it 
with  comparatively  little  friction,  or  the 
chain  may  !«•  secured  to  the  saddle  and  the  saddle  supported  u|)on 
rollers  which  work  over  a  |H'rfectly  true  and  horizontal  bed  formed 
by  a  sjidd'c-plati-  fixed  to  tlie  pier. 

3.  Suspenders.— The  susjK-nders  are  the  vertical  or  inclined 
rods   uliich  carry  the  platform. 

In  Fijr.  8"»  tUo  K.|qj«..!ii<>r  rc«t«  in  the  gnxn-e  of  a  c.-i.^t-iron  yoke 
which  straddles  tlie  cnb'i".     Fig.  8')<]  shows  the  susjxMider  bolted  to 


Fiu.  S.>4. 


SUSPENDERS. 


845 


a  wrought-iron  or  steel  ring  whicli  embraces  the  cable.  When  there 
are  more  than  two  cables  in  the  same  vertical  plane,  various  methods 
are  adopted  to  insure  the  uniform  distribution  of  the  load  amongst 
the  set.     In  Fig.  857,  for  example,  the  suspender  is  fastened  to  the 


Fio.  855. 


Fig.  856. 


Fio.  857. 


Fio.  858. 


centre  of  a  small  wrought-iron  lever  PQ,  and  the  ends  of  the  lever 
aro  connected  with  the  cables  by  the  eciually  strained  rods  PR  and 
(^S.  In  the  Chelsea  bridge  the  distribution  is  made  by  means  of  an 
irregularly  shajx'd  plate  (Fig.  858),  one  angle  of  which  is  supported 
by  a  joint-pin,  while  a  pin  also  pjisses  through  another  angle  and 
rest.s  upon  one  of  the  chains. 

The  siL'^jx'uders  carry  the  ends  of  the  cross-girders  (floor-beams) 
and  are  spaced  from  5  to  20  ft.  apart.  Thev  should  be  provided 
witii  wroiiglit-iron  screw-boxes  for  purfxv'fes  of  adjustment. 

4.  Curve  of  Cable.— Case  A.— An  arbitrarily  loaded  flexible  cable 
takes  the  sliaix'  of  one  of  the  catenaries,  but  the  true  catenary  is 
the  curve  in  which  a  cable  of  uniform  section  and  niaterial  hangs 
under  its  owti  weight  only. 

Ix't  .1  1k<  the  lowest  point  of  the  cable.  Fig.  859,  and  take  the 
vertical  through  .1  jis  the  axis  of  y. 

Take  the  horizontal  through  O  as  the  axis  of  x,  the  origin  0  being 
cho-scn  so  that 

pAO^H^mp, (1) 

/)  Ix'ing  the  weight  of  a  unit  of  length  of  the  cable  and  //  the  hor- 
zontal  pull  at  .1. 

7)1  or  AO  is  the  parameter,  or  modulus,  of  the  catenary,  and 
0(i  is  the  directrix. 

Let  T,  If  l)e  the  co-ordinates  of  any  ptnnt  P,  the  length  of  the 
arc  .1/'  Ining  .v. 

Draw  the  tangent  PT  and  the  ordinate  PN,  and  let  the  uiitfle 
PTN  =  0. 


p 

il 

n 

'1  ' 

n; 

m 

''f. 

wB- 

w 

r.,n.|S 

I'ill 

If 'Mi 

\i 


lii 


846 


THEORY  OF  STRUCTURES. 


The  triangle  PXT  is  evidently  a  triangle  of  forces  for  the  portion 
AP,  PN  representing  the  weight  of  AP  (viz.,  ps),  PT  the  tangential 


pull  T  at  P,  and  AT  the  horizontal  pull  H  at  A. 


Therefore 


dy  PA     ps     s 

:7- =  tan  <? = ^Tf  -  =  T7 = — . 
dx  T^     H    m' 


which  gives  the  differential  equation  to  the  catenary. 
It  may  be  easily  integrated  as  follows: 


I=n1i-^(|)'=\|i+54v«^+< 


or 


di*  dx 


r~ 


\  a-  +  /h2     »t 


(2) 


^3; 


Therefore 


log(s+\/s2  +  w2)=-+c, 


c  being  a  constant  of  integration. 

When  x  =  0,  s  =  0,  and  logm=r.     Hence 


log 


SJ-N   .s'^4-)*(^       X 


III 


m 


or 


Therefore,  also, 


S  +  V  .>-2  4-  »>2    =  yy^gtn  . 

—  s  +  V  .s^  +  m'^  —  ine   «. 


CURVE  OF  CABLE.  847 

Hence  8=f  (e=-e-=)  =^2, (4) 

and  integrating  between  0  and  x, 

y=™(eS+e~5)  =vs2+m2 (5) 

Eq.  (5)  is  the  equation  to  the  true  catenary  and  eq.  (4)  gives  the 
length  of  the  arc  AP. 

Draw  NM  perpendicular  to  PT,  and  let  the  angle  PTN^'PNM 
=  0.    Then 

PM  =PN  sind^y     J. =  s,     ...  m 

and  MN  -  PAT  cos  0=y       "^     ■  =m  m 

Thus,  the  triangle  PMN  possesses  the  property  that  the  side  PM  is 
equal  to  the  length  of  the  arc  AP,  and  the  side  MN  is  equal  to  the 
modulus  ni(=AO). 

The  area  APXO 

=y^  l/dx  =^(e«_e"")  =ms  =2Xtriangle  PMN. 
The  radius  of  curvature,  p,  at  P 

— -^ y-=S=^ (9) 

PG  being  perpendicular  to  PT. 

At  .4,2/  =  m,  and  the  radius  of  curvature  is  also  w.     .    .    .     (10) 

A     •  T     PT  V 

Again,  -;Z=\Tj  =  ^^^'^  ^=-- 

Therefore  T=py; /-,\ 

**"'*  H  =  pm  =  pflu;    ....:,    J    (jj) 

po  being  the  radius  of  curvature  at  .4. 


848 


THEORY  OF  STRUCTURES. 


Tlioso  pjitonary  forinuhr  are  of  little  if  any  use  in  the  design 
liiul  construction  of  su.s]X'nsion  bridges,  as  they  are  htxsod  upon  the 
assuniption  of  a  purely  the(jrotical  load  which  never  occurs  in  prac- 
tice, viz.,  the  weight  of  a  chain  of  uniform  section  and  density. 

Ex.  1.  .1  floating  landing  stage  is  held  in  position  by  a  number  of  4-in.  steel- 
virr  cables  anehored  to  the  shore,  a  shoreward  movement  being  prevented  by  rigid 
iron  booms,  pivottd  at  the  ends  and  Mretehing  jrom  shore  to  stage.  The  differ- 
enee  of  level  between  the  shore  and  stage  attachments  of 
the  cal)lfs  is  iiO  It.,  and  the  horizontal  distance  between 
these  points  is  !•%  Jt.  The\horizontal  pull  upon  each 
cable  is  1360  lbs.  Find  the  length  of  the  cable  and 
the  tensions  at  the  points  of  attachment.  (Weight  of 
cable  =  490  /6s.  per  cubic  foot;  form  of  cable  a  common 
catenary.) 

1360  =  V-  ,''iif490-m. 
Therefore  m"31.8. 


and 


Therefore  50 


which  reduces  to 


m  / 


ISO 

?£i       100       f  m  £1  li? 

«"*        n      ISO  ~    c^-H-e"" 

C  m  -I 


JM 


But  e  "•  °"112ij  and  the  la.-t  equation  becomes 


2t,  _t, 

c"'-3.173t'»  =  112.5 


and 
so  that 


e«- 12.31, 


a-, 


■2.511, 


or 


Again, 


Xi  - 79.85  ft.     and     x,  -=  l.-iO  -x,  - 70.15  ft. 

ni  /  i       --\ 
''i-;^[<""~f    '") -194.40ft. 


Also, 
Therefore 


cr-RVE  OF  CABLE. 
em -e'-'" -9.08. 
i,~j{e'"-e    «)- 142.64  ft., 


849 


and  total  length  of  cable -337.1  ft. 

^^  50(!/.+y,)-s,'-s,' -51.82X337.1, 

*"■  y,+y, -349.38  ft. 

"^"<*  y.- 199.69  ft.     and    y,- 149.69  ft. 

Then,  if  T,  and  T,  are  the  shore  and  stage  tensions  on  the  cables, 

^>       Vi    « 
136o~w-6-28,    or     r,-85411b8. 


and 


l^-f-4.71,     or     r, -6402  lbs. 


Case  B.  Let  the  platform  be  suspended  from  chains  composed 
of  a  number  of  links,  and  let  W  be  the  whole  weight  between  the 


N     O 


Fio.  ."61. 


lowest  point  0,  Fig.  861,  of  the  chain  and  the  upper  end  P  of  any 
given  link.  Let  the  direction  of  this  link  intersect  that  of  the 
horizontal  pull  (H)  at  0  in  E.  Drop  the  perp-ndicular  PN  The 
triangle  PNE  is  evidently  a  triangle  of  forces;  and  if  the  anele 
PEN  =  e,  ** 


tan  9 


PA'    W 


and  hence 


tan^a  W. 


S50 


THEORY  OF  STRUCTURES. 


Thus,  by  treating  each  Unk  separately,  commencing  with  the  lowest, 
the  exact  profile  of  the  chain  may  be  easily  traced. 

Ex.  2.  .4  light  suspension  bridge  earrics  a  fool-path  8  ft.  wide  over  a  river 
90  ft.  iridc  by  means  of  eight  equidistant  suspending  rods,  the  dip  being  10  /(. 
Each  cable  consists  of  nine  straight  linhs.  Find  their  several  lengths.  If  the 
load  upon  the  platform  is  TJO  lbs.  per  square  foot,  and  if  oiio  sixth  of  the  load 
is  borne  by  the  piers,  find  the  sectional  areas  of  the  several  links,  allowing  10,000 
/6s.  per  square  inch. 

Load  concentrated  at  each  of  the  points  E,  F,  G,  and  H  =4500  lbs.    The 

three  links  BC,  CD,  DE  are  kept  in  equi- 
librium by  the  horizontal  pull  H  at  E, 
by  the  pull  T,  along  BA,  and  by  the  uni- 
formly distributed  load  between  E  and  H. 
Hence  /.,,  the  intersection  of  EL  and  AB 
produced,  is  the  middle  point  of  EH,  and 
ALK  is  a  triangle  of  forces  iorH,  T\  and 
Therefore 


the  weight  4x4500-18,000  lbs. 


H- 18000 


LK 
AK' 


■18000x^^  =  45,000  lbs. 


and 


AL    . ,— - 


r,  -  ISOOOjT-  -1800v'725=4500Vll6  lbs. 


The  tuw  links  CD,  DE  are  kept  in  equilibrium  by  the  horizontal  pull  H 
at  E,  by  the  pull  T,  along  DC,  and  by  the  uniformly  distributed  load  between 
E  and  G.  Hence  the  directions  of  H  and  T,  must  intersect  in  F  the  middle 
point  of  EG,  and  BFH  is  a  triangle  of  forces  for  //,  T,  and  the  weight  3  X  4500 
- 13,500  lbs. 


Also 


BH    15 
10  "25 


or    BH-eh. 


Tip  .  

Therefore  T,  - 13500  ^  -  2250v^20'  +  G'  -  4500^109  lbs. 

The  link  DE  is  kept  in  equilibrium  by  the  horizontal  pull  H  at  E,  by  the  pull 
r,  along  DC,  and  by  the  uniformly  distributed  load  between  E  and  F.  Hence 
♦  he  directions  of  //and  T.,  must  intersect  in  M  the  middle  point  of  EF,  and 
C.MG  is  a  triangle  of  forces  for  H,T,  and  the  weight  2X4500=9000  lbs. 


But 
Therefore 


—  '^',    or    CG-3h. 
10     20' 


r,  =  900o|.^/  =  3000  V234= 4.500  v^l04. 


m 


CURVE  OF  CABLE. 


851 


Finally,  the  weight  of  4500  lbs.  at  E  is  equilibrated  bv  the  pull  H  and  b* 
the  pull  r.  along  ED.    Hence  DEF  is  a  triangle  of  forces.  ' 


But 


Therefore 


DF     3 

T-T5'  •'^  ^^-'^t- 

T,  =4500-^, -4500  vToi. 


The  sectional  areas  of  the  .several  links  in  square  inches  are 
•^'         ^2         T,         r,         H 


10000'  10000'   10000'  10000'  10000'   '•''■'  '*-^^'  ^•^^^'  ^-59.  'i-522,  and  4.5^ 

C.^SE  C.-Generally  speaki.ig,  the  distribution  of  the  load  may 
be  assumed  to  be  appro.ximately  uniform  jx^r  horizontal  unit  of 
length  the  load  being  suspended  from  a  number  of  points  alonr 
each  Cham  or  cable  by  means  of  rods.  The  weight  upon  the  cable 
between  the  lowest  point  A,  Fig.  863,  and  any  other  point  Pis  equili- 
brated by  the  horizontal  pull  at  A  and  by  the  tangential  pull  at  P 


Fio.  863. 


The  directions  of  these  two  tensions  intersect  in  the  point  T  which 

IS  necessarily  the  middle  point  of  AX,  since  the  weight  is  uni'formly 

bola"  ''''"'        ■  '  ''  ""  characteristic  property  of  the  para- 

If  X,  y  are  the  co-ordinates  of  P  wth  respect  to  A,  then  PTM 

nnn  r'T  P  Tn '  ^"  'K  ^""'"''^^'  ^'"^'"^  ^  •'"''''  "-  tangential 
pull  T  at  P,  and  the  ^.e,ght  wx  uniformly  distributed  over  .4.V  w 
being  the  intensity  of  the  load.    Hence,  ' 


H      2  ,      2H 


(1) 

the  equation  to  a  parabola  with  its  vertex  at  O,  its  axis  vertical, 
and  its  parameter  equal  to  ^^-=P,  suppose. 


\  ■  ii 


'i 


I 


11 


852  THEORY  OF  STRUCTURES. 

Also,  taking  the  angle  PTN=0, 


ay' 


(2) 


and  the  horizontal  pull  at  trery  point  of  the  cMe  is  the  same  as  that  <U 
tfie  lowest  fioint. 


Again, 


7  =  V  (trj)2  +  //2  =  wxs\\  +  ^2- 


The  radius  of  curvature  at  P 

„2\  i 


(ill/Hi) 

xA    HI 


w 
H 


so  that  the  radius  at  0  is 


and 


H 

H  =  tvpo. 


Parameter,  etc.— Lei  Ai,  /i2  lie  the  elevations  of  C  and  D,  respec- 
tively, above  the  horizontal  line  FAE. 

I>et  AE  =  ai,  AF  =  a2,  and  let  ai+a2  =  a  =  EF. 
By  equation  (1), 


IW Q,      _    02     ^ 


a\  +02 


Then 


w      Vhi     Vh2    Vh\+Vh2    Vhi+VlTz 

"'      Vv'/ii  +Vh2/ 

„    2x1    n-x    2x     ^   \y 
Also,  tan»  =  -^  =  ^=p=2^j-,. 

If  <?i,  <?2  be  the  values  of  0  at  C  and  D  respectively, 

tan  di  =  2>J  77    and    tan  O2  =  2>J  77. 


CURVE  OF  CABLE. 


If  hi  =h2  =  h,  then 


and 


a 


a* 


Oi=a2=:j,  P=jr, 


4h 


tantfi ^tan52. 

a 


Length  of  Arc  of  Cable.— Let  AP  =s. 
«nce  tan  5=^, 


w  .      w 


Hence 


sees  Odd  =^x=^s  cos  d,    or    ds=--^ 


H  P»    dd       H 

'^w^o   ^^^=2^'*^"  ^«^«  ^  +  log.  (tan  »+secd)j 


853 


An  approximate  value  of  the  length  of  the  arc  which  may  be 
used  in  practice  may  be  obtained  as  follows: 


ds^=dx2^dy^^dx^{l  +  (^J^'^  -dx^{l+'^ 

Therefore  ds  =dx(l  +3  ^ )>  approximately. 

Integrating  between  0  and  P, 


'\ 


__  I  w^x''  2  y2 

s=OP=x-l =1-1--— 

*^6  H2         ^3x' 


Deflection  of  a  Cable  due  to  an  Elementary  Change  in  its  Length.— 
The  total  approximate  length  (S)  of  the  cable  AOB  is 


iii 


.b=ai+02+o— +Q— • 

O  Ol        O    02 


854 


THEORY  GF  STIWCTURES. 


Now  oi  and  oa  are  constant,  and  hi  —hz  is  also  constant,  therefore 
dhi  =dh2.    Hence 


dS 


3  \ai      &2' 


If  the  alteration  in  length  is  due  to  a  chcuye  of  t°  in  the  lempercUure, 

dS=etS, 

€  lK>ing  the  coefficient  of  hnear  expansion. 

In  England  the  effective  range  of  temperature  is  about  60"  F., 
while  in  other  countries  it  is  usual  to  provide  for  a  range  of  from 
100°  to  150°  F. 

If  the  alteration  is  due  to  a  pull  of  intensity  /  per  unit  of  area, 

E  being  the  coefficient  of  elasticity  of  the  cable  material. 
If /j,=;i2=;i, 

ai=a2  =  -,    and    dS  = dh. 

2  3  a 

10.  Pressure  upon  Piers,  etc.,  Fig.  865. 

Let  Ti  be  the  tensic^n  in  the  main  cable  at  A) 
T2"    "        "        "    "    back-stay  at  ^; 
a,  ,3  Ix"  the  inclinations  to  the  horizontal  of  the  tangents  at 
.4  to  liie  main  cable  and  back-stay,  respectively. 
Tlie  total  vertical  pressure  upon  the  pier  at  A 

=  r,  sina  +  r:jsin,9  =  P. 

The  total  resultant  iiorizontal  force  at  .1 

=  7,  costt~r2COS(9=Q. 

If  the  cable  is  secured  fo  a  sa<ldle  which  is  free  to  move  horizontally 
on  the  top  of  the  pier  (Via.  SG'i), 

y^the  frictional  resistance  to  the  tendency  to  motion, 


PRESSURE  UPON  PIERS. 

^  Q<tixP, 

Ml  being  the  corresponding  coefficient  of  friction 

.0 


855 


Fio.  864. 


Let  D,  Fig.  865,  be  the  total  height  of  the  pier,  and  let  W  be  its 
weight. 

Let  FG  be  the  base  of  the  pier,  and  A'  the  limiting  position  of 
the  centre  of  j)ressure. 

Let  p,  2  be  the  distance  of  P  and  17,  respectively,  from  K. 


Then 


for  stability  of  position  Q^^^-j; — ^, 


and  for  stability  of  friction,  when  the  pier  is  of  masonry, 
pXj|>^the  coefficient  of  friction  of  the  masonry . 

If  /ii  is  sufficiently  small  to  lx>  disregarded,  Q  is  approximately 
nil,  and  T,  cos  a  =7^2  cos  /?  =H.  The  pressure  ujK>n  the  pier  is  now 
wholly  vertical  and  is   =//(tan  «  +  tan  ,?). 

When  the  cable  slides  over  smooth  rounded  saddles  (Fig.  847), 
the  tensions  Ti  and  T^  are  approximately  the  same. 

Thus 

P  =  7'i(sina+sin/9)    and    Q='Ti(cosa-cos  Ji). 

If  a-^,  <?-0,  and  the  pressure  upon  the  pier  is  wholly  vertical, 
its  amount  Ix'ing  '2Ti  .^in  a. 

Th<'  piors  are  made  of  timlx'r,  imn,  .itwl,  or  nia-sonry,  and  allow 
of  great  scope  in  architectural  design. 


'fil 

B? 

ii 

1 

■'  '^ 

'i 

,  1 

•  t 

■|! 

\ 

*    H 

: 

'  H 

i 

''m 

^ 

,wt. 

H 

•  vm 

e 

i^i 

u 

I   !i  1 


SoO 


THEORY  OF  STRUCTURES. 


The  cable  should  in  no  case  \ic  rigidly  attached  to  the  pier,  unless 
the  lower  end  of  the  latter  is  f.v-e  to  revolve  through  a  small  angle 
about  a  horizontal  axis. 

7.  Weight  of  Cable. — The  ultimate  tenacity  of  iron  wire  is 
90,()0()  lbs.  per  square  inch,  while  that  of  steel  rises  to  200,000  lbs., 
and  oven  more.  The  strength  and  gauge  of  cable  wire  may  be 
insured  by  specifying  that  the  wire  is  to  have  a  certain  ultimate 
tenacity  and  elastic  limit,  and  that  a  given  number  of  lineal  feet 
of  wire  is  to  weigh  one  pound.  I-^jich  of  the  wires  for  the  cables  of 
the  Brooklyn  Bridge  was  to  have  an  ultimate  tenacity  of  3400  lbs., 
an  eUustic  limit  of  KiOO  lbs.,  and  14  lineal  feet  of  the  wire  were  to 
weigh  one  pound.  A  very  uniform  wire  having  a  coefficient  of 
elasticity  of  29,0(X),000  lbs.,  h;us  In-en  the  result,  and  the  process  of 
straidhiemng  has  rai.sed  the  ultimate  tenacity  and  elastic  limit  nearly 
8  JK'    .ent. 

I.et  n'l  be  the  weight  of  a  length  ai{=AK)  of  a  cable  of  suf- 
ficient sectional  area  to  bear  safely  the  horizontal  tension  H ; 

Let  H'a  be  the  weight  of  the  length  ni(=.IC')  of  the  cable  of  a 
sectional  area  sufficient  to  Ix-ar  safely  the  tension  Ti  at  C; 

Let  /  be  the  .safe  inch-stress; 

Let  q  be  the  specific  weight  of  ll  •  cable  material. 


H  .      //  sec  Oi 

Then  U'l^^-rai^    and     11*2  =  — . — .s'lj. 

Hence     U  2  =  H  1— sec  0i  =  —Ui +rr -'-I  1+    -7  + 

I      8  hi2\ 

or  W2=Wi(i+ r,  I,  nearly. 

V      381*/ 


•)• 


A  saving  may  be  effected  by  proportioning  any  given  section  to  the 

pull  across  that  section.     Thus  at  any  jKunt  (x. ;/)  the  pull  ■=//  .sec  d, 

ll  .s<^c  /? 
and    the    corresfxindiiig    necessary    sectional    area^ — —. — .      The 

//  sec  0 
weight  |»er  unit  of  length  -  — -. — 9,  and  the  total  weight  of  the 

length  Ni(=.lC)  is 


WEIGHT  OF  CABLE. 


857 


„,        f'"H seed  ds  , 


sec2  ddx 


But 
Therefore 


Hence 
and  also 


The  weight  of  a  cubic  inch  of  steel  averages  .283  lb. 

The  weight  of  a  cu^ic  inch  of  wrought  iron  averages  .278  lb. 

The  volume  in  inches  of  the  cable  of  weight  Tr,  =  l2ai-. 


Therefore 


W 


12a,y 


'^  =  .283  lb.  or  .278  lb., 


according  as  the  cable  is  made  of  steel  or  iron. 

Let  the  safe  inch-stress  of  steel  wire  be  taken  at  33,960  lbs  of 
the  best  cable  iron  at  14,958  lbs.,  and  of  the  best  chain  links'  at 
9972  lbs.    Then 


W,-i/a,X.283x 


12 


Ha 


33(500^  wioi^°''«^^'''a*''e8; 


12 


Ha, 


W,  - Hai  X .278 X  ^^  =  -^  for  iron  cables; 


W,-/fo,X.278x 


12        Ha, 

~r„^  for  link  cables. 


9972 


m 

V 

■ 


IT  1  ■ '' 


3000 


858 


THEORY  OF  STRUCTURES. 


Note. — About  one  eighth  may  be  added  to  the  net  weight  of  a 
chain  cable  for  eyes  and  fastenings. 

Ex.  3.  A  bridge  444  ft.  long  consists  of  a  central  span  o/  180  //.  and  two 
side  spans  each  of  132  ft.:  each,  side  of  the  ptatform  is  su-ipended  by  vertical  rods 
from  two  iron-mre  cables;  each  pair  of  cables  pa.sses  over  two  masonry  abutments 
and  tuo  piers,  the  former  being  24  ft.  and  the  latter  39  //.  above  the  surface  of  the 
ground;  the  lowest  point  of  the  cables  in  each  span  is  19  ft.  above  the  ground  sur- 
face; at  the  abutments  the  cables  are  connected  with  straight  wrought-tron  chains, 
by  means  of  which  they  are  attached  to  aneliorages  at  a  horizontal  distance  of  66  ft. 
from  the  foot  of  each  abutment;  the  dead  weight  of  the  bridge  is  3500  ifts.  per  lineal 
joot,  and  the  bridge  is  covered  with  a  proof  load  of  4500  lbs.  per  lineal  foot. 


V-2    and     i, -2x,-88  ft. 


4.'iOO  +  .3500 
The  load  per  lineal  foot  carried  by  each  cable  --^ -. -2000  lbs. 

Then        H,  -  T,  y^L T,-^P^^  -  88  x  2.      ^J^  -  387,200  lbs. 

'v'TT'+20'       ^22 '  +  5"  -0 

Therefore     «, -387,200  lbs.,  T,  =  425,323  lbs.  and  T, -396,074  lbs., 


"^— ^-«r-4- 


Fio.  860. 

AH  45 

Again,  H.-T    .--—^- 90X2000 X.^- 405,000  lbs. 

v45'+20  -" 


Therefore 


//,- 405,000  lbs.    and     J", -443,200  lbs. 


2   5'  2  "H)* 

The  :ength  of  the  C(Me  ^flC  -44  +g- 1^  +88  +  -    —  -  135A  ft. 


r!so 


/        2  20'\ 
2-Klength  o/  CD -2 (90  +3  —j  -  185H  ft. 


Change  of  dip  corrcs[)onding  to  a  variation  of  60°  F.  from  the  mean  tern- 
perature: 


EXAMPLE  OF  SUSPENSION  BRIDGE. 

60 


859 


or 


(a)  Side  Spon.-Change  in  lengthof  ^^C- ^^135^ -.0564  ft.    Hence 
change  of  dip  x  j-  ( A  +  20^  _  o564, 
change  of  dip -.1241  ft. 
(6)  Centre  5pan.-Change   in    length   of   cable -j^l85jf -.0775   ft. 

Hence  change  of  dip  x|  (|)  - .0775. 

*°^  change  of  dip  - .  1308  ft. 

Change  of  dtp  due  to  load,  15,000  Ibs./sq.  in.,  being  the  safe  working  stress: 
(a)  Side  Span.-Sectional   area   of   ABC   (assumed  uniform)  -     ^'     . 

28.355  sq.  ins.  ^^^^ 

135,V 


and 


Extension  of  ABC -15, 000 X^-^^,.o677  ft., 
change  of  dip  Xj  (^  +^  -  .0677  ft., 
change  of  dip  -  .1489  ft. 


(6)  Centre  Span.— Sectional  area  of  CD  (assumed  uniform)  — ^  -29  55 
«j.  ins.  ^^*^ 

Extension  of  cable  of  centre  span  - 15,000  g^^Q-  -.093  ft.. 


and 


or 


change  of  dip x|^ -.093, 
cAan^o/dip-.157ft. 


Weight  of  cables: 


Weight  of  cable  for  side  span     -135A  x^^j|;^^X1728x.278-12,809  lbs. 


<«  U       II         «| 


29.55 


centre  span  - 185^  x  -~-  X 1728  X  .278  - 18,.329  '  • 


'i 


IS' 


til 

'J. 


860  THEORY  OF  STRUCTURES.  -c  ,— - 

If  cables  are  proportioned  at  each  point  to  the  pull,  then 

(88  +  3   j^+44  +  -|^)l2  =  ll,6611b8. 


loOi'K) 


,  .,       ^,      387200 X .278 /„„     2   20' 
wt.  of  6ide  cable  = 

,,    ,.                 ,,      „    40r,000x.278 
"    "  centre  cable  =  2  X 


l.iOOO 


(90+ J  yy-j  12  =  16,747  lbs. 


Piers. — Overturning  moment  is  greatest  when  the  proof  load  covers  centre 

span  only.     Total  vertical  load  on  pier  =8000x90 +3.")00X  88  =  1,028,000  lbs. 

<  +8 
The  weight  of  the  pier  =39-:j-14|X  128,  (  being  the  thickness  of  the  base  and 

14j  ft.  its  uniform  width. 

A^ain,       //,' =  8000  X  90  X  ||     and    W/ =  3500  X  88  X  If. 
Therefore  the  horizontal  pull  at  the  top  of  the  pier 

=  ///-/// =942,000  lbs. 
Hence,   if  centre    of   resistance   at    base    is   j<   from  the 
middle  point, 

942400  X 39  =  overturning  moment 

-39^14jX128Xj/  +  l,028,000xi<, 


and 


t- 30.6  ft. 


.3fi  fi  +  8 
Hence,  too,  the  weight  of  the  pier -39-^;^ — 145X128-1,6.51,271   lbs.,  and 

the  total  pressure  on  the  ba.se -2,679,271  lbs. 
Anchorage. — Vertical  pull  on  anchorage 

-W,,',-.3S7,200XA 

-  29,334  lbs.  for  each  cable. 

Let  IF -weight  of  masonry  required  in  anchorage  for  each  cable  to  resist 
horizontal  displacement. 

Then  ,,1V  -  .76  X IV  -  387,200  lbs. 

and  ir- 509,474  lbs. 

Thus  the  total  weight  required  to  resist  the  upward  pull 

-4  X  29,3.34  - 117,.336  lbs, 
and  the  total  weight  required  to  resist  the  horizontal  pull 

-4  X.'»09,474  -2,0.37,896  lbs. 


The  tension  in  the  anchorage  bar-//,  -  -     -^-388,310  lbs. 

66 


SLOPINO  SUSPENDERS. 


861 


9.  Curve  of  Cable  from  which  the  load  is  suspended  by  a  series 
of  sloping  rods. 


r      0       E 

Fio.  868. 

Let  0  be  the  lowest  point  of  such  a  cable.  I^t  the  tangent  at  0, 
and  a  line  through  0  parallel  to  the  suspenders,  be  the  axes  of  x  and 
y  respectively. 

Let  w'  be  the  intensity  of  the  oblique  load.  Consider  a  portion 
OP  of  the  cable,  and  let  the  co-ordinates  of  P  with  respect  to  OX, 
OY  be  X  and  y. 

Draw  the  ordinate  P.V,  and  let  the  tangent  at  P  meet  ON  in  E. 

As  before,  P.V^  is  a  triangle  of  forces,  and  E  is  the  middle  point 
of  ON.    Then 

v/x    PN    2y  2H 

'~~  or    x2  =  — ; 

w 


H  ~NE~~i'  ""^  ""  =i:?y< 


the  equation  to  a  parabola  with  its  axis  parallel  to  OY  and  its  focus 

oil 
ut  a  pomt  .S,  where  4.SfO  =  ^. 

In  bridges  with  sloping  rods  longitudinal  stresses  are  developed 
which  vary  in  iuteasity  in  different  parts  of  the  platform.  Such 
bridges  are  nmch  stiffer  vertically  than  when  the  rods  are  vertical. 

Parameter. ~\A^i  the  axis  meet  the  tangent  at  0  in  7",  and  let 
its  inclination  to  OX  he  i. 

Ui  A  be  the  vertex,  antl  ON'  a  i)eri)endicular  to  the  axis. 


Then 
But 


SO  =  Sr  =  S\+AT'  =  SA  +AN'. 
^SAN'^ON'^^N'T'^  tan2  i^-AAN'^  tan*  t. 


Therefore  AS^A  N'  tau^  i,    and    SO  ^AS(1+  cot»  0  -  ~ 

sm^t 


IH 


Hence 


THEORY  OF  STRUCTURES. 
the  parameter  =  4AS  =  iSO  sin^  i. 


Stresses.— Ifit  P  be  the  oblique  load  upon  the  cable  between  O 
and  P. 

Let  Q  })o  the  total  thrust  upon  the  platform  at  E; 

10  "    "   loail  i)er  horizontal  unit  of  length; 

q  "    "  rate  of  increase  of  stress  along  platform; 

t   "    "    length  of  P^; 

a  =  length  of  each  bay  of  platform. 

Then  tc'  =  w  cosec  i  and  pull  on  each  rod  =  iv'a  =  wa  cosec  i. 

Also    q  =  w  cot  i,  and  the  horizontal  component  per  panel  =  wa  cot  J, 

which  represents  the  increment  of  force  tleveloped  in  the  platform 
at  the  foot  of  the  sloping  rod. 

Again,  H  is  the  horizontal  force  at  0,  and  therefore  the  horizontal 
pull  on  the  chain 

=11— wa  cot  t  =  jTy-  — uu  cot  t, 

if  I  is  the  span  and  d  the  dip. 

The  tension  at  P  =  wxsecO,  0  being  the  angle  between  the  tan- 
gent at  P  and  the  horizontal. 


Also,      H 


n-'x^- 


-  2x&  ■  SO  =  2AS-^  =  2  45-T^ 


sin^  i 


Assuming  that  the  iM)rtions  of  the  cable 
between  the  upixT  ends  of  consecutive  rods 
are  straight,  the  stresses  dcveloiH-d  may  be 
easily  determined  graphically. 

Fig.  S()()  represents  the  half-span  and 
Fig.  870  is  the  stress  (lijigriiiii  when  tlie  ends  of  the  p!;itform  are 
not  attached  to  the  piers,  the  horizontal  member  being  hi  ten.sion. 


Fio.  8<)!>. 


SLOPING  SUSPENDERS. 


863 


This  member  will  be  in  compression  if  its  ends  are  attached  to  the 
piers  and  the  stress  diagram  is  then  Fig.  871. 


Fio.  870. 


Fig.  871. 


Length  of  Cahle.-Let  s  be  the  length  of  OP,  and  let  6  be  the  inclina- 
tionoi  PE  to  OY.    Then 

s-'AP-AO 

H  sin^  i  r 
=    2w'     i  **"  (^°  ~  ^)  ^^  (90°  -  d) 

+log,|tan  (90°-5)  +sec  (W-ff) j  -tan  (90°-t)  sec  (90O-t) 

-log^jtan  (90°-t)  +sec  (gC-i)}  l 

^ ''"- loot  e  cosec  5-cot  t  cosec  i  +  log.  ^2L£±£2?£1?1 
'      ^  ^coti+coseci  J' 


2w;' 
and  approximately, 


.     2    yZsin^i 

S  =  X  +  WCOSl+-;r-  ~ -.. 

O    X  +  J/COSl 


II.  AuxUiary  or  Stiffening  Truss.  — The  object  of  a  stiffening 
truss  (Fig.  872)  is  co  distribute  a  passing  load  over  the  cable  in  such 


DEC         b" 

Fio.  872. 


a  manner  that  it  cannot  be  distorted.    The  pull  upon  each  suspender 
must  therefore  be  the  same,  and  this  virtually  assumes  that  the 


i 


>    ^ 


I"  fi 


S64 


THEORY  OF  STRUCTURES. 


effect  of  the  extensibility  of  the  cable  and  suspenders  upon  the 
figure  of  the  stiffening  truss  may  be  disregarded. 

The  eniis  0  and  A  must  be  anchored,  or  held  dotvn  by  pins,  but 
should  be  free  to  move  horizontally. 

Let  there  be  n  suspenders  dividing  the  span  into  (n  +  1)  equal 
segments  of  length  a. 

Let  P  be  the  total  weight  transmitted  to  the  cable  and  z  the 
distance  of  its  centre  of  gravity  from  the  vertical  through  0. 

Let  T  be  the  pull  u]K)n  each  suspender. 

Taking  moments  about  0, 

Pz=T{a  +  2a  +  3a  +  .  .  .  +  na)  =  Ta^^^^  =  7^^, 

I  being  the  length  of  0.4. 

Also,  if  t  is  the  intensity  of  pull  per  unit  of  span, 

P 
•  tl^uT,    and  hence    Pz  =  i—. 

Let  there  be  a  central  suspender  of  length  s.     There  will  there- 


fore be 


n-1 


suspenders  on  each  siile  of  the  centre. 


P 


The  parameter  of  the  parabola  =  77 
Hence  the  total  length  of  all  the  suspenders 

.s+2Ji-^,+fa{l'+2>+3'+.  .  .+^]} 
n'^  «(h2-1)        /       h  n-V\ 

If  there  is  no  central  susjx>iider,  i.e.,  if  n  is  even, 
the  total  length  =  («  - 1)  ( .s  +  ^  — — ) . 
Denote  the  total  length  of  saspendors  by  L.    Then 
the  streiis  length  =  TL  =  -jPL. 


STIFFE.yiNO  TRUSS.  gg^ 

Let  "'be  the  uniform  intensity  of  the  dead  load. 
t.vsE  I.  The  bridge  partially  loaded. 

Let  w'  b(>  tlie  maximum  uniform  intensity  of  the  live  load    anH 
let    Ins  lojul  advance  fron.  .1  an.l  cover  a  length  j^        '  ^ 

tivelv  '  ""'  '''  ^"  ^'  ^  '^'  P''^^^"'-*"^  ^*  ^  «"d  A  respec 

For  equilibrium, 

Ri+R2  +  tl-u-l-rv'(l-x)^0;      .     .    .    .     (i) 

Ril  +  t^-uj-'-^(l-x)2^0 (2) 

thel'u"  j:;;^'  "'^'^  ''  "^«  -^^^ht  is  to  be^transmitted  through 

tl  =  wl  +  w'(l-x) /gj 

From  eqs.  (1),  (2),  and  (3), 

-R^=-2j^l-^)=R2,      (4) 

which  shows  that  the  reactions  at  0  and  A  are  equal  in  magnitude 
but  opposite  in  kind.     They  are  evidently  greatest  when  x=  1,  i  e 
when  tl^^live  load  covers  half  the   bridge,  and  the  common'value 
is  thon  — . 

o 

The  shearing  force  at  any  point  between  0  and  B  distant  x'  from  0 

which  becomes  |:|(,_,)_^^^^^  ^,,^„  ^,  ^^^^^,  ^      ^^  ^^^ 

ach  uui  and  is  an  absolute  maxinmm  when  the  live  load  covers 
•.alf  the  bndge.  Tl.  web  of  the  truss  must  therefore  be  desTgLed 
to  bear  a  shear  of  ~  at  the  centre  and  ends. 


.1 


il 


Ml 


it- 

} 


8G6 


THEORY  OF  STRUCTURKS. 


Again,  the  bending  moment  at  any  point  between  0  and  B  distant 
j'  from  0 

=  K,x'  +  '-y^>  =  -m^(x'2-x:c'),      ....     (6) 


which  is  greatest  when  x'  =  -,  i.e.,  at  the  centre  of  OB,  its  value  then 
8     I 


U-'  l-x 
iK'iiig  --^  ~7~^^'    Thus  the  bending  moment  is  an  absolute  maximum 


when  3-(/j2-j3)=o,  i.e.,  when  x  =  -^l,  and  its  value  is  then-— F. 
ax  o  54 

The  bending  moment  at  any  point  between  B  and  A  distant  x^ 
from  0 


=  ftlX'+4^x'2-|(x'-x)2  =  ~y(x'-x)(i-x'),     .      .      (7) 


J  /-4-T 

which  is  greatest  when  -^,!(x'-x)(Z-x')l  =0,  i.e.,  when  x'  =  -^,  or 
at  the  centre  of  AB,  its  value  then  being  —  y(i-x)2.     Thus    the 


bending  moment  is  an  absolute  maximum  when  ^\x{l-x)^]=0, 

I  w* 

i.e.,  when  x^-x,  and  its  value  is  then  +—1^. 
o  54 

Hence  the  maximum  bending  moments  of  the  unloaded  and  loaded 
divisions  of  the  truss  are  equal  in  magnitude  but  opposite  in  direction, 
and  occur  ot  the  points  of  trisection  (/),  C)  of  OA  when  the  lire  load 
rovers  one  third  (AC)  and  two  thirds  (AD)  of  the  bridge  respectively. 

luich  chord  must  evidently  be  designed  to  resist  both  tension 
and  coniprcssioii,  and  in  order  to  avoid  unnecessary  nicety  of  calcu- 
lation, the  section  of  the  truss  may  1'^  kept  uniform  throughout 
the  middle  half  of  its  length. 

Cask  II.  .1  sinf/lc  concctttratcd  hud  W  at  any  point  B  of  the  truss. 
W  now  takes  the  place  of  the  live  load  of  intensity  w' . 

The  remainder  of  the  notation  and  the  method  of  procedure 
being  precisely  the  same  as  before,  the  corresponding  equations  are 


STIFFES'INO  TRUSS. 

Ri+R2  +  (t-u')l~W=0. 


867 
(10 


t  —  w 


Ril+-Y-P-W{l~x)=0.  ..;:..    (20 


TV 


t-w  =  -^ 


(3') 


W/       I 


--7(^4)- 


R9 


(40 

which  shows  that  the  reactions  at  0  and  A  are  equal  in  maimitude 

but  oppasite  in  kind.    They  are  greatest  when  x  =  0  and  wS^x=t 

i.e.,  when  W  is  either  at  0  or  at  A,  and  the  common  value  is  then  ^. 

The  shearing  force  at  any  point  between  0  and  B  distant  x'  from  0 

=  «.+(^-uOx'  =  f(x'-x4),        ....      (5,) 

which  is  a  maximum  when  x'=x,  and  its  value  is  then  -. 

The  web  must  therefore  be  designed  to  bear  a  shear  of  ^  through- 
out the  whole  length  of  the  truss. 

X'  from'o  *^'  ^"'""^  "'°"''"*  ^*  ^"^'  ^'""^  ^*'''^°  ^  ^^"^  ^  di«t«"t 

F/r..^  lot  x<^.  The  bonding  moment  is  ymUive  and  is  a  maxi- 
mum  when  x'=j,  its  value  then  being 

11' 

+  2^(/j--x2). 

-Vex/,  lot  X  >^.    The  bomling  moment  is  then  negative  and  is  a 
maximum  when  x'=x-^,  its  value  then  bein^ 


i 


ft 


W 


i.!'^ 


I 

\ 

!  :flB 

lii 

1 

V 

S68 


THEORY  OF  STRUCTURES. 


■^('-2)' 


Tlio  Ix'iKliiif;  inomcnt  at  any  point  iM'twoon  B  and  .t  distant   x' 
frotn  0 

^R,x'  +  {t-w)j-]V(x'-x)=j{x'-I)(^-x),.     .     (7') 
which  is  a  uiaxiinuni  wiicn 


5?!"'^"' 


I.e.,  whon  j  =x  +  -.,  and  its  valiu-  is  tlion   ~ .jT\^~:i} 


\nte. — Tho  stiffening  truss  is  most  cffcclivc  in  its  action,  but 
adds  considcrahly  to  the  wciglit  and  cost  of  tlic  wiiolc  structure. 
l'ro\i<i(iii  has  to  1k'  made  Ixilii  for  the  cxtia  truss  and  for  the  extra 
iiintcria!  re<juircd  in  the  calile  10  carry  this  extra  load. 

SliiJfinnii  Tr'fv.'N'  Uinijvil  nt  the  (\'iilrv. — Pr<nision  may  ]v  made 
for  coiiiileracting  tiie  straining  (hie  to'dianges  of  tenijx»rature  by 
liinglng  tiie  truss  at  the  centre  /V. 

Let  a  liv(>  load  of  intensitv  ir'  ailvance  from  A. 


First,  let  tiie  hvi-  load  cover  a  length  .1 


«-'(>i)- 


Let  R\.  A*2  Ix'  the  pressures  at  O.  A.  res[H'ctivoly. 
The  eijuations  of  eciuiiihrium  are 


hi  t  l{2  +  (l->r)l-ir'X'^();      . 

^^'j  •''-"' s- TV-' -2/  ^"; 


(1) 

(2) 
(3) 


s     s 

Eqs.  (2)  and  (.'5)  Ix'ing  ohfained  by  taking  moments  alnmt  E,      Hence 


t-K 


\i-  -  lu 


r)i  . 


.    (4) 


869 


(5) 


ST/FFE\\rXG  TRUSS. 
^      1  w' 

Ji2-r,j(l~xy 

Next,  let  the  live  load  cover  the  length  BoU]^. 

Ix>t  Mi  -X  as  before,  and  let  /?,,  R,',  f  be  the  new  values  of  R, 
hi,  t,  r<'s|x'ctively.  ' 

The  equations  of  equiHbriuni  are  now 

R,'  +  R,'  +  (t'-u')l-,r'(l-x)=0;     ....     (7) 
«.'^  +  a'-«')^-^i-(/-x)=0;      .    :    .    :    .    (8) 

^  +  (/'-"0~-O;     ........     (9) 

and  hence 

f-w  =  2j,(l-xr-[^~(t-u'-w')];    .     .     ,     :     (10) 
«.'=-r,r(/^-4/x  +  3a-^)(= -/?.);    .    .     .     („) 

Diajjran,  of  Maxinuun  Sheorin,,  Forre.~-Th.  sh.-ar  at  anv  ,x,int 
distant  ^  from  .1  ,„  the  unloaded  iK.rtion  BO  wh.-u  the  live  load 
covers  AB 

-n^+{t-,r)il-z) (,3^ 

=  -|/^'4(/'-jr-,r')(/-e)| 

^mimis  the  sh(-ar  at   th(.  .nm,   ,K.int  when  AH  is 
unloaded  and   the  Hve  load  covers  Ii(f. 

For  a  Kiven  >ah,e  of  r  the  n.axionMo  shear,  ,„sili,r  or  neq^Uive 
at  any  point  of  Oli  is  found  In'  .nakinsr  fs,H.  o,/  (13))  ' 

dRi  +  {l-z)dit-u')'^0, 


870 


THEORY  OF  STRUCTURES. 


or 


or 


j{-2l+3x)—^^il-zX-il  +  ix)=Q, 
Az-2l 


Hence,  by  eqs.  (4),  (5),  (13),  (14), 

the  maximum  shear  =  i  Iw'x 


(14) 


l-2» 

1-x' 


.     (15) 


Fig.  873. 


and  may  be  rep  resented  by  the  ordinate  (posi- 
tive or  ncQative)  of  the  curve  mn})q. 

For  example,  at  the  points  defined  by 

z=     I,         ¥,      ii, 

the  shears  are  greatest  when 

J=    ¥,  |/.       hi, 

and  their  values  are,  resjjectively, 

^J«''/.     T/o"''/,    0. 


Again,  the  shear  at  any  point  (hstant  z  from  A  in  the  loaded 
portion  BE  when  tlie  live  loail  covers  AB 

^Ri  +  {t-tr)il-2)-w'ix-z) (J6) 

=  Ri  +  {t-v'-w')(l-z)+iv'{l-x) 

=  -\Ri'  +  {t'-wXl-z)-w'il-x)\ 

-=»»ini<.v  the  shear  at  the  xame  fxiint  when  AB  is 
unloaded  and  the  live  load  covers  BO. 


Hence,  by  eqs.  (4),  (5),  (IG), 


the  s 


]ir' 


hear=TT,-^(/-42)(/-j-)2,      ....    (17) 

increasing  for  a  given  value  of  z  with  l-x,  and,  therefore,  a  maximum 
when  x  =  7.     Thus 

the  maximum  sliear=  T--.j-(l-4x)(l-x)S      .     .     (IS) 
and  occurs  immediately  in  front  of  the  load  when  it  covers  AB,  and 


STIFFENING  TRUSS. 


871 


immediately  behind  the  load  when  it  covers  BO.    It  may  be  repre- 
sented by  the  ordinate  (positive  or  negative)  of  the  curve  arsq. 
For  example,  at  the  points  defined  by 

^  =  ^  =  ^  ¥,  il,  II,  H 

the  maximum  shears  given  by  eq.  (18)  are,  respectively, 

0,         ±jUWI,        ±j\Wl,         ±j%\w%         ±iw'l. 

Diagram  of  Maximum  Bending  Moment.— The  bending  moment 
at  any  point  in  BO  distant  z  from  A  when  the  live  load  covers  AB 


•Ri{l-z)  +  {t-w)^=!-j^ 


(19) 


-  [Ri'il-z)  +  it' -w-W^-^ 

-•■/?.'(/-.)+a'-.)^    '^M 


=  minus  the  bending  moment  at  the  same  point  when 
the  live  load  covers  BO. 

Hence,  by  eqs.  (4),  (5),  (19),  the  bonding  moment 

1  w'  .  1  ,/.' 

-'±^j(/2-4te+3x2)(/-2)T^-^(/2-4/x  +  Jx2)(Z-2)2. 
For  a  given  value  of  z  this  is  a  maximum  and  equal  to 

W'2l-Zl-2Z  2lZ 

±V    (1-az)       ^h^"    '=iTaz- 

Thus  the  maximum  bending  moment  may  be  represented  by 
the  ordinate  (positive  or  negative)  of  a  curve. 
For  example,  at  the  points  defined  by 


M'l, 


v. 


V, 


H 


I 


r 


,. 


t 


Pi 


■  '•«! 

i^H 

1 

^  '  ($ 

r 

III 

1 

Ml 


872  THEORY  OF  STRUCTURES. 

t'.ic  iK'iidiiig  moments  are  greatest  wlien  x  = 

tlieir  values  being,  resjK'ctively, 

0,  "fdhu^'P.     Titu"'/^      "f^hw'P,  0. 


/- 


X.      «/       hi 


V 


/E 


^ 


Fig.  874. 


The  absolute  maiimum  bending  moment  may  be  found  as  follows: 
For  a  given  value  of  x  the  Ix'nding  moment  (see  eq.  (19))  is  a 
maximum  when 

Iii  +  it-u')(I-z)=0, 


or 


l-z- 


'l-w' 


Hence,  the  maximum  lK>iiding  moment 

^1  R,2       ^w;(12-4lx  +  3x2)* 

It  will  \yo  an  absolute  iiiiiximum  for  a  value  of  x  found  by  putting 
ils  difTcreiitial  with  res|i(M't  to  .r  (■(|ii;il  to  nil. 
This  differential  easily  reduces  to 

.•i.r'-<>/.r2  H)/-V -/;,-(). 

X     -jl   is  ;iri   aitproximate  solution  of   this  equation,   and   the  cor- 
responding iiiMxiiiiuiii  U'lidinf:  Midinent         *  w'F. 

'ihe  prfCfMling  eiilculations  show  Ihdt  nt  ctrni  jinint  in  its  lentjlh 
till'  Ifiss  Diiiif  III'  siihifclcil  III  rr/iifl  iiKixiiiiinii  shears  anil  cijiiitl  tnnxiniiim 
livmliiii/  niDiiii'Hls  iij  itiiiHisilc  sitins. 

Apiiii.  it  iiKiy  lie  cM-ily  -ilinwn  in  n  similar  matmer  that  when 
n  sin.<j;!e  wci^lit  11*  inivi'Is  over  tin-  trux-^. 

the   maxitnum  I'osHiir  shear  at   a  dislaiiee  z  from  .1 


SL'SPENSIOX  BRIDGE  LOADS. 


873 


or 


and 


=  -j^(2l2-5lz  +  422); 

the  maximum  neijatire  sliear  either 
W  „ 

=  -|j(l2-5lz  +  4z2) 

I  W 

=jy(3i-4z), 

the  maximum  Ixnuling  moment 

W 

=  ±-p-z(l-z)(l-az). 

12.  Suspension-bridge  Loads.— The  heaviest  distributed  load  to 
which  a  Iiighway  bridge  may  be  subjected  is  that  due  to  a  dense 
crowd  of  people,  and  is  fixeil  by  modern  French  practice  at  82  lbs 
P'r  s(,uare  foot.  Probably,  however,  it  is  unsafe  to  estimate  the 
loa.l  at  less  than  from  100  to  140  lbs.  per  squar(>  foot,  while  allowance 
has  also  to  be  made  for  the  concentration  upon  a  single  wheel  of 
as  much  as  36,000  lbs.,  and  jNThaps  more. 

A  moderate  force  repeatedly  applied  will,  if  the  interval  l)etween 
the  blows  c()rresiM)nds  to  the  vibration  interval  of  the  chain,  rapidlv 
produce  an  excessive  o.scillation  (Chap.  IV,  Art.  ;{).  Thus  a  i)rj- 
ces.si,)u  marching  in  step  across  a  susiM'ii.sion  bridge  mav  strain  it 
far  more  intcMisely  than  a  dead  load,  and  will  set  up  a  svnchnmous 
vibration  which  may  prove  absolutely  dangerous.  For  a  like  n-ason 
the  wind  usually  sets  up  a  wave  motion  from  end  to  end  of  a  bridge. 

The  fndor  of  safety  for  the  dead  load  of  a  susix-nsion  bridge  should 
not  Ik'  !,.ss  than  2]  or  3,  and  for  the  live  load  it  is  advisable  to  make 
It  C.  W  ith  ivsiH-ct  to  this  point  it  may  1h>  remarked  that  the  enicienry 
of  a  cable  does  not  (Ic|m'ii,1  so  much  u|M)n  its  i-ltimate  .strength  iis 
U|K.n  its  limit  of  elasticity,  aii.l  so  long  as  the  latter  is  not  excee.led 
«!"•  «ab!..  remains  uninjured.  For  example,  the  hrcakinq  im,,ht  of 
"lie  of  the  I.Vinch  cables  of  the  Brock  yn  Bridge  is  estimated  to 
l«e  IL>,(KK)  tons,  its  I, mil  of  claslirily  In-ing  81  IS  tons;  so  that  with 
n  "i'i.v  as  a  factor  of  safety,  tiie  stres.s  would  still  fall  Ix-lnw  the 
elastic  limit  and  have  no  injurious  effect.     The  continual  application 


']! 


>  1 


m 


M 


im 


874 


THEORY  OF  STRUCTURES. 


of  such  a  load  would  doubtless  ultimately  lead  to  the  destruction 
of  the  bridge. 

The  dip  of  the  cable  of  a  suspension  bridge  usually  varies  from 
tV  to  xS  "f  the  span,  and  is  rarely  jus  much  as  j^  except  for  small 
spans.  Although  a  greater  ratio  of  dip  to  span  would  give  incrcjLsed 
economy  and  an  increa.'ied  limiting  span,  the  pa.ssage  of  a  live  load 
would  1k'  accompanied  by  a  gnviter  distortion  of  the  chains  anil  a 
larger  oscillatory  movement.  Steadiness  is  therefore  secured  at  the 
cost  of  economy  by  adopting  a  comparatively  flat  curve  for  the 
chains. 

13.  Modifications  of  the  Simple  Suspension  Bridge. — The  (iis- 
advantages  comi(>cted  with  sa'ijx'nsion  bridges  are  very  great.  The 
position  of  the  platform  is  restricted,  ma.«!sive  anchorages  and  piers 
are  generally  required,  and  any  change  in  the  ilistribution  of  the 
load  i)ro(luces  a  .sensible  deformation  in  the  structure.  Owing  to 
the  want  of  rigidity,  a  considerable  vertical  and  horizontal  oscillatory 
motion  may  l>e  causeil,  and  many  efforts  have  been  made  to  modify 
the  bridge  in  such  a  maimer  as  to  neutralize  the  tendency  to  o.scilla- 
tion. 

((/)  The  simplest  improvement  is  that  sliown  in  Fig.  875,  where 
the  point  of  the  cable  most  liable  to  deformation  is  attached  to  the 
piers  by  short  straight  chains  AB. 


¥vi   s-.i. 


(h)  A  scries  of  inclined  stays,  or  iron  ropes,  radiating  from  the 
pier-saddles,  may  Ik-  made  to  sup|M)rt  the  platform  at  a  number  of 
f  (juidistant  |M)ints  (Fig.  870).  Such  rojies  were  used  in  the  Niagara 
Bridge,  and  still  more  recently  in  the  Brooklyn  Bridge.  The 
lowrr  ends  of  the  rniH's  are  generally  made  fast  to  the  top  or  Ixittom 
chord  of  the  bridge-truss,  so  that  the  corres]K)nding  chord  .stress 
is  increased  and  the  neutral  axis  pro|K)rtionately  displaced.  To 
reuiedy  this,  it  ha.-  iH-en  projK/sed  to  connect  tht-  roj)t-.s  with  a  liori- 
zoiital  tie  coincident  in  position  with  the  neutral  axis.     Again,  the 


MODIFIED  SUSPENSION  BRIDGE. 


875 


cables  of  the  Niagara  and  Brooklyn  bridges  do  not  hang  in  vertical 
planes,  but  are  inclined  inwards,  the  distance  between  them  being 


Fig.  876. 

greatest  at  the  piers  pnd  least  at  the  cenire  of  the  span.  This  draw- 
hig  in  adds  greatly  to  the  lateral  stability,  which  may  be  still  further 
increased  by  a  series  of  horizontal  ties. 

(c)  In  Fig.  877  two  cables  in  the  same  vertical  plane  are  diaft- 
onally  braced  together.    In  principle  this  method  is  similar  to  that 


Fio.  877. 

adopted  in  the  stiffening  truss  (discussetl  in  Art.  11),  but  is  probably 
less  efficient  on  account  of  tlie  flexible  character  of  the  cables, 
although  a  slight  economy  of  material  might  doubtless  be  realized. 
The  braces  act  both  as  strut.?  and  tie.s,  and  the  stresses  to  whici» 
they  are  subjected  may  be  ea.sily  calculated. 

(d)  In  Fig.  878  a  single  chain  is  diagonally  braced  to  the  plat- 
form.   The  weight  of  the  bridge  must  be  sufficient  to  insure  that 


Fio.  878. 

no  suspender  will  \yo  subjected  to  a  thrast,  or  the  efficiency  of  the 
arrangement  is  destroyed.  .An  objection  to  this  as  well  a.s  to  the 
preceding  metluxl  is  that  the  variation  in  the  curvature  of  the  chain 
under  changes  of  teinixrature  tends  to  loosen  and  strain  the  joints. 
The  principle  has  Invn  adopted  (Fig.  879)  with  greater  prfec- 
tion  in  the  construction  of  a  foot-bridge  at  Frankfort.  The  girder 
is  cut  at  the  centre,  the  chain  is  hinged,  and  the  rigidity  is  obtained 


R 


M 


,1'  lii 


,  1 ,1 

1  ■■■ 

i'il 

III' 

}-fflii 

^^H 

S76 


THF.OuY  OF  STRICTURE.^. 


by  moans  of  vertical  anil  inclined  braces  which  act  both  as  struts 
ami  ties. 


Fio.  879. 


(e)  In  Fig.  SSO  the  girder  is  supported  at  several  points  by 
straight  chains  running  directly  to  the  pier-saddles,  and  the  chains 


Fkj.  8S0. 

are  kept  in  place  by  being  hung  from  a  curved  chain  by  vertical  rods. 

(/)  It  has  l«'en  i)roposed  to  employ  a  stiff  inverted  arched  rib 
of  wrought  iron  instead  of  the  flexible  cable.  All  straining  action 
may  be  eliminated  by  hinging  the  rib  at  the  centre  and  piers,  and 
the  theory  of  the  stresses  develo|x>d  in  this  tension  rib  is  preciselv 
similar  to  that  of  the  archeil  rib,  except  that  the  stresses  are  reversed 
in  kind. 

(f/j  The  jilatforin  of  every  susiK'nsion  bridge  should  be  brace<l 
horizontally.  The  floor-lx-ams  are  sometimes  laid  on  the  skew  in 
order  tliiit  the  two  ends  of  ;i  boam  may  1m'  susiH-nded  from  jMunta 
which  do  not  (isciHate  coiici.rdantly,  and  also  to  ili.stribute  the  load 
over  a  greater  length  of  cable. 


EXAMPLES. 

1.  The  si>;in  of  u  susiK-nsi(.n  l)ri(ljr<'  is  ->(K)  ft.,  the  dip  of  the  rhains  is  80  ft. 
and  the  wcifiht  of  the  ro;i(i\vay  is  1  ton  jx-r  foot  run.  Find  tlu-  tensions  at 
the  middle  and  ends  ot  eaeli  cluiin.  .\ns.  ;{|  }  tons;  .VS.'.U  Ions. 

2.  .\ssnininj;  that  a  steel  ro|M-  (or  a  sinpl,.  wi,-,.)  will  Lear  a  tension  of  15 
tons  iHT  s.|ii.ire  iii.-h,  (.how  that  it  will  s.afely  Imnip  its  own  wei-lit  over  a  span 
of  ;il).>ul  one  mile,  the  dip  heiiijr  one  fourteenth  of  the  span. 

.I'lN.   .Maximum  tension     .'i;{ Ors  !l)s. 

3.  Show  that  a  steel  rope  of  the  Ix'st  .pialiiy,  with  a  dip  of  one  .seven'h 
r.f  the  >^Y.■A^^,  will  vM  !)re.ik  Uhid  I  he  .-pan  i-^ra-^U  7  mile.-,  (he  idlunate  -tM-nRth 
of  the  rojie  U-ing  *>0  tons  [ler  square  ineh.     (I  ton '•_'_•  10  Ih.s.  i 

Alls.  .Ma.\iniuin  tension  =  59, ")4.Jton.s/s(|mirt-  ineh. 


EXAMPLES. 


S77 


if 


4.  If  the  span  =  /,  the  total  uniform  loa(l  =  ir,  and  the  dip=  j^  show  that 

the  maximum  tension  =  1  .oSir,  the  mit;imum  tension  =  1. .-,»',  the  length  of 
the  chain  =  1.018/,  and  find  the  iuerease  of  dip  corresponding  to  an  elonea- 
tioii  of  1  in.  in  the  chain. 


5.  A  cable  weighing  p  lbs.  per  lineal  foot  of  length  is  stretched  between 
sup[)orts  in  tlie  same  horizontal  line  and  20  ft.  apart.  If  the  maximum  deflec- 
tion IS  i  ft.,  determine  the  greatest  and  least  tensions. 

Am.  Parameter  m  =  lOO  ft.;  maximum  tension  =  lOOin;  minimum  ten- 
sion =  100/>. 

6.  The  dead  weight  of  a  suspension  bridge  of  1600  ft.  .span  is  J  ton  per 

lineal  foot;    the  dip  =  '"^3".     Find  the  greatest  and  least  pulls  upon  one  of 

the  chains.    The  ends  of  the  chains  are  attached  to  .saddles  on  rollers  on  the 
top  of  piers  50  ft.  high,  and  the  backstays  are  anchored  .JO  ft.  from  the  foot 
of  each  pier.     Find  the  load  upon  the  ])iers  and  the  pull  upon  the  anchorage. 
'^n«-  2-)5  tons;  24;j?  tons;  0.37^  tons;  344.G  tons. 

7.  The  platform  of  a  suspension  foot-bridge  o.  100  ft.  span  is  10  ft    wide 
and  supports  a  load  of  l-,0  lbs.  p<.r  .square  foot,  including  its  own  weight 
The  two  suspension  chains  have  a  dip  of  20  ft.     Find  the  force  acting  on  each 
chain  close  to  the  tower  and  in  the  middle,  assuming  the  chain  to  hang  in  a 
parabolic  curve.  ^m.  44,;575  lbs. 

8.  The  river  span  of  a  suspension  bridge  is  930  ft.  and  weighs  5970  tons 
of  winch  1439  tons  are  borne  by  stays  radiating  from  the  summit  of  each 
pier,  while  the  remaining  weight  is  distributed  between  four  1.5-in.  steel-win- 
cai)les,  producing  in  each  at  the  piers  a  tension  of  2064  tons.  Find  the  dii) 
of  the  c;ibles. 

Th<.  estimated  maximum  traflie  upon  the  river  span  is  1311  tons  uniformlv 
distributed.     Dt>termme  the  increased  stress  in  the  cables. 

To  what  extent  might  the  traffic  be  safely  increased,  the  limit  of  elasticity 
f  a  cable  being  SI  16  tons,  and  its  breaking  stress  12,300  tons? 

Arts.  63.884  ft.;  596.4  tons  to  13,303  tons  uniformly  distributed 

9.  The  uniform  load  on  each  of  the  cables  of  a  suspension  span  800  ft  in 
length  IS  4000  lbs.  per  lineal  foot  of  span.  The  dip  of  the  cables  is  60  ft. 
tiiid  the  stress  in  the  cables  at  the  centre  and  at  the  towers. 

.      ,  ,  •>«•'*•  5,333,333  lbs.;  5,5{)8, 130  lbs. 

10.  .\  telegraph  wire  ,\y  m.  diameter  is  supported  on  poles  170  ft.  apart 
and  dips  2  ft.  in  the  middle.     Find  the  pull  on  the  wire.  Am.  49  lbs. 

11.  A  trolley-wire  has  to  l)e  carried  on  jM)les  round    a  curve  of  I'^OOfl 
nulius.     The  poles  are  spaced  40  yards  apart,  and  in  the  middle  of  each  span 
the  wire  sags  down  6  ins.  below  the  point*  of  support.     If  the  wire  weighs 
1 J  lbs.  per  yard,  show  that  the  resultant  horizontal  pull  on  each  pole  is  verv 
nearly  180  lbs.  ' 


m] 


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878 


THF'my  OF  STRUCTURES. 


la.  A  copper  telegraph-wire  weighing  J  lb.  per  lineal  yard  is  suspended 
between  poles  on  level  ground  so  that  the  greatest  dip  of  the  wire  is  2  ft.  The 
tension  at  the  lowest  point  of  the  wire  is  100  lbs.  Find  the  distance  between 
the  poles.  Ans.  168.48  ft. 

13.  A  flexible  cable  weighing  -fn  lb.  per  lineal  foot  is  suspended  between 
two  poles  .4  and  B,  400  ft.  apart,  and  hangs  in  a  catenary  having  a  modulus 
of  2000  ft.    The  poles  are  of  the  same  height  as  the  crown  level. 

A  second  cable,  of  the  same  weight  per  lineal  foot,  is  supported  between 
B  and  a  third  pole  C  of  the  same  height,  the  angle  ABC  being  150^  If  the 
horizontal  pull  of  this  cable  is  1.155  times  that  on  the  first  cable,  find  its  dip 
and  modulus  and  show  that  the  intermediate  pole  should  be  supported  by 
a  stay  in  a  plane  at  right  angles  to  B  and  ^4.  If  the  stay  slopes  at  45°,  find  the 
stress  to  which  it  is  subjected.    Atm.  Dip=8.6  ft.;  w=2310  ft.;  167.385  lbs. 

14.  A  suspension  bridge  has  a  dip  of  10  ft.  and  a  span  of  300  ft.  Find 
the  increase  of  dip  due  to  a  change  of  lOO'*  F.  from  the  mean  temperature, 
the  coefficient  of  expansion  being  .00125  per  180°  F. 

Also  find  the  corresponding  flange  stress  in  the  stiffening  truss,  which  is 
12J  ft.  deep,  the  coefficient  of  elasticity  being  8000  tons. 

"      Ans.  1.17  ft.;  6.24  tons. 

15.  The  ends  of  a  cable  are  attached  to  saddles  free  to  move  horizontally. 
If  Ja  is  the  horizontal  movement  of  each  saddle  due  to  the  expansion  of  the 
cables  in  the  side  spans,  and  if  JS  is  the  extension  of  the  chain  between  the 
two  saddles,  show  that  the  increment  of  the  dip  (A)  is  approximately 


1« 
I6h 


JS+Ja 


/3  a  _^\ 
V8  h     a} 


16.  Show  that  the  total  extension  of  a  cable  of  uniform  sectional  area  A 
under  a  uniformly  distributed  load  of  intensity  w  is 


8E.U 
I  being  the  span  and  d  the  dip. 

17.  A  suspension  bridge  has  a  dip  of  .30  ft.  and  the  span  is  900  ft.  If  the 
coefficioiit  of  opatision  is  .000007  por  1°  F.  and  if  15,000  tons  per  square  inch 
is  the  coefficient  of  elasticity,  find  (a)  the  change  in  dip  corresponding  to  a 
fluctuation  of  .VP  from  the  mean  temperature,  (6)  the  corresponding  flange 
Stress  in  an  auxiliary  truss  25  ft.  deep. 

Ans.  (fl)  1.777ft.;  (6)  3.29  tons/sq.  in. 

18.  The  cables  of  a  suspension  bridge  for  a  span  of  200  ft.  have  a  dip  of 
40  ft.  Find  (a)  the  length  of  a  cable,  and  al.so  fin!  (b)  the  change  of  dip  cor- 
responding to  a  change  of  .30°  ('.  from  the  mean  tcmjwrature,  the  coeflScient 
of  linear  expansion  l)eing  .000012  jx-r  degree  centigrade. 

The  suspension  bridge  is  supplied  with  an  auxiliary  truss  4  ft.  deep.     Show 


EXAMPLES. 


879 


(c)  that  the  intensity  of  flange  stress  at  the  centre  developed  by  the  change 
of  temperature  is  717  lbs.  per  square  inch,    (ff  =  24,000,000  lbs.) 

If  the  live  load  is  4000  lbs.  per  lineal  foot  of  span,  determine  (rf)  the  maximum 
S.F.  and  B.M.  to  which  the  truss  is  subjected. 

Am.  (a)221§  ft.;  (b)  .0740  ft.;  (t)  710  Ibs./sq.  in.;  (d)  .50,000  lbs.; 
1,481,482  ft.-lbs, 

19.  A  suspension  bridge  of  240  ft.  span  and  20  ft.  dip  has  48  su-«nuers  on 
each  side;  the  dead  weight -3000  lbs.  per  lineal  foot;  the  live  load  -2000  lbs. 
per  lineal  foot.  F.nd  the  maximum  pull  on  a  suspender,  the  maximum  bend- 
ing moment  and  the  maximum  shear  on  the  stiflFening  truss.  Also  find  the 
elongation  in  the  chain  due  to  the  live  load. 

.4ns.  Max.  pull -12,500  lbs.;  max.  shear-30,000  lbs.;  max.  B.M.- 
1,066,666J  ft.-lbs.;  elongation- 92,760,000 ->£^,  A  being  sectional  area  of  a 
cable  and  E  the  coefficient  of  elasticity. 

20.  Each  side  of  the  platform  of  a  suspension  bridge  for  a  span  of  100  ft. 
is  carried  by  nine  equidistant  suspenders.  Design  a  stiffening  truss  for  a  live 
load  of  1000  lbs.  per  lineal  foot,  and  determine  the  pull  upon  the  suspenders 
due  to  the  live  load  when  the  load  produces  (1)  an  absolute  maximum  shear; 
(2)  an  absolute  maximum  bending  moment. 

Ans.  Max.  shear -6250  lbs.;  max.  B.M.-92,592H  ft.-lbs.;  pull  on  sus- 
pender =(1)  2777i  lbs.,  (2) -18511?  lbs.  or  3703H  lbs. 

21.  The  platform  of  a  suspension  bridge  of  300  ft.  span  is  suspended  by 
vertical  rods  spaced  10  ft.  apart  and  the  platform  is  also  provided  with  an 
auxiliary  truss.  Determine  the  pull  on  a  suspender  when  oiie  half  the  bridge 
carries  a  live  load  of  half  a  ton  per  foot  run.  Also  find  the  maximum  bending 
moment  and  maximum  shear  to  which  the  auxiliary  truss  is  subjected. 

Ans.  1.21  tons;  416?  ft.-tons;  9j  tons. 

22.  A  foot-path  8  ft.  wide  is  to  be  carried  over  a  river  100  ft.  wide  by  two 
cables  of  uniform  sectional  area  and  having  a  dip  of  10  ft.  Assuming  the 
load  on  the  platform  to  be  112  lbs.  per  square  foot,  find  the  greatest  pull  on 
the  cables,  their  sectional  area,  length,  and  weight.  (Safe  stress  =  8960  lbs. 
per  square  inch;  specific  weight  of  cable -480  lbs.  per  cubic  foot.)  Find  the 
depression  in  the  cables  due  to  an  increment  of  length  undfir  a  change  of  60°  F. 
from  the  mean  temperature.     (Coefficient  of  expansion  =  l-f- 144,000.) 

Ans.  56,000 lbs.:  60,312  lbs.;  6.73  sq.  ins.;  102tft.;  2302.65  lbs.;  '.0802  ft. 

23.  In  a  suspension  bridge  (recently  blown  down)  each  cable  was  designed 
to  carry  a  total  load  of  84  tons  (including  its  own  weight).  The  distance 
between  the  piers-  1270  ft.;  the  deflection  of  the  cable -91  ft.  Find  (a)  the 
length  of  the  cable;  (6)  the  pull  on  the  cable  at  the  piers  and  at  the  lowest 
point ;  (c)  the  amounts  by  which  these  pulls  are  changed  by  a  variation  of  40°  F. 
from  the  mean  temperature;  (rf)  the  tension  in  the  backstays,  assuming  thetn 
to  l)e  approximately  straight  and  inclined  to  the  vertical  at  the  angle  whose 
tangent  is  f. 

rhe  platform  was  hung  from  the  cables  by  means  of  480  suspenders  (240 
on  each  sideK  Find  (e)  tho  pull  nn  each  suspender  and  (/)  the  total  length 
of  the  suspenders,  the  lowest  point  of  a  cable  being  14  ft.  above  the  platform. 

Ans.  (0)  1287.4  ft.;  (/>)  146i',  and  152.4  tons;  (c)  1.5  and  1.45  tons'; 
(d)  394.55  tons;  (c)  .35  ton;  (/)  10,565.6  ft. 


Ill 


1-' 


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lUh.UliY  OF  STIiUCTLRES. 


24.  Ill  a  su.s|)oii.si(jii  bridge  of  KKH)  ft.  span  the  platform  is  carried  by 
two  cables,  one  on  each  si.Ie.  The  bridj^e  weiglis  i  ton  ]H'r  hneal  foot  of  span 
and  the  thp  is  uuc  iluiUiiUlt  of  the  span.  l''ind  tlie  greatest  and  least  pul's 
on  ;i  cable.  If  the  cables  are  attached  to  saddles  or  rollers  on  the  top  of  pieis 
M  ft.  high  and  the  b.ickstays  are  anchored  at  .")U  ft.  from  the  foot  of  each  pit  1, 
lind  the  load  upon  tl".    piers  and  the  pull  on  the  anchorage. 

Ann.  (ISO  tons;  (mO  tons;  ISOO  tons. 

25.  Find  the  length  of  the  cable  in  tlie  preceding  example,  ana  al.-o  Una 
the  <'hange  of  dip  corresponding  to  a  cliange  of  ;l()^  (".  from  the  ineaii  tem- 
perature, the  ro;  tticient  of  linear  expulsion  iM'ing  .(MKKJIL'  per  degree  centigrade. 

If  an  auxiliary  truss  4  ft.  deep  is  ;idiled,  iirid  tlu;  intensity  of  flange  stress 
at  the  centre  lieveloped  by  the  change  of  temperature. 

Ana.   U125.2  ft.;  1.  IL'G  ft.;  Of.0,0<)0  :  /. 

26.  The  cables  for  a  susiH-nsion  bri<lgc  of  "J  10  ft.  clear  span  are  suspended 
from  i)iers  which  are  "Jo  ft.  and  4  ft  rcs|H'<'tively  above  the  lowest  point  of 
th.'  cables. 

The  top  of  the  lowest  pier  i.s  anchored  by  a  backstay  inclined  at  ()()°  to  the 
verti<'al,  while  the  higher  ])ier  is  anchored  by  a  backstay  inclined  at  4.")^  to  the 
vertii-il.  nctermine  la)  the  length  of  the  cable  between  the  piers,  (/))  the 
horizontal  pull  on  the  cable,  (ci  the  tensions  in  the  calile  at  the  tojjs  of  the 
piers  when  the  load  on  each  cable  is  hall  a  ton  jht  lineal  foot  of  span. 

Ans.   (a)  -212.0  ft.;  (/>)  L'L>5  tons;  (r)  aiS.lO.')  tons,  L'.TO.Sl.")  tons. 

27.  .\t  Houghton  a  sus|K'nsion  bridge  of  14;}  ft.  span  and  1L'.3  ft.  dip 
failed  on  account  of  si  -"ous  oscillations  caused  by  the  marching  of  a  company 
of  soldiers  across  the  bridge.  The  .•iiichorage  was  a  2-in.  bolt  at  right  angles 
to  the  sus|>ension  link  and  having  a  bearing  of  3i  ins.  The  estimated  weight 
of  the  soldiers  was  4.S  tons  (long),  and  this  combined  with  the  dead  weight  01 
the  bridge  produced  a  force  of  .37.-  tons  on  each  anchor-bar.  Taking  IS  tons 
as  th  •  ultimaie  M>iiacity  of  good  iron,  determine  the  factor  of  .safety  and  the 
deadweight  of  the  bridge.  An.i.  '.i;  48.4  tons. 

28.  The  platform  of  a  suspension  bridge  of  l.jO  ft.  sjian  is  suspended  from 
the  two  cables  liy  NS  vertical  rods  (44  on  each  side);  the  dip  of  the  cables  is 
15  ft.;  there  are  two  stilTening  trusses;  the  dead  weight  is  2240  lbs.  ]K'r  lineal 
foot,  of  which  (III''  hdlj  is  divided  equally  between  the  two  piers.  Find  the 
stresses  ;it  the  middle  and  ends  of  the  cables  when  a  uniformly  distributed 
load  of  7S,7.")0  llis.  covers  one  half  of  the  bridge.  .Also  find  the  ma.vimum 
shears  and  bending  moments  to  whi<'h  the  stitTening  trusses  arc  subjected 
when  a  live  load  of  10.")0  His.  per  lineal  foot  cros.ses  the  bridge. 

Ans.  Pull  on  su.sponder  =  2803};  lbs.;  f/  =  -^'  7=  1.54,'_'18J  lbs. 

\  2!) 

Max.  shear  on  each  truss  at  centre  and  due  to  7S,7.")0  lbs.  =9843i  lbs. 

=  that  due  to  10.")0  lbs.  [ht  lineal  foot. 
Max.  H..M    due  to  78  7.")0  lbs.  is  at  centre  of  loaded  and  unloaded 

halves  and=  lS4,.")70,'s  ft. -lbs. 
Abs.  ma.x.  15. M.  diic  to  10."!)  lbs.  per  lineal  foot  is  at  points  of  trisec- 

tion  and  =  218,750  ft.-lbs. 


EXAMPLES. 


881 


29.  Solve  the  preceding  example  when  the  trusses  are  hinged  at  the  centre. 
An.s.  I'uUon^-    lender  =  2803} J  lbs.;  //= -^,-r=  1^,218}  lbs. 
Max.  shear  due  to  78,750  lbs.  =  9843J  lbs.  at^'centre  of  span  and  at 
end  of  loaded  half  of  bridge;    max.  .shears  due  to  10?,0  lbs    p. 
hneal  n,ot  =  13,12.5  oOOtii,  4921;,  830.3,V„and  9843^  lbs.  at  ends 

equiT 'S^Jr  "    '*  '''^  ^'"'^  '"''"^  ^'^•^  ""'''  ^P^"  '"^«  ^°- 
Max.  B  M.  due  to  78,7.50  lbs.  is  at  centre  of  half  truss  and  =  184..570,V 

lus^^s^  :  f:^'-  '^"^  '°  '"'^^  '^'^-  P*"-  ""««'  foot  =  17G,180rii' 
2-!l,484;,  and  lo3,.80S.|  ft.-lbs.  at  ,H.ints  dividing  the  half  truss 
into  tour  equal  segments. 

Trp:'.i  LM3'  ,-ir;  :*r  ':,i™  ;trt„x 

.JL  "^  .7'P""^'°"  ^'•^'^gf  360  ft.  long  consists  of  a  central  span  of  160  ft 
and  two  s,de  spans,  each  of  100  ft.  The  lowest  points  of  the  cable  a  e  in  the 
same  horizontal  plane,  70  ft.  above  mean  water  level,  30  ft.  below  the  sv^n^t 

whS  TTT    ''"  '^'-'  f  ^  "'  ^^•^  f'"^*^"™  ^'^e'-^  -«  two  cablesTom 
which  the  platform  is  suspended  by  vertical  rods  10  ft    anart      Thl  k\? 

oad  .s  4000  lb.,  per  lineal  foot  of  spL.    The  trafn   load  isS  1 J^  ,S 

oot  of  span      When  the  train  load  covers  the  whole  of  the  centre  sSnrd 

(a)  the  maximum  and  minimum  stresses  on  the  cables;    (6)  S  overtLte 

moment  on  the  pier  at  the  mean  water  level.     When  the  trlin  ?y^'^"™"'« 

the  bridge  between  the  lowest  points  of  the  cent.;\tn  and'r "dXnTnd 

Arm.  (a)    Cfnlrc  span,   //  =  266,666J1    and    T  =  l-i-i -it-H   m.  .       j 
//=106«,30J  and  r=, .33.3.33*  lbs.';   (^  mUmSlVtuZt  TZn 
17,500  lbs.;  .sirfc  .spaw,  22,000  lbs.  '   \<^)  centre  span, 

32.  An  island  divides  a  river  into  two  channels.    The  river  is  crossed  hv 
a  susixMision  bridge  of  two  spans,  the  one  of  210  ft.,  the  other  of  ^8^  ft     Th^ 
lowest  points  of  the  cable  are  in  the  same  horizontal  nhne   36  f     J  "     TJ 
summit  of  the  pier  erected  on  the  island,  9  ft    Sw  the  too  of  fh;  ^K 7      ! 

(h)  the  horizontal  pull  upon  each  cable,  (c)  the  total  len^h  of  the  cable        ' 
The  pior  IS  a  me.aaur.,.1  prism  of  12  ft.  uniform  width  m  (-  »,    k      T     ■  , 


f  ^  .1 


H 


■) 


I 


i 

I 


'11 


,  Hi » 


S8i 


THEORY  OF  SriiUCTUHES. 


inp  that  the  deviation  of  the  centre  of  pressure  from  the  middle  point  of  the 
baso  .s  not  to  exceed  '24  ft. 

The  platform  is  proviiled  with  a  stiffening  truss,  the  verticnl  suspenders 
being  l.'  ft.  apart.  If  the  dead  load  for  the  whole  truss  is  1000  lbs.  per  lineal 
foot  of  truss  and  if  one  half  of  each  span  is  covered  with  an  additional  load 
of  1000  lbs.  per  lineal  foot  of  truss,  find  (e)  the  stress  iu  a  sus)  -luicr  in  eaeh 
span,  (/)  the  maximum  B.M.  and  S.F.  on  the  truss  in  each  .'pan,  (g)  the  total 
length  of  the  susix-ndcrs,  the  susixMidiTs  at  the  centre  lyini;  1-'  ft.  in  Icnjith. 

Ans.  (n)  39S.89,  116.62,  3t)8.4.S.  104.4  tons;  (h)  3t)4J,  100  tons;  (r)  409.63 
ft.;  (d)  14.8,63.3  ft.;  (t)  8.52,  8.44  tons;  (/)  432,300 ft.-ton.s;  13.5, 11.25ton8; 

ig)  350.4  ft. 

33.  A  bridge  444  ft.  long  consists  of  a  central  -span  of  ISO  ft.  and  two  side 
spans  each  of  132  ft.;  each  side  of  the  platform  is  suspended  by  vertical  rods 
from  two  iron-wire  cabl.  <:  each  pair  of  cables  passe.<  over  two  masonry  abut- 
ments and  two  piers,  the  former  IxMng  24  ft.  and  tho  latter  39  ft.  ai)ove  the 
surface  of  the  ground;  the  lowest  i>oint  of  the  cables  iii  .-adi  span  is  19  ft. 
al>ove  the  ground  surface:  at  the  abutments  the  cables  are  coiuiected  with 
.straight  WTought-iron  chaitis,  l)y  means  of  which  they  are  attached  to  anchor- 
ages at  a  horizontal  distance  of  (Ki  ft.  from  the  foot  of  each  abutment ;  the  dead 
weight  of  the  bridge  is  3500  lbs.  |x>r  lineal  foot,  and  the  bridge  is  covered  with 
a  proof  load  of  4r)00  lbs.  |)er  lineal  foot.  The  piers  are  wrought-iron  o-scillating 
columns,  and  if  equilibrium  under  an  imequally  distributed  load  is  main- 
tained by  connecting  the  heads  of  the  columns  with  each  other  and  with  the 
abutments  by  iron-wire  stays,  determine  the  proper  dimensions  of  the  stays, 
assuming  them  approximately  straight.  Assume  that  the  proof  load  covers 
(o)  a  side  span;   (b)  two  side  spans;   (c)  the  centre  span. 

Ana.  (a)  Pull  on  stays  in  centre  span -840,050  lbs. 

(5)    "     <<     <<      ' -double  that  in  (a)> 

(c)      '     "     "     "     aide  span  -948,466  lbs. 


CHAPTER  XII. 

ARCHES  AND  ARCHED  RIBS, 

I.  Arches.— Fig.  881  represents  a  string  stretched  over  a  number  of 
smooth  pegs  and  carrying  certain  specified  loads,  ab,  be,  cd,  de.    Tak- 


Fio.  881. 


Via.  882. 


ing  a€  aa  the  line  of  loads  and  0  as  the  pole  (Chapter  I)  in  Fig  882 
the  horizontal  component  of  the  tension  along  each  element  betweea 
consecutive  pegs  is  constant  and  equal  to  Oil. 


Fio.  883.  Fro.  884. 

Invert  Fig.  881  and  assume  that  each  element  remains  riirid 
The  several  forces  remain  the  same  in  magnitu'le  but  are  reversed  in 

883 


884 


THEORY  OF  STRUCTURES. 


kind,  and  each  eloinont  is  now  acted  upon  by  a  constant  horizontal 
thrust,  OH.  With  a  new  set  of  forces  the  frame  will  assume  a  new 
shajjo  and  there  will  l)e  a  new  constant  horizontal  thrust.  This  is 
a  general  pro|x<rty  and  the  rigid  portions  between  consecutive  pegs 
form  a  line  of  resistance.  Again,  KL  in  Fig.  S85  is  the  closing  line 
of  the  funicular  |K)lygon  for  a  numl)er  of  given  vertical  loads  acting 


Fio.  885 


upon  the  beam  AB,  and  by  Chapter  I  the  B.M.  at  any  point  G  is 
^-OHxDE,  where  OH  is  parallel  to  the  cio-sing  line.  Taking  OH 
to  Ik'  unity  and  inverting  the  funicular  polygon  so  that  the  clo.sing 
line  is  horizontal,  it  Ixronies  the  actual  B.M.  diagram  for  the  loads 
in  question.    The  axis  AEB  of  an  arch,  Fig.  887,  rises  to  meet  the 


Fkj.  s«7. 

B.M.  diagram  rj/)'».  ami  the  B.M.  on  the  arch,  a.«  distinguished  from 
that  on  the  liorizotitjil  \tcnu\,inOH  xDK.  The  following  is  a  .simple 
general  pnntf  <»f  this  result: 

In  Fig.  HH7-AEIi  is  the  axis  of  the  arch,  nl)li  is  the  B.M.  curve, 
and  DHF  is  a  vertical  ordinate  at  any  [X)int  H  of  the  axis. 

Let  //,  I'  be  n-sjH'ctively  the  horizontal  and  vertical  com|X)nenta 
of  the  reaction  at  .1; 


ARCHES. 


885 


Let  M,  be  the  B.M.  at  E  of  the  vertical  load  on  the  arch  between 
A  and  F. 


Then 

But 

Therefore 


B.M.  at  E=VAF-HEF-M^. 
0=V  AF-H  DF-M,. 
B.M.  at  E  =  M  =  HDE. 


Again,  let  the  normal  at  E  meet  the  B.M.  curve  in  ZX  and  let  r 
be  the  thrust  along  the  axis  at  E.    Then 


D'E 


or 


T  cos  DEiy  =  H^  T~,  approximately, 
HDE  =  T  D'E  =  M. 


The  great  difficulty  in  developing  the  analj-sis  of  the  arch  is  due  to 
the  uncertainty  as  to  the  true  positions  of  the  ends  of  the  B.M.  curve. 
If  It  IS  possible  to  introduce  hinges  at  the  ends  of  the  arch  or  at  any 
other  pomt,  as,  e.g.,  its  centre,  the  B.M.  curve  must  pass  through 
■such  pomts.  From  the  nature  of  the  construction  it  is  impracticable 
to  use  hmges  with  masonry  arches,  although  attempts  have  been 
made  to  provide  for  a  partial  rotation  bv  introducing  flat  iron 
plates  at  the  skewbucks  and  at  the  key.  Tlieso  arches  neceasarily 
belong  more  or  less  to  what  may  be  called  an  indeterminate  class 


Fia.  888. 


The  different  parts  of  an  arch  are  indicated  bv  the  t<>rn»  in  Fig 
S8.S.    Tl,e  rise  of  the  arch  is  the  vertical  distance  between  the  spruv- 
mg  and  the  crown,  "^ 


1 

iU 

r 

.1 

A 

8 ' 

i  5-   1 


m 


ft 

u 

8S6 


THEORY  OF  STRUCTURES. 


In  tho  figure,  A  BCD  represents  the  profile  of  an  arch.  The  under 
surface  AD  is  called  the  soffit  or  intrados.  The  upper  surface  BC 
is  sometimes  improixrly  called  the  extrados.  The  highest  point  K 
of  the  soffit  is  the  croini  or  key  of  the  arch.  The  sprinrfings  or  skew- 
hnrkx  are  tho  surfaces  AB,  DC  from  which  the  arch  springs,  and  the 
fio'inrhes  are  the  portions  of  the  arch  alx)ut  half-way  lietween  the 
sprinsjiiigs  and  the  crown.  Upon  each  of  the  arch  faces  stands  a 
.-'IKitidrel  wall,  and  the  space  between  these  two  external  spandrels 
may  1k'  occupied  by  a  series  of  internal  spandrels  spaced  at  definite 
ilistances  apart,  or  may  be  filled  up  to  a  certain  level  with  masonry 
(i.e..  hnckin(j)  and  ixhove  that  with  ordinary  ballast  or  other  rough 
iiiatcrial  (i.e.,  filling). 

A  masofiry  arch  consists  of  courses  of  wedge-shaped  blocks  with 
tlie  iH'd-joiiits  |)erjH'ndicular,  or  nearly  so,  to  the  soffit.  The  blocks 
are  called  routisoirs,  and  the  roussoirs  at  the  crown  are  the  keystones 
of  the  arch. 

.\  brick  arch  is  usually  built  in  a  numlx-r  of  rings. 

Consider  the  jwrtion  of  the  arch  bounded  by  the  vertical  plane 
KE  at  tlie  key  and  by  the  plane  AB. 


Fio.  889. 

It  is  kept  in  eqiiiiibrium  by  the  reaction  /?  at  KE,  the  reaction 
/t'l  at  AH.  and  the  weiKlif  I'l  of  the  portion  under  consideration 
;uid  its  sii|H'rineumlH'tit  load. 

Let  N  itiid  T  Ix'  the  [Mjirits  of  a[)plicatinn  of  Ri  and  H  resj)ec- 
tivfly. 

l-cl  tlic  directions  of  H\  and  If  inter-xect  in  a  point.  The  direc- 
tion «il    )  1  must  also  pass  through  the  same  jKunt. 

Taking  moment^  a'«nit  .'^', 


ARCHES. 


887 


pi  and  j/i  being  respectively  the  perpendicular  distances  of  the 
directions  of  R  and  l',  from  S. 

Similarly;  the  portion  KECD  of  the  arch  gives  the  equation 

Y2  being  the  weight  to  which  it  is  subjected,  and  7)3,  1/2  the  perpen- 
dicular distances  of  the  directions  of  R  and  Fz  from  the  point  of 
application  V  of  the  reaction  at  the  plane  DC. 

If  the  arch  and  the  loading  are  symmetrical  with  respect  to  the 
plane  KE, 

^1  =  ^^^2'    2/1  =  J/2,    and  therefore    p,  =  pj. 

Hence  the  direction  of  R  will  be  horizontal,  which  might  have  been 
inferred  by  reason  of  the  symmetry. 

The  magnitudes  of  the  reactions  are  indeterminate,  as  the  posi- 
tions of  the  points  of  application  {S,  T,  V)  are  arbitrary,  and  can 
only  be  fixed  by  a  knowledge  of  the  law  of  the  variation  of  the  stress 
in  the  material  at  the  bounding  planes  AB,  KE. 

Suppose  the  arch  to  be  dinded  into  a  number  of  elementary 
portions  ke',  k'e"  .  .  .  (e.g.,  the  voussoirs  of  a  masonry  arch)  by  a 
series  of  joints  ke,  k'e'  .  .  . 


;m^-«' 


Fio.  890. 


Fig.  891. 


Let  Wu  Wj,  .  .  .  Ik'  the  loads  directly  supported  by  the  several 
l»ortions.  These  loads  generally  consist  of  the  weight  of  a  portion 
Ce.g.,  ke'j-rtUv  weight  of  the  su|M>riMf'uml)ent  mass  +  the  load  u|)on 
the  overlying  rrmdway;  the  lines  of  action  of  the  loads  are,  therefore, 
nearly  alwnys  vertical. 

IL'irh  fifnuiitiuy  [Kirlion  nmy  Ik-  considered  m  acted  upon  and 
kept  in  eciuilibrium  by  //(rec  forces,  viz.,  the  external  load  and  the 


?.  s 


m 


888 


THEORY  OF  STRUCTURES. 


prcssuros  at  the  joints.  If  the  pressure  and  its  point  of  application 
at  any  given  joint  have  Ix'en  determined,  the  pressures  and  the 
oorres])on(hng  points  of  appHcation  at  the  other  joints  may  also  be 
found. 

For,  let  1234  .  .  .  Ix'  the  line  of  loads,  so  that  12=11'!,  23- 

W2,  .  .  . 

Assume  that  the  pressure  P  and  its  {wint  of  application  r  at  any 

given  joint  /.f  are  known. 

Draw  0\  to  represent  P  in  direction  and  magnitude. 

Then  02  fvideiitly  represents  the  resultant  of  P  and  TFi  in  direc- 
tion and  magnitude,  and  this  resultant  nmst  be  equal  and  opposite 
to  the  pressure  Pi  at  the  joint  k'e'. 

Hence  a  line  n'n  drawn  through  n,  the  intersection  of  Pand  TTj, 
parallel  to  02,  is  the  direction  of  the  pressure  Pi,  and  intersect* 
k'e'  in  the  jwint  of  application  r'  of  Pi. 

Again  03  represents  the  resultant  of  Pi  and  W2  in  direction  and 
magnitude,  and  this  resultant  must  be  equal  and  opposite  to  the 
pri'ssure  P2  at  the  joint  k"e". 

The  line  n"n'  drawn  through  n',  the  intersection  of  Pi  and  W2, 
parallel  to  03,  is  th.e  «lirection  of  the  pressure  Pj  and  intersects  k"e" 
in  the  jKHut  of  application  r"  of  P2. 

Proc(>eding  in  this  manner,  a  series  of  points  of  application  or 
centre-'^  of  rcxistaiice  r',  r",  r'" ,  .  .  .  may  be  found,  the  corresponding 
pressures  being  represented  by  02,  03,  04,  .  .  . 

The  jxdygnn  of  jrrexsures  formed  by  the  lines  of  action  of  P,  Pi, 
p£,  ...  is  termed  an  equilibrated  })ohj(jon,  and  is  a  funicular  polygon 
of  the  loads  ujx)n  the  several  jx)rtions. 

The  ix)lyg(»n  formed  by  joining  the  points  r,  r',  r",  .  .  .  suc- 
cessively, is  called  the  line  of  reKintanre. 

In  the  limit,  when  the  joints  are  supposed  indefinitely  near, 
thes<'  polygons  Ix'come  curves,  the  curve  in  the  case  of  the  equili- 
brated |M)lygon  Ix'ing  known  jis  a  linear  arch. 

The  two  curve.s  may,  without  sensible  error,  be  supposed  identical, 
and  they  will  exaetly  coincide  if  the  jtnnts  (of  course  imaginary  in 
such  a  ra.se)  are  made  parallel  to  the  lines  of  action  of  the  external 
loads.     This  may  Ix'  easily  proved  jus  follows: 


Let  the  fifsiirtr  reprpsrn)   i\   |M)rtioii  of  an  arch  bounded  by  the  joints 


ARCHES. 


8S» 


(imaginary)  KE  .l/.V  parallel  to  the  lines  of  action  of  the  external  loads,  which 

will  be  assumed  vertical. 

Iteduce  the  suixrincumbent  loads  to  an  equivalent  mass  of  arch  material 
Let  h,  e.g.,  be  the  depth  of  material  of  specific  weight  w.,  overlying  the 

arch  at  any  given  point,  and  let  Q  be  the  load  per  unit  of  area  of  roadway. 
Also,  let  u-  be  the  specific  weight  of  the 

arch  material. 

Then  x,  the  equivalent  depth,  is  given 
by 

If   the   value   of  x  is  determined   at 

different  points  along  the  arch,  a  profile 

tn  may  be  obtained  defining  a  mass  ENne 

of  arch  material  which  may  Ix;  substituted 

for  the  sufxjrincumbent  load.    Denote  the 

weight  of  the  mass  MKen  by  IF. 

Let  the  pressure  P  and  its  point  of  application  0  at  the  joint  KE  be  given 
Take  O  as  the  origin,  the  line  OA  in  the  direction  of  P  as  the  axis  of  x  and 

the  vertical  through  0  as  the  axis  of  y,  and  let  0  be  the  angle  between  the  two 

axes. 

Let  the  lines  of  action  of  P  and  W  intersect  in  G.  The  line  of  action  of 
their  resultant  will  intersect  MN  in  the  centre  of  resistance  0,. 

Let  .Y,  }'  be  the  co-ordinates  of  0,. 

Let  «  be  the  depth  of  an  elementary  slice  of  thickness  dx,  parallel  to  OAT 
at  any  abscissa  x.    Its  WQight  ~wzdx  sin  0. 


Fia.  892. 


Then 


WOG~  f^  wzdx  Bin  Ox-WiX -ACT). 


n  ,P     AG     AG    . 

W  ~  AO, "  T '  *'"•*  *"*  triangle  AGOi  is  evidently  a  triangle  of  forces 

for  the  forces  actitig  upon  the  mass  under  consideration. 
Also,  ^'-fo  vozdxttine. 

Therefore        ^    ^'^^dx  aine-WX-w'^^Y-xfJ^wzBinedx-PY. 

This  is  the  equation  to  the  line  of  resistance. 
Taking  the  differential  of  this  equation, 

trz'X  sin  OdX-Xwz'  sin  ddX+  WdX-PdY, 

z'  being  the  depth  corresponding  to  the  abscissa  X. 

TKo~r„-„  ^^'     "'     •"^'     "in  AGa 

Therefore  -^  ^  .^^-  -^.^  ^O.^-^'*"  '^^^  '^  ""W. 


i|^|| 


ui 


^  III 


kM 

m 

■i 

n 

'■i  j 

h 

•890 


THEORY  OF  STRUCTURES. 


Fio.  893. 


Thus  the  tangents  to  the  curve  of  pressures  and  to  the  curve  of  centres  of 
pressure  at  any  given  point  coincide,  and  the  curves  must  also  coincide. 

2.  Conditions  of  Equilibrium.— Let  the  Fig.  893  represent  a  por- 
tion of  an  arch  of  thickness,  unity,  between  any  two  bed-joints  (red 
or  immjinary)  MX,  PQ. 

Let  ir  be  its  weight  together  with  that  of  the  superincumbent 
load.     Let  the  direction  of  the  reaction  R'  at  the  joint  ^fX  intersect 

MX  in  m  and  the  direction  of  \V  in  n. 
V)",,  :■  For  equilibrium,  the  reaction  R"  at 
the  joint  PQ  must  also  pass  through  n. 
I^t  its  direction  intersect  PQ  in  0. 
•  In  order"  that  the  eciuilibrium  may  be 
stable,  three  conditions  must  be  ful- 
filled, viz.: 

First.  The  point  0  must  lie  between 
P  and  Q,   so  that   there   may  be  no 
tendency  to  turn   about  the  edges  P  and  Q. 

Second.  There  must  he  no  .^tiding  alonq  PQ,  and  therefore  the 
angle  between  the  direction  of  R"  and  the  norma!  to  PQ  must  not 
exceed  the  angle  of  friction  of  the  material  of  which  the  arch  is  com- 
posed. 

(N.B. — The  angle  of  friction  for  stone  upon  stone  is  alwut  30°.) 

TJiird.  The  v)aiimum  intensity  of  stre-^-'i  at  any  point  in  PQ  mmt 
not  exceed  the  .^afe  re.'<istance  of  the  material. 

Further,  the  stress  should  nf)t  eliange  in  character,  in  the  case 
of  masonry  and  brick  urohes.  but  should  Ije  a  eomjm'ssion  at  every 
point,  as  these  materials  are  not  suited  to  withstand  tensile  forces. 

The  Ix'st  position  for  O  would  !«■  the  middle  point  of  PQ,  as  the 
pressure  would  then  In-  uniformly  distributed  over  the  area  PQ. 
It  is,  however,  impracticable  to  insure  such  a  di.stribution,  and  it 
has  b'en  sometimes  assumed  that  the  stress  varies  uniformly. 

With  this  assumi)tion,  let   .V  Im'  the  normal  component  of  R". 

Let  /  be  th(>  maximum  compressive  stress,  i.e.,  the  stress  at  the 
most  coiiiiiresscd  edge,  e.g.,  P. 

Let  (/S  ijj'tj.  S  JH'ing  the  midille  point  of  PQ,  and  q  a  cocfii- 
cient  wlioso  value  is  to  be  determined. 


CONDITIONS  OF  EQUILIBRIUM. 


891 


ThenifPO<^, 
o 


if  PO>f, 


N  =  ifPO  =  ifPQ(ii-q); 


N  = 


fPQ 

l+6q' 
PQ 


and  in  the  limit  when  PO^-f,  i.e.,  when  the  intensity  of  stress 
varies  uniformly  from  /  at  P  to  nil  at  Q, 

?=i    and    iV  =  ^^. 

(See  Art.  12,  Chap.  V.) 

Similarly,  if  Q  is  the  most  compressed  edge,  the  limiting  position 
of  0,  the  centre  of  resistance  or  pressure,  is  at  a  point  0'  defined  hv 

Hence,  as  there  should  be  no  tendency  on  the  part  of  the  joints 
to  open  at  either  edge,  it  is  inferred  that  PO  or  QO'  should  be  >-^, 
i.e.,  that  the  point  O  should  lie  within  the  middle  third  of  the  joint. 

Experience,  however,  shows  that  the  "middle-third"  theory 
cannot  be  accepte.l  as  a  solution  of  the  problem  of  arch  stability 
and  that  its  chief  use  is  to  indicate  the  projjer  dimensions  of  the  abut- 
ments. Joint  cracks  are  to  h'  found  in  more  than  90  per  cent  of 
the  arches  actually  co.istructed,  and  cases  may  be  instanced  in  which 
the  jomts  have  ojiened  .so  widely  that  the  whole  of  the  thrust  is 
transmitted  through  the  edges.  In  Telfor.l's  ma.sonrv  arch  over 
the  Severn,  of  1.50  ft.  span,  Haker  discovered  that  there  had  been 
a  settlement  (15  ins.)  sufficient  to  induce  a  .slight  reverse  curvature 
at  the  crown  of  the  soffit.  Again,  the  jwsition  of  the  centre  of  pres- 
sure at  a  joint  is  indotorininate.  and  it  is  (horeforr  impossible  as 
well  as  useless  to  make  any  calculations  as  to  the  maximum  intensity 
of  stress  due  to  the  pressure  at  the  joint.    What  seems  to  happen 


llrl 


!  ; 


892 


THEORY  OF  STRUCTURES. 


in  practice  is,  that  the  straining  at  the  joints  generally  exceeds  the 
limit  of  elasticity,  and  that  the  pressure  is  uniformly  .'.l^tributed 
for  a  certain  distance  on  each  side  of  the  curve  of  pressuvi-.-;.  Thus, 
the  proper  ditncnsions  of  a  stable  arch  are  usually  determined  by 
empirical  rules  which  have  been  deduced  as  the  results  of  experience. 
For  example.  Baker  makes  the  following  statement: 

liCt  r  be  the  thrust  in  tons  or  pounds  per  lineal  foot  of  width  of 
arch. 

I^et  /  be  the  safe  working  stress  in  tons  or  pounds  per  square 
foot. 

An  arch  will  be  stable  if  an  ideal  arch,  with  its  bounding  surfaces 

1  T 

at  a  minimum  distance  of  :j  y  from  the  curve  of  pressures,  can  be 

traced  so  as  to  lie  within  the  actual  arch.  An  advance  would  be 
made  towards  a  more  correct  theory  if  it  were  possible  to  introduce 
into  the  question  the  ela.sticity  and  compressibihty  of  the  materials 
of  construction.  These  elements,  however,  vary  between  such  wide 
limits  that  no  reliance  can  be  placed  upon  the  stresses  derivable 
from  their  values. 

Joint  of  Rupture.— As  already  shown,  the  B.M.  at  any  point  in 
the  axis  of  an  arch  is  HDE,  H  being  the  horizontal  thrust  and  DE 
the  vertical  intercept  at  the  point  between  the  axis  and  the  B.M- 
curve.  For  a  constant  B.M.,  H  diminishes  as  DE  increases,  and 
for  a  minimum  thrust  t'le  B.M.  curve  should  be  as  high  as  possible, 
consistent,  of  course,  wi...  stability.  It  is  evident  that  in  a  masonry 
or  brick  arch  the  line  of  resistance  should  fall  within  the  thickness 
of  the  arch,  or  failure  may  occur  by  the  opening  of  the  joints,  as  in 
Fig.  894.    To  avoiil  such  a  result  it  is  a  common  practice  to  require 


Fio.  894. 


Fig.  895. 


that  the  line  of  re.<?i.'<tanre  shall  fall  within  the  middle  third  of  the 
thickness  of  the  arch,  with  the  object  also  of  insuring  that  the  stress 
at  every  point  shall  be  a  compression.    Thus,  for  a  minimum  thrust 


896. 


JOINT  OP  RUPTURE.  §93 

tho  line  of  resistance  should  take  a  position  nr  where  the  centre  of 
resistance  r  is  as  high  as  possible  consistent  with  stability,  coinciding 
for  example,  with  the  upper  end  of  the  mvidU  third  of  the  depth  of 
tlie  arch.    The  point  n  should  not  fall  below  the  lower  end  of  the 
middle  third  of  the  arch  thickness  at  the  skew-back. 

In  general,  let  12,  34  be  the  bound- 
ing surfaces  between  which  the  curve 
of  pressures  must  lie  and  let  4  be  the 
centre  of  resistance  at  the  crown.  A 
series  of  curves  of  pressure  may  be 
drawn  for  the  same  given  load,  but 
with  different  values  of  the  horizontal  thrust  h. 

Let  Axy  be  that  particular  curve  which  for  a  value  H  of  the  hori- 
zontal thrust  is  tangent  to  the  surface  12  at  x;  the  joint  at  x  is  called 
the  joint  of  rupture. 

The  position  of  this  joint  in  any  given  arch  may  be  tentatively 
lound  as  follows:  ^ 

Let  /  be  any  joint  in  the  surface  12. 
Let  IF  be  the  weight  upon  the  arch  between  /  and  1. 
Let  X  be  the  horizontal  distance  between  J  and  the  centre  of 
gravity  of  W. 

Let  r  be  the  vertical  distance  between  J  and  4 
It  ^vill  also  be  assumed  that  the  thrast  at  4  is  horizontal 
If  the  curve  of  pressure  is  now  supposed  to  pass  through  J  the 
corresponding  value  of  the  horizontal  thrust  h  is  given  by      ' 

hY^WX. 


m 


1 3 


I:  IM 


fi'.i 


J   . 


By  means  of  this  equation,  values  of  h  may  be  calculated  for  a 
number  of  joints  in  the  neighborhood  of  the  haunch,  and  the  greatest 
of  these  values  will  ho  the  horizontal  thrust  //  for  the  joint  x  This 
IS  evident,  as  the  curve  of  pressure  for  a  smaller  value  of  h  must 
necessarily  fall  below  Axy. 

When  this  happens,  the  joints  will  tend  to  open  at  the  lower 
edge  of  the  joint  14  and  at  the  upper  edges  of  the  joints  at  x  and 
23,  so  that  the  arch  may  sink  at  the  crown  and  spread,  unles«  the 
abutments  and  the  lower  portions  of  the  arch  are  massive  enouirh 
to  counteract  this  tendency. 


I 


i 


i'     1 


:  '111 


h 


894 


THEORY  OF  STRUCTURES. 


If  the  curvp  of  pressure  fall  above  ixy  an  amount  of  backing 
sufticient  to  transmit  the  thrust  to  the  abutments  nuLst  be  provided. 
The  same  resuU  may  be  attained  by  a  imiform  increase  in  the  thick- 
ness of  the  arch  ring,  or  by  a  gradual  increase  from  the  crown  to 
the  abutments. 

For  example,  the  upper  surface  (extrados)  of  the  ring  for  an  arch  with  a 
semicircular  soffit  .lA'B,  having  its  centre  at  0,  may  \k  delineated  i;i  the  fol- 

l(/\ving  manner: 

Let  z  define  the  joint  of  rupture  in  the 
soffit. 

The  anitle  AOi  is  approximately  30°  for 
a  semicircular  soffit  and  45°  for  an  elliptical 
soffit.     Ill  the  present  case  take  .4Oj;  =  30°, 
and  in  Ox  produced  take  zj'  =  2xA'D,  KD 
Fig    897.  being  the  thickness  at  the  crown. 

The  arc  Dx'  of  a  circle  struck  from  a  centre  in  DO  produced  may  be  taken 
as  a  part  of  the  upper  boundary  of  the  ring,  and  the  remainder  may  be  com- 
pleted by  the  tangent  at  .r'  to  the  arc  Dx'. 

Minimum  Tliirkncs.t  of  Ahiitmnit.—Lot  T  be  the  resultant  thrust  at  the 
horizontal  joint  liC  of  a  rectangular  abutment  ABCD. 

Let  y  be  the  distance  of  its  point  of  apjilication  from  B. 
Let  H  and  V  be  the  horizontal  and  vertical  components 
of  T. 

Let  w  be  the  specific  weight  of  the  material  in  the  abut- 
ment . 

Let  h  be  the  height  AR  of  the  abutment. 

Let  t  ha  the  wi(lth  AD  of  the  abutment. 

In  order  that  there  may  Ijc  no  tendency  to  turn  about 

th'  toe  D,  the  moment  of  the  weight  of  the  abutment  with 

respect  to  D  plus  the  moment  of  V  with  resi)ect  to  D  must 

be  greater  than  the  moment  of  //  with  respect  to  D.     Or, 

uht~  +  V{t-y)>Hh, 


Fio.  898. 


or 


V        \1H    2V        V" 

wh     \  if      uh       w'h' 


This  relation  must  hold  good  whatever  the  height  of  the  abutment  may 
be ;   and  if  h  is  made  equal  to  ac , 


which  defines  a  minimum  limit  for  the  thickness  of  the  abutment. 


li! 


EMPIRICAL  FORMULA. 


■>!,•;> 


3-  Empirical  Formula— In  practice  the  thickness  t  at  the  crown 
is  often  found  in  terms  of  s,  the  span,  or  in  terms  of  p,  the  radius 
of  curvature  at  the  crown,  from  the  formulae 

^=^5,    or    t=Vcp, 
t,  s,  and  p  being  all  in  feet,  and  c  being  a  constant. 
According  to  Dupuit,  <  =  .36\/s"fora  full  arch; 

<  =  .27v'.s  for  a  segmental  arch. 


According  to  Rankine,  =Vl[2p  for  a  single  arch; 

t=V Alp  for  an  arch  of  a  series 

Ex.  1.  A  masonry  arch  of  90  //.  span  and  30  ft.  rise,  with  a  parabolic  intrados 
and  a  horizontal  extrados,  springs  from  abutments  with  veHical  faces  and  10  ft. 
thick,  the  outside  faces  being  carried  up  to  meet  the  extrados.  The  depth  of  the 
keystone  is  3  ft.  The  centre  of  resistance  at  the  springing  is  the  middle  of  the 
joint,  and  at  the  crown  12  ins.  below 
the  extrados.  The  specific  weight  of 
the  masonry  may  be  taken  at  150  lbs. 
per  cubic  foot.  Determine  (o)  the 
resultant  pressure  in  the  vertical  joint 
at  the  crown;  (b)  the  resultant  pres- 
sure in  the  horizontal  joint  at  the 
springing;  (c)  the  maximum  stress 
in  the  vertical  joint  aligning  with  the 
inside  of  an  abutment. 

Let  X,  Y  be  the  vertical  and 
horizontal    distances,    respectively, 


Fig.  899. 


,  — I —     ---  — J  J 

from  the  point  C,  of  the  C.  of  G.  of  portion  of  the  structure  bounded  by  the 
surfaces  CO,  CE,  EF,  FA,  and  AO.    Then 

r(45X3  +  33X  10+1X30X45)-  7X915 

-45X3X22J  +  33X 10X50+ JX30X45X33} 
and  X(45x3+33X10+iX30X45)-XX915 

-45X3Xli  +  33X10xl6J  +  iX30X45xl3» 
Therefore  r-37'-95    and    X-12'08. 


1:1. 

;.     1 

t     \ 

i 

;     ;i 

: 

,    1 


i         ■    1 


l|.^ 


if 


1       I 

-J 

;    'J 

■v     >i 

!  * 

'■ 

i  • 

1 

\  ^= 

;■  ! 

|i 

1 

If' 

!■ 

;| 

|; 

1 

Ti 

1 

S96 


THEORY  OF  STRUCTURES. 


The  weight  R' of  th-  .         •  under  conaideration^ 91 5x150 

=  137,250  lbs. 

The  resultant  force  /?  at  the  centre  of  resistance  S,  the  middle  point  of 
the  horizontal  plane  AF,  passt-s  through  the  iH)int  of  intersection  K  of  the 
line  of  action  of  \V  and  of  the  horizontal  thrust  H  at  the  centre  of  resistance 
T,  CO  being  the  key  line  and  tT  =  irO  =  1  ft.  The  triangle  KLS  is  therefore 
a  triangle  of  forces  and  if  0  is  the  angle  KSL, 

„    SL     .■)() -.37.95      „,„„ 
cot  0  =  ^  = 3'^ "  ••^^^• 


Therefore 
Hence 


6"  "  09°  •  22      and    couec  6  - 1 .069. 


ft-W'jrj  -  137,2.50  cot  0 - 51,689  Ibe. 


and 


.SK 


R-W— ^  =  137,2.50  cosec  <?-  146,721  lbs. 


Again  let  I'y/  be  the  vertical  and  horizontal  distances  of  G',  the  C.  of  G. 

of  the  portion  between  AB  and  CO,  from  C. 
Then 

y\45x3  +  iX30-<  4.5) -i/'y. 585-45x3x221 
+  iX30X  45X33} -18,225 

and         z'X.585-45X3XlJ  +  §^30X45X12 

-5602). 
Therefore    y'-31'i',    and    i'-9'il. 
Also  w',  the  weight  of  ABCO 

-.585x150-87.7.50  lbs. 


Fio.  900. 


The  rpsultant  of  H  and  W  acts  along  K'X  and  must  Ix-  equal  and  opposite 
to  the  resultant  pressun-  A'  upon  the  face  Mi,  i*o  that  .V  is  the  centre  of  resis- 
tances in  this  plane.  Thus  K'NL'  is  a  triangle  of  forces  and  if  a  is  tbe  angle 
K'NL', 

H     51«>S9      ., 
cot  a  -  jy,  -  ^^._.j,  =>  ..*.M<(     iinil     n  -  .-.O}". 


Henr 


«77."«0 
ft'  -  W  c«j«c  a  -  93,(M2  Iba. 


If  the  intensity  of  the  (jressurc  varies  from  a  maximum  /  at  A  to  nil  at  B, 
then 

2  2 


MOSELEYS  PRISCIPLE.  897 

4.  Moseley's  Principle.-If  the  forces,  which  act  upon  or  within 
a  body  or  structure,  are  in  e,,uilil,num  they  may  be  classified  as 
active  anc  paj.rve  forces,  standing  to  each  other  in  the  relation  of 
cause  and  effect.  The  passive  forces  are  then  the  hast  which  are 
capable  of  balancing  the  active  forces  consistently  with  the  physical 
condition  of  the  body  or  structure. 

For  the  passive  forces  are  due  to  the  application  of  the  active 
forces  to  the  structure  and  do  not  increase  after  they  have  balanced 
the  active  forces.  They  mil,  cns^-quently,  not  increase  beyond  the 
smallest  amount  capable  of  balancing  the  active  forces.  It  may 
therefore,  be  concluded  that  as  the  force  which  one  member  of  a 
structure  exerts  on  another  is  a  minimum  with  any  given  specified 
loading,  then  the  horizontal  component  of  the  thrust  in  the  arch 
nng  must  be  a  minimum  with  that  loading,  and  hence  the 
me  of  resistance  m\\  be  that  which,  consistent  with  stability  gives 
the  horizontal  component  a  minimum  value. 

To  draw  a  force  polygon  through  any  three  points  K,  L,  and  M 


Fio.  901. 


Fio.  go2. 


Pint.  Draw  any  force  polygon  with  a  closing-line  km,  0  beimr 
the  pole  and  OJ  parallel  to  Arm.  ^  ,  w  oemg 

.Seronrf  Draw  70'  parallel  t  .  KM  and  with  pole  (T  draw  a  new 
Ix)ly«on  which  will  ,^  through  a  ,K.int  P.  It  will  not  rgJnera^ 
l««  through  L,  but  will  be  above  or  below  this  ,H,i„t.  Jf  Tl^v^ 
increase  the  polar  di«tan«.  in  tlie  proportion  of  -^,  and  take  a 


11 


i  4'' 


898 


THEORY  OF  STRUCTURES. 


new  jrole  0"  at  the  proper  horizontal  distance  from  the  line  of  loads^ 
alonp  the  line  JO'. 

Of  course  it  is  only  necessary  to  draw  the  first  trial  polygon. 

5.  Fuller's  Method. -The  following  is  the  Fuller  method  of 
drawing  the  line  of  least  resistance  for  an  unsymmetrically  loaded 

arch: 

I^^t  ir,,  ir2.  .  .  .H'9  l>f^  the  loads  on  the  several  voussoirs.  Draw 
the  load-line  and  also  the  corres|)onding  funicular  polygon  Jklmno. 


•»!     W|    W|    <%    W^ 


Fio.  904. 


Fig.  903. 


From  the  point  n  draw  any  line  oZ  intersecting  the  l)a8c-line  in  Z. 
Let  horizontal  lines  through  k\  I,  m  .  .  .  intersect  the  line  oZ,  and 
from  the  jH)ints  .,f  intersection  drop  the  verticals  I,  II,  III      .  . 

From  the  |)oints  (»f  mtersection  of  the  lines  of  action  of  W\,  IFj,  \\\ 
.  .  .  with  the  upixT  and  lower  lK)undaries  of  the  middle  third,  draw 
horizontals  intersecting  the  corresjKinding  verticals  I,  II,  III,  etc. 

Join  the  |K)ints  so  obtained  and  complete  the  irregular  figure 
ir.Vr^.  Within  the  area  WXYZ  select  a  straight  line  iZ'  contain- 
ing the  smallest  po-ssihle  angle  at  x— and  contained  wholly  within 
the  iMumdaries  of  the  area  ir.Vl'^. 

The  straight  line  xZ'  Itears  the  same  r -lations  to  the  [xilygon 
nHiuire«l  t<»  Ik'  drawii  within  the  centnil  third  of  the  arch  ring  iu» 
the  straight   litn'  oZ  d<H'S  to  the  iM)lygon  Jkhnno. 

From  the  ixnnfs  «>f  intersection  «»f  the  line  Z'x  with  the  verticals 
I,  II,  etc.,  <lraw  liorizohlal  lines  inters<'cting  the  lines  of  action  of 
H'l,  II2,  "'.1  •  •  •  Hy  joining  the  jKiinls  thus  obtained,  the  reciuircd 
j)olygon  Ijing  within  the  central  third  of  the  ureli  ring  is  drawa. 


FULLER'S  METHOD. 

zontal  thrust  of  tho  arch  required  to  meet  all  the  conditions 

.f  It  IS  not  found  passible  to  draw  any  straight  line  within  the 
lx,.mdanes  o     the  area  WXYZ  intersecting  the  middle  third  1  ne 

t^^^  ''"  ''-'  '-'  '"-^  ^  ^— •  -"'  t'>i^  -- 

The  alK,ve  method  is  of  course  equally  applicable  to  an  arch 

^.th  a  symmetrical  load,  and  it  avoids  the  neces.sitv  of  .Ira^n^ 

repeated  tnal  hnes  of  least  resistance,  which  is  verv  tedioui  ^ 

6.  Examples  of  Linear  Arches  or  Curves  of  Resistance.-Li„«r 

Arch  ,«  the  Form  of  a  Catenary.     If  the  cable  in  Art.  4,  ChapIS^ 

Case  A,  is  inverted  and  stiffene'»  so  as  to  ' 

resist  distortion,  a  linear  arch  ;^    .btained  »      t        q  „ 

smtuble  for  a  real  arch  which  has  to  suj)-  |        Vf-^-^" 

j)ort  a  load  distributed  in  such  a  manner 

that  the  weight  \i\m\  any  jKirtion  .!/>  is 

proportional  to  the  length  of  .!/>,  and  is^' 

in   fact  =  /).s   the  area  0.\PX  Ijoing  w.s. 

Thas,  a  lamina  of  thickness  unity  and 


Fio.  905. 


ne  ur  icals  AG  /.\.  weighs  ,m.v,  and  may  l)e  taken  to  represent 
the  load  upon  the  arch  if  „.,«.v  =  ;.,  1.,,,  if  .,«  =  ;,,  i,.,  jf  Hu- weS 
of  m  units  of  the  lamina  is  w  ^ 

cur.'^tuittw  M  ''"■"^'  "\  ''''  -"-"  =  '^-"-'  =  "y.  the  ra.li,H  of 
cur\ature  (.«)  at  the  crowii  being  ecjual  to  m 

A  .lisadvantage  attached  to  a  linear  arch  in  the  form  of  a  catenarv 

.OS  in  the  fact  that  only  one  catenary  can  ,.»ss  through  tuT^^ 

Hunts    while,  m  practice,  it  is  often  necessary  that  a^i  arch^hlil 

|m..s  through  three  jK.ints  in  order  to  me<.t  the  requin-nien     o    a 

«.ven  rise  ami  span.     This  difHculty  may  be  obviald  bv  t  <«  L^ 

"f  the  tranxfornml  catenary.  ' 

l>)n  the  lamina  /M/'A.Va.  W  erect  a  solid,  mth  it.  horizontal 
-Cion-s  al  the  same,  ami,  for  simplicity,  with  its  gi-n.-rati  g  M  ^ 
|»er|»r'iidicular  to  the  Imw.  *^    '"^ 

rut  this  s,.lid  by  a  plane  through  .V.V  incline,!  at  any  rr.r.ire,I 
-.gle  to  ,hH«.....  The  int.n..ction  of  t}„.  plane  and  solid  wil  2 
a  iran.lor,ned  catenary  I^AT'.  or  u  new  linear  arch,  an.l  t  «    ha^ 


;  M 


M 


i 
f 

i 


ill 


wo 


THEORY  OF  STRICTURES. 


of  a  uew  lamina  P'A'P'W,  under  which  the  arch  will  he  balanced. 
This  is  evident,  as  the  new  arch  and  laiuina  are  merely  parallel  pro- 
jections of  the  original. 

The  projections  of  horizontal  lines  will  remain  the  same  in  length. 
The  projection  of  the  vertical  Unes  will  be  c  times  the  length 
of  the  lines  from  which  they  are  projected,  c  being  the 
secant  of  the  angle  0  made  by  the  cutting-plane  with 
the  base.    Thus 

.  OA'   ;  - 

c=8ec  I 


Fio    906. 


OA 


Let  I,  Y  be  the  co-ordinates  of  any  point  P  of  the  transformed 
catenary; 
"   X,  y  bo  the  co-ordinates  of  the  corresponding  point  P  in  the 

catenary  proper; 
"   A'0  =  A/(>m). 
Then 

Y    FN  A'O    M 


y'  PS''^     AO     m' 


The  equation  to  the  catenary  proper  is 


m/  i 


Therefore 


%  ■ 


(1) 

(2) 
(3) 


which  is  the  equation  to  the  transformed  catenary. 

With  this  form  of  linear  arch  the  <lepths  ^f  over  the  crown  and 
Y  over  the  springings,  for  a  ."pan  '2x,  may  Ix*  lussumeil,  and  the  cor- 
re8|X)nding  value  of  m  determined  from  vq.  (3),  which  may  be  more 
conveniently  written  in  the  form 


log. 


Nr^-4 


in       -  iM" 

Tlie  slope  t'  at  P*  Is  given  by 

^    dY     M  {  '-      -^\     Ms 


(4) 


\   i? 


TRA  NSFORMED  CA  TEN  A  RY. 


901 


s  being  the  length  AP  of  the  catenary  proper,  corresponding  to  the 
length  A'P'  of  the  transformed  catenary. 


The  area  OA'FN^  fjdx^^(Ji-e-^^ 


'Ms. 


The  tangents  at  P  and  P'  necessarily  intersect  in  the  point  T, 
and  P'TN,  PTN  are  triangles  of  forces  for  the  portion  AP*  and  AP, 
respectively. 

Let  H',  H  be  the  horizontal  thrusts  &t  A'  snnX  A,  respectively. 

V 

\         P',P  "     "  weights  ujKin  AT  and  AP,  respectively. 
R'        *'    "  tangential  thrust  at  P*. 


Then 


\ 


and  therefore 
Also, 

and 


F    areaOyl'PA-    Ms    M 
P  "  area  OAP.\  ~  nis  ~m' 

P'  =  — P  =  —wm.«  =  wMs. 
mm 


W-F  cot  i'  -  wMs^  =  wm»  -  H. 
Ma  ' 

R'-H'8ect'-urm"^'l  +  — j--ir\/m«+lPta. 


f    .    i 


The  radius  of  curvature  p'  at  the  crown -tt.    Therefore 

H'-wAfp'-'H-wp, 

and  the  radius  of  the  "catenary  proper"  is  Af  times  the  radius  of 
the  transformed  catenary. 

The  t.  1  "equilibrated  arch"  has  generally  been  applied  to  a 
linear  arch  with  a  horitontal  extrados. 

Ex.  2.  Determine  the  tmnnformwi  eatmarif  /or  an  arch  of  30  //.  ipan  and 
22i  /(.  rite,  the  height  o/  manonry  over  the  croum  being  13J  //.;  weight  nf  the  masonrjf 
- 125  lb».  per  cubic  foot.  AUo  find  the  thrtut  at  the  springing  and  the  curvatw 
at  the  crown  and  the  tpringing. 


Ill^'i 


902 


THEORY  OF  STRUCTURES. 


r-22i+13i-36';    M-13J';    ^->og.(|  +  ^(|)'-l)  -1.6369; 

15  _li 

therefore    m  =  9.1637,    c"-5.1387    and    e    •"-.1946. 


Hence 


Also 


s  -  ilP  -  ^^^(5.13.87  -  .1946)  -  22'.653. 


Fio.  907. 


.,    Ms     13ix  22.6.53    „«,,„         .     ..    ..^«„^ 
'^'-^—kmiy-'^-^^^    '^^    i'-74».39'; 

//' _u,TO»  =  125 ;9. 16.37)'=  10,497  lbs.; 
fi' -«' sec  i'- 39,6.53  lbs. 

At  the  crown     />'  -  "w  -  6-22  ft. 
Again,    r-6i(^^^^-^). 


The,^fore  ^-'^(i      -^)     and    ^'-«i{  ^^  -^). 


Hence  at  the  springitig 


dV    6ix  4.9441 
dx "     9.163 


^i»-3.6418.     (S)'-13.263. 


dnr   6^x■5.3.^^3 

dx'       (9.1637)'  ■•■*^'»'* 


Therefore  a/  tA«  springing, 

.'-^-i^t^^  125.65  ft. 

(6)  /.iVar  y4reA  tn  Me  Form  o/  a  Parabola. — Suppose  that  the  cable  In 
Art.  4,  Chap.  XI,  Case  C,  is  exactly  inverted,  and  that  it  is  stiffened  in  such 
•  manner  as  to  resist  distortion.  Suppose  also  that  the  load  still  remains 
a  uniformly  distributed  weight  of  intensity  w  per  horizontal  unit  of  length. 
A  thrwit  will  now  be  dovelo|)ed  at  every  point  of  the  inverted  cable  equal  to 
the  Ipiision  at  the  corresponding  point  of  the  original  cable.  Thus  the  inverted 
paralxilii  is  a  linear  nrch  suitable  for  a  real  arch  which  has  to  support  a  load 
of  iiitetmity  w  per  horizontal  unit  of  length. 

The  horizontal  thrust  at  the  crown  -  //  -  v^, 

^  being  the  radius  of  curvature  at  the  crown. 


CIRCLLAli  AND  ELLIPTIC  ARCHES. 


903 


Fio.  908. 


(c)  Circular   and  EUiptic  Linear  Arches— X  linear    arch   which   has  to 
support  an  external  normal  pressure  of  uniform  intenaity  should   be  circular. 

Consider  an  indefinitely  small  element 
CD,  which  may  be  assumed  to  be  approx- 
imately straight. 

Let  the  direction  of  the  resultant  pressure 
upon  CD,  viz.,  pCD,  make  an  angle  d  with 
OB. 

Let  CE,  DE  be  the  vertical  and  horizontal 
projections  of  CD. 

The  angle  DCE-0. 

The  horizontal  component  o{  pCD  =  pCDcoa0-pCE. 

This  is  distributed  over  the  vertical  projection  CE,  and  the  horizontal 
intensity  of  pressure -=pCJ?-fC£- p. 

Similarly,  it  may  be  shown  that  the  vertical  intensity  of  pressure -p. 

Thus,  at  any  point  of  the  arch, 
the  horizontal  intensity  of  pressure  "  vertical  intensity  -  normal  intensity -p. 

Again,  the  total  horizontal  pressure  on  one  half  of  the  arch 

-  S(p.CE)  -pl(CE)  -pr-H, 
and  the  total  vertical  pressure  on  one  half  of  the  arch 


-  £{p-DE)-pI(DE)  -pr-P. 

Hence  at  any  point  of  the  arch  the  tangential  thrust -ft. 

Next,  upon  the  semicircle  as  base  erect  a  semi-cylinder.  Cut  the  latter 
by  an  inclined  plane  drawn  through  a  line  in  the  plane  of  the  base  parallel  to 
OA.    The  intersection  of  the  cutting-plane  and  the  semi-cylinder  ia  the  semi- 


Fio.  909. 


ellipse  B'AB',  in  which  the  vertical  lines  are  unohanp>d  in  length,  while  the 
lengths  of  the  horizontal  lines  are  c  timiw  the  lengths  of  the  corresponding  lines 
in  the  semicircle,  c  being  the  secant  of  the  angle  made  by  cutting-plane  with  the 
base.  A  semi-elliptic  arch  is  thus  obtained,  and  the  forces  to  which  it  is  sub- 
jei'ted  are  }>arallel  projections  of  the  fnrfT*  acting  upon  the  semicircular  arch. 

These  new  forces  are  in  equilibrium. 

I*t  P-the  total  vertical  pressure  upon  one  half  of  the  arch; 
//'-the  total  horiiontal  pressure  upon  one  half  of  the  arch; 


■  i 

i     ■■     I 


ii^ 


904 


THEORY  OF  STRUCTURES. 


pV- vertical  intensity  of  pressure  =^Tg-,; 


H' 


pi  — horizontal  intensity  of  pressure  =  ^. 


Then 


p.        p 


Pi' OB' 


PI 
cr 


cOB 
H'-cH-cP~cP'; 

P'"0.4'~0.l  "  r 


P.. 

c' 


•cp. 


(1) 

(2) 
(3) 
(4) 


Henc«,  by  eq.  (3), 


HI 
P' 


OB' 
OA 


or  the  total  horieontal  and  vertical  thrusts  are  in  the  ratio  of  the  axes  to  which 
they  are  respectively  parallel,  and,  by  eqs.  (2)  and  (4), 

f^" c'"  OB'" 

or  the  vertical  and  horizontal  intensities  nj  pressure  are  in  the  ratio  of  the  squares 
of  the  axes  Ut  which  they  are  respectively  parallel. 

Any  two  rertangular  axes  OG,  OK  in  the  circle  will  project  into  a  pair 
of  conjugate  radii  0(t',  OK'  in  the  ellipse. 
Let  OG'-r„  OK' -r,; 

Q- total  thrust  along  elliptic  arch  at  K-; 
p  „   "        "        "         "        "     "  G. 


Then 


H 


H_ 
K 


=  -,    and 


9. 

R 


or  the  total  thrusts  along  an  elliptic  arch  at  the  extremitiet  of  a  pair  of  conjugate 
radii  are  tn  the  ratio  of  the  radii  U<  which  they  are  respectively  panUlel. 

The  preceding  results  show  that  an  elliptic  linear  arch  is  suitable  for  a 
load  distributed  in  such  a  manner  that  the  vertical  and  horizontal  intensities, 
eqs.  (2)  and  (4),  at  any  |K>int  of  the  arch  are  unequal,  but  are  uniform  in 
direction  and  maf^nitude. 

Again,  it  can  *  ea!<ily  shown  that  the  projected  forces  acting  upon  the 
elliptic  arch  are  in  equilibrium. 

The  equations  of  equilibrium  for  the  forces  acting  upon  the  circular  arch 
may  be  written 

d(l^)+Xds'0, 


i{ifl^y<is-o. 


I 


HYDROSTATIC  ARCH. 


90& 


T  being  the  thrust  along  the  arch  at  the  point  xy,  and  X,  Y  the  forces  acting 
upon  the  arch  parallel  to  the  axes  of  x  and  y  respectively. 
If  T',  X',  Y'  be  the  corresponding  projected  forces, 

~-~,    Xds-cX'da',    Yds-YW, 
as'     da 

Hence  the  above  equations  may  be  written 

d(^^cdx'\  +cX'da'-0, 


and 


or 


and 


+  Y'ds'-O. 


Hence  the  forces  T',  X',  and  Y'  are  also  in  equilibrium 

(d)  Hydrostatic  Arch. — Let  the  figure  represent  a  portion  of  a  linear  arclt 
suited  to  support  a  load  which  will  induce  on  it  a  normal  pressure  at  every 
point.     The  pressure  being  normal  has  no  tangential 
component,  and  the  thrust  (T)  along  the  arch  must 
therefore  be  everywhere  the  same. 

Consider  any  indefinitely  small  element  CD. 

It  is  kept  in  equilibrium  by  the  e(|ual  thrusts  {T) 
at  the  extremities  C  and  D,  and  by  the  pressure 
p-CD.  The  intensity  of  pressure  p  being  assumed 
uniform  for  the  element  CD,  the  line  of  action  of  the 
pressure  pCD  bisects  CD  at  right  angles. 

Let  the  normals  at  C  and  D  meet  in  0„  the  centre 
of  curvature. 

Take  Ofi-O^D-p,  and  the  angle  C0,D-'2JS. 

Resolving  along  the  bisector  of  the  angle  CO,D, 


910. 


\:,i\ 


or 

and  hence 


2T  sin  J0-pCD-pp-  240, 
2TJ0-pp2JO; 
T-pp-a  conctant  . 


(1> 


Th»w,  a  wrica  nf  mjrvoa  may  hr  obtained!  in  which  p  varies  inversely  as  p, 
and  the  hydrostatic  arch  is  that  curve  for  which  the  pressure  p  at  any  point 
is  directly  proportional  to  the  depth  of  the  poitU  bdow  a  given  horitonlal  plane. 

Denote  the  depth  by  y,  and  let  w  be  the  specific  weight  of  the  8ubstano» 


906  THEORY  OF  STRUCTURES. 

to  which  the  pressure  p  is  due.    Then 

p=wy, ••(2) 

and  T=pp=wyp=*  comUnl (3) 

The  curve  may  be  delineated  by  means  of  the  equation 

yp= const W 

It  maybe  shown,  precisely  as  in  case  (c),  that  the  horizontal  intensity  of 
pressure  (p^) 

-the  vertical  intensity  (;)»)=  p (5) 

Take  as  the  oripn  of  coordinates  the  iwint  0,  Fig.  911,  vertically  above 
the  crown  of  the  arch  in  the  given  horizontal  i)lane. 
Let  the  horizontal  litie  through  0  l)e  the  axis  of  j; 

"    vertical       "  "        "  ?/. 

Any  pcjrtion  AM  of  the  arch  is  kept  in  equilibrium  by  the  equal  thrusts 

(T)  at  A  and  M,  and  by  the  resultant  load  P  u|K)n  AM,  which  must  necessarily 

O  act  in  a  direction  bisecting  the  angle  ANM. 

i  Complete  the  parallelogram  AM,  and  take 

N  '<_S!a  _,^^^  SN-'NM  to  represent   T. 

/C''^C'/  """"^N.  The  diagonal  "NL  will  therefore  represent 

j^  L  •        Let  0  be  the  inclination  of  the  tangent  at 

Pjq    gjj  3/  to  the  horizontal. 

The  verticalload  upon  ^M— vertical  com- 
ponent of  P 

-  LK  -  r  sin  5  -  p/>  sin  fl  -  v:yp  sin  0  -  viyj)^  sin  0,        ...    (6) 

j/o,  p^  Ixiing  the  values  of  y,  p,  respectively,  at  A. 

The  horizontal  load  upon  A.W- horizontal  component  of  P 

-  jv/c  -  5iv  -  A's = r  -  r  cos  0  -  2r /sin  ^y 

-2pp/8in-j  -2u'i//>^sin  jj  -2u'3/opj8in-j  .       .    .    (7) 
Again,  the  vertical  load  upon  AM 

"  I   pdx'-wj  ydx-wy^oSind; (8) 

•'o  *^o 

the  horijiontAl  load  upon  AM 

/y  fy  w  /      5  \' 

pdy-wj  ydy--^(y^-yo')-2wy^J9in-]  .    .     .    .    (9) 


I':. 


.S    '1 


1113 


GEOSTATIC  ARCH. 


907 


Eq.  (8)  also  shows  that  the  area  bounded  by  the  curve  AM,  the 
verticals  through  M  and  A,  and  the  horizontal  through  0  ia  equal  to  y^„  sin  d 
and  IS  tliorefore  proportional  to  sin  0.  At  the  points  defined  by  0  =  90°  the 
tangents  to  the  arch  are  vortical,  and  the  portion  of  the  arch  between  these 
tangents  is  alone  available  for  supix)rting  a  load.  The  vertical  and  horizontal 
loads  upon  one  half  the  arch  are  each  equal  to  wy^,. 

The  relation  given  in  eq.  (1)  holds  true  in  any  arch  for  elements  uion 
which  the  pressure  is  wholly  normal. 

This  ha.s  been  already  proved  for  the  parabola  and  catenary,  m  cases 
(a)  and  (6). 

At  the  point  A'  of  the  elliptic  arch, 


P  = 


OB" 
OA' 


Hence,  the  horizontal  thrust  at  A' 


mil 


~PwP='-p-pcr-cH. 

(e)  Geostatic  Arch.— The  geostalic  arch,  Fig.  912,  ia  a  paraUel  projection  of 
the  hydrostatic  arch. 

The  vertical  forces  and  the  lengths  of  vertical  lines  are  unchanged. 

The  horizontal  forces  and  lengths  of  horizontal  lines 
are  changed  in  a  given  ratio  c  to  1. 

Let   B'A   be  the   half-geostatic   curve   deri\(d  from 
the  half-hydrostatic  curve  BA. 

The  vertical  load  on  AB' 


(1) 


"P""  P=  thrust  along  arch  at  B'.      . 
The  horizontal  load  on  AB' 

-ff'—c//-=  thrust  along  arch  at  A 
The  new  vertical  intensity 


Fio.  912. 


—Jo 


-Pi/'---- 


-^— ^=Pi_P 


OB'    cOB     c     c 
The  new  horizontal  intensity 

,     H'     cH 

'^''dA'n.r'P'-'P- 


(2) 


(3) 


(4) 


Thus,  the  geostatic  arch  is  s  lited  to  support  a  load  so  distributed  as  to 
produce  at  any  point  a  pair  of  conjugate  pressures;  pressures,  in  fact,  similar 
to  those  developed  according  to  the  theory  of  earth-work. 


I   1 


\:\ 


'   I 


908 


THEORY  OF  STRUCTURES. 


Let  Ri,  R,  be  the  radii  of  curvature  of  the  geostatic  arch  at  the  points  .4, 
B',  respectively,  and  let  n,  r,  be  the  radii  of  curvature  at  the  corresponding 
points  .4,  B  of  the  hydrostatic  arch. 

The  load  is  wholly  normal  at  A  and  B'.    Thus, 

H'-py'R.-^^R.-cH^cpr, (5) 

c 

and  /?,=-cV, (6> 

Also,  F  =  p/R,'cpR,''P^pr (7) 

and  cR.,  =  r^ (8) 


(/)  General  Case. — Ix't  Fig.  913  rpprp.«!ent  any  linear  arch  suited 
to  support  a  load  which  is  symmetrically  distributed  wth  respect 

p       to  the  crowii  -1,  and  wliich  irroduces  at  every 

jKiint  of  the  arch  a  pair  of  conjugate  pressures, 
the  one  horizontal  and  the  other  vertical. 

Take  as  the  axis  of  y  the  vertical  through 

the  crowTi,  ami  tus  the  axis  of  x  the  hoiizontal 

through  an  origin  0  at  a  given  distance  fron;  A. 

Any  jxjrtion  A.\f  of  the  arch  is  kept  in 

Fio.  913.  equilibrium  by  the  horizontal  thrust  H  at  A, 

the  tangential  thrust   T  at  M,  and  the  resultant  loatl  u|X)n  AAfy 

which  mast  necessarily  act  through  the  point  of  intersect: 'H  iV  of 

the  lines  of  action  of  H  and  T. 

Since  the  load  at  A  is  wholly  vertical,  H  at  A  is  given  by 


Ho  =  PoPo, 


(1) 


po  and  po  being,  respectively,  the  vertical  intensity  of  pressure  and 
the  radius  of  curvature  at  A. 

Let  MX  =  T,  and  take  XS^Ho- 

Complete  the  parallelogram  SM;  the  diagonal  XL  is  the  resultant 
load  u|X)n  AM  in  direction  and  magnitude. 

The  vertical  (KL)  and  the  horizontal  (KX)  projections  of  XL 
are,  therefore,  respectively,  the  vertical  and  horizontal  loads  upon 
AM. 


GESERAL  CASE.  909 

Denote  the  vertical  loatl  by  V,  the  horizontal  by  H.    Then 

T sin  d^KL='V^fJj,^x, (2) 

and    H  =  KN  =  SN-SK  =  Ho-V  cot  0,    ...    (3) 
0  being  the  angle  between  AfN  and  the  horizon. 

dV 
p^,  the  vertical  intensity  of  pressure  =  ;7-.    ...     (4) 

p„  the  horizontal  intensity  of  pressure 

^d^^-d;/^'"^''^ (5) 

Ex.  3.  A  semicircular  arch  of  radius  r,  n'ith  a  horizontal  extradoa  at  a  vertical 
■distance  R  from  the  centre. 

The  angle  between  the  radius  to  .V  and  the  vertioal-O.    Then, 

i-rsinO,    yR-rcosO      ....        .  (1) 

dx-rcosOdO    dy-rainOdO (2) 

Pf-wy-w(R-r  cos  6) (3) 

w  being  the  specific  weight  of  the  load.    Hence 

V-wf  (R —r cm fi)r COB dd9 
Jo 

/„  .     -     rO       rsin29\ 
-u-r^ftsinff-- ^j.    .....    (4) 

Eqs.  (3)  and  (4)  give  H;   to- 

p,~w{R-r), (5) 

and  hence  H^-ivr(R-r) (6) 

Pxi  the  horizontal  intensity  of  pressure 

*'/i/     i  o\        /d    r  9 -sin  0  cos  9  \ 

--  ~(Vcot9)~w[R---~-  ^.^-^^  ^-rcose).   ...    (7) 

Rankine  gives  the  following  method  of  determining  whether 
a  linear  arch  may  be  adopted  .a.-?  the  intra<!os  of  a  real  arch.  At  the 
crown  a  of  a  linear  arch  ab  measure  on  the  normal  a  length  ac,  so 


&    A  ■■ 


1  i  I!  i 


i  ■    SI 

I.;  ?■: 


I  ■. 


il 


I  .1. 

li 


910 


THEORY  OF  STRICTURES. 


that  c  may  fall  within  the  limits  ro(]uirpil  for  stability  (o.g.,  within 
tho  middle  third). 

At  r  two  equal  and  opposite  forces,  of  the  same  magnitude  aw 
the  horizontal  thrust  //  at  n,  and  acting  at  riglrt  angles  to  ac,  may 
b«>  introduced  without  altering  the  e<|uilibrium. 

Thus  the  thrust  at  a  is  replaced  '  y  an  eijual  thrust  at  c,  and  a 
right-handed  couple  of  moment  Ilxnr. 

Similarly,  the  tangential  thrust  T  at  any  point  d  of  ah  may  Ix' 
replaced  by  an  e<iual  and  parallel  thriL«t  at  e.  and  a  couple  of  moment 
TxJe. 

The  arch  will  Ik-  stable  if  the  length  of  de,  which  is  normal  to 
ah  at  d,  is  fixed  by  the  condition  Txde=Hxnc,  and  if  the  line  whic'i 
is  the  locas  of  e  falls  within  a  certain  area  (e.g..  within  the  midtlle 
third  of  the  arch  ring.) 

7.  Arched  Ribs  in  Iron,  Steel,  or  Timber.— The  term  arched  ril» 
is  ap|)lied  to  arches  constructed  of  iron,  stei-l,  or  timlx-r.  The  coef- 
ficients of  el.'Lxticity  are  (juantities  which  are  found  to  lie  l)etw(>eii 
certain  not  very  wide  limits,  and  their  values  may  lie  introduced 
into  the  calculations  with  the  result  of  giving  to  them  gn'ater  accuracy. 
There  are  other  considerations,  however,  involved  in  the  problem 
of  the  stability  of  arched  ribs  which  still  n-nder  its  solution  more 
or  less  indeterminate. 

'i'he  total  stress  at  any  [Kiint  is  made  up  of  a  nunjl)er  of  si'beidiary 
stresses,  (,f  which  the  most  ini|Hirtant  are.  (1)  a  direct  compreafiive 
stress;  (2)  a  stress  due  to  (le.xure;  (.{j  a  stress  duo  to  a  change  of 
temiK'ratnre  IJich  of  these  may  l»e  investigated  seiwrateiy,  and 
the  results  su|HT|>«)s<'d. 

8.  Rib  with   Hinged  Ends;    InvariabUity  of   Span.-Ut  ABC  be 


Fi<i    911 


thr"  axis  of  a  rib  supinirled  at  the  ends  on  pins  or  on  cylindrical 
bearings.     The  resultant  thrusts  at  .1  ami  ('  must  necesBarily  pius 


RIB  WITH  HINQED  ENDS. 


911 


through  tho  pontrps  of  rotation.  The  vertical  components  of  the 
thrusts  are  equal  to  the  correspomUng  reactions  at  the  ends  of  a 
girder  of  tiie  same  span  and  similarly  loaded,  and  H  is  to  be 
determined  as  in  Art.  1  when  DE  has  Ix'en  found. 

I^t  A IX'  be  the  linear  arch  for  any  arbitrary  distribution  of 
the  load,  and  let  it  intersect  the  axis  of  the  rib  at  S.  The  curvature 
of  the  more  heavily  loaded  portion  AES  will  be  flattened,  while  that 
of  the  remairnler  will  be  sharpened. 

The  l)ending  moment  at  any  point  E  of  the  axis  tends  to  change 
the  inclination  of  the  rib  at  that  point. 

I^t  the  vertical  through  E  intersect  the  linear  arch  in  D  and 
the  horizontal  through  A  in  F. 

leX  0  be  the  inclination  of  the  tangent  at  E  to  the  horizontal. 

Let  /  be  the  moment  of  inertia  of  the  section  of  the  rib  at  E. 

Let  ds  be  an  element  of  the  axis  at  E. 

Change  of  mchnation  at  E  =  dO=  „,      — ^i — . 

El  El 

If  this  change  of  curvature  were  effected  by  causing  the  whole 

curve  on  the  left  of  E  to  turn  about  E  through  an  angle  dd,  the 

horizontal  ilisplacement  of  ..1  wouUl  be 

AA' ~  AL  co^  LAA'^AEdd  sin  EAF-AEdd^ 

AE 

-EF  dO^~I)EEFdH. 

This  is  evidently  e(|ual  to  the  horizontal  displacement  of  E,  and 
the  algebraic  sum  of  the  horizontal  di.-*placement8  of  all  points  along 
the  axis  is 


t,II  DE  EF  ds 

— Er~  ' 


fHDEEFd.'i 
J  Ei =0'  •     •    (1) 


since  the  length  AC  is  assumed  to  l)p  invariable. 

Thus,  the  actual  linear  arch  must  fulfil  the  condition  expressed 
by  eo.  (I),  which  may  Ik-  written 


/ 


DE  EF  ds 
1 »» 


(2) 


Binc<«  //  and  E  are  constant. 


ill 
1 


912 


THEORY  OF  STRUCTURES. 


If  tlio  rib  is  of  uniform  srction,  /  is  also  constant,  and  eq.  (2) 
becomes 


y^DE  EF  d8  =  o. 


(3) 


Also,  since  DE  is  the  difference  Ix'twiM'ii  DF  and  EF, 
/(EF~EF)EF  d8  =  o  =  yDF  EF  d»-/EF2<U. 


(4) 


It  must  lx»  rememlxTfd  that  tlie  integration  ext«'nds  from  end  to 
end  of  the  arch. 

Va\.  1  expresses  the  fact  tliat  th"  sfxm  remains  invariable  when 
a  series  of  iM'iiding  moments,  Hf)E,  act  at  |)oints  along  the  rib. 
Thes«'.  however,  are  accompanied  by  a  thrust  along  the  arch,  and 
the  axis  of  the  rib  varies  in  length  with  the  variation  of  thrust. 

Let  //o  Im-  the  horizontal  thru.>*t  for  that  symmetrical  loading 
which  makes  the  linear  arch  coincitic  with  the  axis  of  the  rib. 

Ix't   To  •»<>  the  corres|)onding  thrast  along  the  rib  at  E. 

The  shortening  of  the  element  ds  at  E  of  unit  section 


T~To 


E 


V/.V. 


liCt   a  numlM'.   of  weights  M'l.  M'^,  H's 
different  points  along  the  arched  rib. 


be  concentrated  at 


Fio.  01  .V 


Kki   01  tl. 


Take  1234  .  .   .   n  as  the  line   of  loads,    ll'i   l)eirig  represented 
by  12,  \y 2  by  23,  11*3  by  3t,  etc.,  and  let  the  segments  \r,  nx,  reH|)ec- 


(2) 
(3) 


RIB  WITH  HI  SO  ED  ENDS. 


013 


tively,  rcpK'sont  the  vertical  n>action  at  A  and  C.    Take  the  hori- 
zontal length  xP  to  represent  H,  and  draw  the  radial  lines  Pi,  P2, 

The  equilibrium  polygon  Agig^g,  .  .  .  mu«f  be  the  funicular 
polygon  of  the  forces  with  respect  to  the  pole  P,  and  therefore  the 
directions  of  the  resultant  thrusts  from  A  to  Ei,  Ex  to  E2,  E2  to  E^, 
...  are  respectively  \rtaa\M  to  P\ ,  1%  f%  .  .  . 

The  tangential  (axial)  thrust  and  shear  at  any  point  p  of  the  rib, 
e.g.,  iM-tween  E2  and  E3,  may  lie  easily  found  by  drawing  Pt  parallel 
to  the  tangent  at  />,  an.l  3/  iKTix'n.lieul.ir  to  Pt.  The  direct  tan- 
gential thrust  is  evidently  repres^-nt.-.!  by  /'/,  and  the  normal  shear 
at  the  same  point  by  3/.     The  latter  is  lK)rne  by  the  web. 

If  /*  is  i;  ixiint  at  wliieh  a  weight  is  concentrated,  e.g.,  E^,  draw 
Pt'r  parallel  to  the  tanginit  at   E,,  and  51',  Of  perpendicular  to 

/Y  r..present«  the  axial  thrust  immediately  on  the  left  of  Es, 
and  .')/'  th..  c(.rresiH)nding  normal  sh«'ar,  while  /'/"  represents  the 
axial  thrust  im!Me<liately  on  the  right  .>f  A'j  and  &"  the  correspond- 
ing normal  shear. 

A  vertical  line  through  /'  can  only  meet  the  lin,.  of  |„ad«  ^t 
infinity. 

Thus,  it  would  reciuin-  the  loads  at  .1  and  T  to  be  infinitely  great 
in  order  that  the  thrusts  at  these  |K)intM  might  Im^  vertical.  Prac- 
tically, no  linear  arch  will  even  approximateiv  coinci,!..  with  the 
axis  of  a  rib  rising  vertically  .it  the  springir,,^^.  an,l  |i,.„ce  neitl^r 
a  wiiucircular  nor  a  .s.-mi -elliptical  axis  is  to  U-  recommen.led. 

Kx.  4.  r.rt  the  ax^  <»/  a  nh  or  uniform  section  and  hin^^d  at  b,4h  end»  be 
a  gemmrtl,'  of  ntdiux  r. 

Ui  a  «..«]..  UTi^ht  W  iM.  pla,.c..l  u,«r<  the  ril,  .U  a  point  «ho«  horiiontal 
.liMtjiti.v  from  (t.  \\w  <..itn'  of  tlio  (4|)iui.  iv  ,;. 

/>m"o<'*"""  '*"^*  ^'"^  he,uthH,.mo,un<t  d,„.„„m)  ...twuiu  of  two  •Uaigbt  UoM 

Draw  any  verti.al  line  ini.TWM-Jinjt  th.'  axis,  the  ° 

linear  anh  and  the  aprinKinc-line  AC  in  A",  ly,  F' 


>c»|mitivdy. 


liCt  OF'-T,  ami  U-t  dx  \w  the  \v 


ii'rti 


upon  AC  of  the  eletncnt  d»  at  E 
Then 


(riiontal  projec- 


dt 


••cofec  y.'  )F' 


•i^i,    or    E'F'da-r-dx. 


it 


Ui 


Fig.  91T. 


CJU  THEORY  OF  STKUCTURES. 

Apphing  condition  (4), 

/*'  lyP'Tdx^  f  ryF'rdx-^  f  E'Prdx, 

or  f  lyP'dx  +  fo'F'dx  -  f_  E'F'dx, 

or  area  of  triangle  ADC  =arca  of  semicircle. 

Therefore,  if  z  be  the  vertical  distance  of  D  from  AC, 

>  _ 

one-half  of  lencth  of  rib. 


„._     or    Z-- 


Therefore 


DE  -DF-E'^-'j-  v^;^^', 


ami  if  h  is  the  horizontal  thnist  on  the  arch  due  to  W, 

2r 


h    DF-W 


2T 


Similarly,  if  there  are  a  numl)er  of  \vri)jhts  11'.,  W„  Wt,  .  .  ■  upon  th« 
rih,  and  if  /ii,  h„  ha,  .  .  .  arc  the  corresiiondirig  horizontal  thrust*,  the  total 
horizontal  thrust  //  will  lie  the  sum  of  these  sejiarate  thrusts,  i.e., 

H-h,+h,  + (5) 

Tt  will  lie  olxier^-ed  that  the  apices  (D,,  D„  Di,  .  .  .)  of  the  several  linear 

arches  1  triangles)  lie  in  a  horizontal  line  at  the  vertical  distance  ;j  fn)m  the 

iprin)tinx-linc. 

hx. .'».  .1  neminrrulnr  rih  of  JS  fi.  xpnn  earrirx  a  xtfight  n/i  ton  al  4  ft.  (nu(U- 
urrd  hiirizimlnlln)  jr«m  thf  '•nilrf.  Find  the  thnmt  and  shear  at  ths  centrt  of 
the  rib  and  at  the  ft-nnt  at  iihiri  the    I'eight  is  concentrated. 


Therefore 


,-/)f'_VX7-2r. 

„     ^      1  IS  V 10     4.-.  „     4.'5, 

''"^-4       .H--2S*""'''      ^'     ""616'- 


The  vertical  reaction  at   .I-.',/  and  at   /<-/,.'. 

Let  0  be  the  angle  lM>twci-ii  the  titnKcnt  at  E  and  the  horiionUL    Than 

coeec  #-%-/- 3.5,    and    »-10°.36'. 
Uf       4 


RIB   WITH  HINGED  KNDS 


91& 


A 


/    >  \ 


Fig.  918 


Fio.  919. 


In   Fig.  919  QH  \n  the  line  of  loads,  QA  being  ,»,<  and  RA  J^t.    Tak» 
AP~ff,l  and  draw  /V>  parallfl  to  the  UnRent  at  E.     Ab«i  let  fall  the  pet- 
pendic'ulani  Q/,  and  Rt,  upon   Hll, 
Then,  below  the  weight 

the  axml  thrust  =  P/,  - .  1 1 «( -  ff  cos  «  +  ,»,  ain  fl) , 
' '   shear  -</,<-.  1332<(  -  -  //  sin  0  + ,»,  cos  0), 

and,  aitrtc  tAe  ufiy/i/ 

the  axial  thrust  -  P/,- .0445<(  -W  cos  0 -A  sin  8), 
"  shear  -ft/,-.1064/(-//8in  <>  +  /,  costf). 

PQ  and  P«  are  of  course  parallel  to  the  lines  Dli,  DA,  respectively,  of  tJ» 
Un<!ar  areh. 

Ex.  6.  /><  the  an>  be  a  parabola  of  upon  21  and  rue  k.     From  the  properties 
of  the  parabola, 

E'F'-kd-^),    iyF'-'^l~^\ 
V      /'/  l±a 

Also.  d.«-rfx'(l+4j^x'),    or    da -dx(n-2^^x«),  approximately. 

Applying  condition  (4), 


/; 


a  9 


Kio.  920. 


916  THEORY  OF  STRUCTURES. 

which  easily  reduces  to 


an  equation  Kiviiif;  2  or  DF. 

If  the  arch  is  Hat,  80  that  d.i  may  1»  considered  as  approximately  equal  to 

k' 
dx,  the  term  2^x'  in  the  above  equation  may  be  disregarded,  and  it  may  be 

easily  shown  that 

f        1  /'  +  a'l        16 

33     kl' 


or 


s  51'  -••■ 


Ex.  fi.  Drmr  the  eq\nUhrium  poliigon  for  a  parabolic  arch  of  100  ft.  tpan 
^itul  20  //.  ri.'<r  irhrn  Imuird  irith  irrii/hls  nj  3,  '1,  4,  and  2  <«««,  rcxpectiv^y,  at 
the  nut  of  llir  third,  sislh,  riijhth,  and  ninth  dirisionx  from  the  left  support, 
of  ten  equal  hnriznntal  divixionn.     (Xeijlert  the  ireiijht  of  the  rib.) 

If  the  rih  nmsist  of  a  ireh  and  of  two  flant/en  "JJ  ft.  from  centre  to  centre, 
lietrrminf  the  miuimnm  flange  stress. 


First  to  find  //. 


•20-E,F,    M     -lO-E^F,    20    E,F^    20-E^F, 


20' 


50' 


10' 


30' 


40' 


Heiir 


Fio    9L'I 


Thonfore  E,F,=  Ift'.S,  £,F,-19'.2, 

E,F,  -  12'.8, and  E,F,-T2. 
ApplviniE  the  last  formula, 


:?2<M) , 
1  .1  f'- 


800  - 


H«N) ,  .  3200 , 

e.»,,,,ft.,  and   ^.-y^^ft. 


3200        30 '70         HOO       fiOv40     ^      .    800         80x20 


KM) 


31 


100 


29 


100 


.    ,3200    .novio    .„ 

■^     '""109 '"-100    -•^• 


Therefore     A,-2.;W2/,    /i,-l.H»W,    ,'i,-2.32/,    /i,-^613l, 
and  //  -  A,  -"  S  +  A,  +  A,  -  7. 1 75  tons. 


RIB  WITH  HINOED  ENDS. 

In  Fig.  922  take  the  verliral  through  yl  aa  the  line  of  loada. 

The  vertical  reaction  at  the 
left  support -3.9  tons. 

The  vertioal  reaction  at  the 
ri^t  support -7.1  tons. 

The  polar  distance  AO  — 
7.175  tons. 

Draw  the  radial  lines  Op,  Oq, 
Or,  «.•(,  Ot.  Then  Ad/l^t^„  R  is 
the  linear  arch,  with  its  sides 
parallel  to  the  radial  lines  from 
O.  Denoting  by  a,  ?,  j,  J,  ,  the 
inclinations  of  these  lines  to  the 
horizontal, 


917 


3^ 
H' 

hi 
H 

7.1 
tan  .--. 


t^na-—,    tany9-^, 
tan,-^.    t«..-^. 


-ijj- 


Fio.  022. 


117 


Therefore    rf^.-30tan  a-i^^    dj.-30(tan  «+tan /9)-ii^, 

«'J.-</J,-20tanr-'^",    ftnd    dJ.-lOtan  .-^'. 

The  vertical  intercept  l«.tween  the  linear  arch  and  the  axis  of  the  rib  i« 
greatest  at  e,  and  therefore  the  B.M.  at  this  point  is  a  maximum.     Hem-e 

max.  B.M.  - //-v.- 7.176  (?^- 12.8)  -30.156  ft.-tons. 

The  maximum  bmding  »trt„  at  the  same  point,  /  being  the  moment  of 
mertia  of  the  rib, 


-  ±  JX30.156X  12- ±~- tons  per  »q. 


in. 


Draw  Ot/  parallel  to  the  tangent  at  r„  and  drop  the  perpendiculars  A 
rtr    If  i>  IS  the  inchnation  ..f  the  tangent  to  the  horiiontal, 

*^'"         3(j         -0.48,     and     fl-25'39'. 


^1    \ 


Lkl 


W8  THEORY  OF  STRUCTURES. 

Hence,  just  below  the  weight, 

the  axial  thrust  -  0(,  -  8.676  tona  (-Hco80+5.1sia^ 
and  the  shear  -  gf,. 

Just  above  the  weight, 

the  axial  thrust  -  Ot,  -  6.945  tona  ( -  tf  cos  0  + 1 . 1  sin  9) 
and  the  shear  -  r/,. 

Therefore  the  maximum  stress  in  tons  per  square  inch  -  ■ '  '  ±— — , 

A  i 

A  IxMDg  the  seetional  area  of  the  rib  at  e,. 
Again,  if  F  is  the  total  flange  stress, 

Fx2i-30.156    and    F-12.063  tona. 

Ex.  7.  Let  the  axis  of  the  rib  be  a  circular  arc  of  span  21  and  radivt  r, 
nuhtending  nn  nniflr  '2a  at  the  centre  N . 

Let  the  angles  between  the  radii  NE,  NE',  and  the  vertical  be  p  and  9 

respectively. 
i  9  The  element  rf»  at  £'-rdO.    Also 


'^ 


^^-A--> 


X 


N 
Fia.  923. 


£:'F'-r(eo8fl-cosa);  ylF'-/-r  sin  fl; 
I^F' -  ~~  (I -r  Bin  0). 
Applying  condition  (4), 

/    r'(cose-coso)Vrf» 

/*•    « 

-   #      r^^(/-r8in  fl)r(cosfl-cos  «)r* 


/■ 


,_   (/-rsin  «')r(co«  tf-coa  «)rrf>. 


which  easily  reduces  to 

r|a(cos2a+2)-j8in2a| 

~  .T       .    'Hs'n  «  -  a  cos  a)  +  7"  (coS  2a-  cos  2/9) 
('— o'l  ■* 

-r/cos  a(cos  a— COS/?)— /a(8in  p-ficoB  a)[, 
an  ecjuation  giving  *  or  DF.     Also, 

DE-DF-EF, 


RIB  WITH  FIXED  ENDS.  919 

»nd  the  corresponding  horizontal  thrust  may  be  found,  as  before,  by  the 
equation 

Note.—U  o'-gO", 

9.  Rib  with  Ends  Absolutely  Fixed.— Let  ABC  be  the  axis  of 
the  rib.    The  fixture  of  the  oiuls  introduces  two  unknown  moments 

D 

WL iF_._ ^^L 

Fio.  024. 

at  these  points,  and  since  H  is  also  unknown,  three  conditions  must 
be  satisfied  before  the  strength  of  the  rib  can  be  calculated. 

Repfesoiit  th(  linear  areh  by  the  dotted  lines  KL\  the  points 
K,  L  may  fall  above  or  Ik-Iow  the  points  A,  C. 

I^t  a  vertical  line  EF  intersect  the  Ibear  arch  in  D,  the  axis 
of  the  rib  in  E,  and  tlw  horizontal  through  A  in  F. 

As  before,  the  change  of  inclination  at  E,  or  <W,   ~-Sr-     But 

EI 

the  total  change  of  inclination  of  the  rib  between  A  and  C  must 

ke  nil,  as  the  ends  are  fixed. 

Then.fore  J  -^-0-J  ~^j-, (1) 

which  may  be  written 

JT^"^ (2) 

sinw  H  and  E  are  constant. 

If  the  section  of  the  rib  is  uniform,  /  is  constant  and  eq.  (2) 
becomes 

fl)Eds-0 (3) 


'  ■  \ 


t  • 


1  *j--i 


920  THEORY  OF  STRUCTURES. 

Again,  tht>  total  horizontal  ilisplapomont  botwpen  A  and  C  wil? 
Ik'  nil  if  the  abutments  are  iiniii()val)le.  If  they  yield,  the  nmonnt 
of  the  yielding  tnust  lx>  determined  in  each  case,  and  may  be  denoted 
by  an  expression  of  the  form  /<//,  n  In-iiig  some  coefficient. 

Aa  U'fore,  the  total  horizontal  displacement 

ru  in:  KTds 

"J           El 
Therefore       J  ^^        -=0    or     -/J? (4) 

But  H  and  E  are  constant    so  that 

/•DE  EFds 

/  1 -0    or     =/<E (5) 

If  the  section  of  the  rib  is  uniform,  /  is  also  constant,  and  hence 

fl)EEFdii-=0    or     =/.A7; (6) 

and  since  DE  is  the  difference  Ix-tween  DF  and  EF,  this  last  may 
be  written 

fl)F  EF  (Is^fEF^ds^O    or     -fiEI.     ...    (7) 

Again,  the  total  I'ertical  tlisplacement  between  A  and  C  must 
be  nil. 

The  vertical  displacement  of  E  (Fig.  914) 

^A'L^^ALcos^EAF'-AEda^ 

Att 

"AFdO-AF^-^- 
El 

Hence  the  total  vertical  displacement 

rn  DEAF  ,      ^ 


:fl 


RJB  WITH  FIXED  ENDS. 


wliich  may  be  written 


/"DE  AF, 
J  — f-«i»=o, 


921 


(0) 


since  H  and  E  are  constant.    If  the  section  of  the  rib  is  also  con- 
stant, 


/dE  AF  d8-o=/DF  AF  d8-/EF  AF  da.    . 


(10; 


Eqs,  (2),  (5),  and  (9)  are  the  three  equations  of  condition. 
In  eq.  (9)  AF  must  be  measured  from  same  abutment  throughout 
the  summation. 

The  integration  extends  from  A  to  C. 

Ex.  8.  Let  the  axis  be  a  parabola  of  span  21  and  rise  k  (Fig.  925). 


Fia.  925. 


Alao, 


D'F'~y^  +  (l-x)^  on  the  left  of  DF, 
1+0  ' 


-y*  +  (i-')TZ~  on  the  right  of  DF; 


ud 


AF'-l^x. 


The  equations  of  condition  become 


!    i 


r.i 


ri 


Jli- 


MICROCOPY   RESOIUTION    TEST   CHART 

(ANSI  and  ISO  TEST  CHART  No    2! 


1.0 


I.I 


If  K  m 

^  1^     II  2.2 
t    I4S     1 2£ 


1^25  iu 


1.8 


1.6 


J§ 


d     -APPLIED  IIVMGE     Ir 


'■ui-Mtfr,     IK(»*     r^ri,  '*604  MSA 


922 


THEORY  OF  STRUCTURES. 


+/'I'''-«-"S'1('-f)'('*^f'*)* 

-/>(-?r(-;^")-. 

and 

These  equations  niaj-  Iw  tit  once  integrated,  and  the  resulting  equations 
will  give  the  values  of  (/,,  y„  z. 

If  the  areh  is  very  flat,  so  that  ds  may  l)e  taken  to  be  approximately  the 
same  as  dx,  it  may  b,;  easily  shown  that 

akl-'^Sa  2kl-5«  .  6 

y,-=—     -   ,    7,  =  —. ,    and     2-— k. 

'      i5l+a'    "    15    -a'  S 


Lot  /)',  h",  h'"  ...  be  the  horizontal  thrusts  due  to  loads  p',  p",  p'" 

respectively. 

Let  //,',  !/i", !/,'"  .... ,'//- ;//'. .'//"  .  .  .  ,  be  the  corresponding  values  of  y„  y. 
Let  yi,  ^a  be  the  resultant  values  for  the  total  thrust  H,    Then 


and 


H.y>A'(/,'+''".V,"+''"V"  +  -  •  .-2(hy,) 


^  denoting  nifirbrair  sum.     If  i/,  or  y,  is  negative,  it  merely  indicates  that 
A'  or  ("  falls  U'low  .If. 


Fill.  <t'.'fi. 


10.  Value  of  H.— In  tho  rnso  of  tho  flat  paralmlic  arcli  AC,  fixed 

at  Ixith  ends  and  earryinft  a  weight  \V  at  A', 

distant  o  from  the  middle  iM)int  0,  mensuivd 

*g?'^°^"     T~~yr~^^^^v^g'    liorizdntally,  xiniplc  ('.\|m'ssions  for  H  and 

\  „  "::  "  -o  g  _^S     j,^^,  vertical  ivactions  {/^.4, /^c)  at  .4  and  C 

can  easily  Ik-  found. 

Ut   the   segjuents   AF  ^J  -a-p, 
CF  -Ua-q. 
Viu.  827.  Xaitp  the  verticala 


VALUE  OF  II 


923 


^^'=,.4a'^4,3p_:2_. 


"15* /-a     15 


and 


^"    15'^ /+a      15^^^' 
FI)  =  z  =|-A:  (Ex.  8). 


The  linear  arch  is  composed  of  the  straight  Hnes  A'D  DC' 
Take  the  vertical  LK^W  and  draw  AT,  LP  parallel  to*  ^'A 
t  D,  respectively.    Also  draw  the  horizontal  lines  PQ,  A'M    and 

Then  PQ=/r,    A'0  =  /?^,    and    AQ  =  7?e. 


W'    «.     ff, 


Hence    ^^^r  +  Tr  =  tan  <?+tan  </. 


and 


3a      kl-' 


Hence,  also, 

RA=i/tan  »=,^Nf  A'l/l-^?.:?:q2a  +  p) 
32      A7^  15    ;;2       ^        J.J      , 


and 


Rc=^tan.A-,^:ir',T^/l 


1:"),,.;A/2S    /+J,,     Wp2(i  +  q) 
32      A/3  15''  />2    =^        13—- 


«-       I."  ^    ."".''Jl".  '  "'■'■''  ""'^  'irtrrminr  the  maximum  flange  Mres,c»  fen' 

a<  /  ,<•  rm/  „/  ,l,r  first  ,sro,«,/,  //„„/,  f.,„rth.  n,„i  fijth  divmon.,  of  right  equal  hori- 
...nta  .l,n..o..s  The  riU  m  0/  ,M.ie-tec  .ert,„n  a,ul  30  ins! deep.  'aJ/^ 
the  .hear,  and  the  axial  thr,  -U  at  the  fifth  point  of  division. 

MV      "40»"       20' 

10'      "      10'     • 
Therefore        K,F,  -  7'.        fc-js-,  - 1 2' 

A.fi- 15'- £,/!',.  '  l-io.  928. 


.,  , 


« 


■A  ^ 


924 


THEORY  OF  STRUCTURES. 


At  H'l,      2/1         —       "15,     J/j  1(,5, 


Applying  the  formuljB  obtained  in  Ex.  8  and  in  Art.  9, 

.,             3075          „    ,       _245  p   .  21 

"              1U24''      ''•*              16''  "'-  16'» 

8192''     "^         ~i2s''  ^C  ''1^') 


^2. 


!/.' 


■1? 
.■) ' 


-;s-.    2/j 


£,'    2/.'"  =  --' 


^S'        J/l 


45' 
32 
15' 
96 
25' 


2211 
"45  ' 

Hi  25' 

!/2  15' 


"  81<)2''     "^ 


"         32'' 


2/2 


32 


45'    "  "8192''     "•*  128''       "^  los* 


45' 


Thereforethe  total  //  =  /.'+A"+A"'+/i""  +  fc""'  =  8.2306< 


845, 


"    Rc=Rc'  +  Rc"  + . . .  +  Rc'""  =  fi'- 


Again,  if  y,,  y,  are  the  true  ordinates  of  the  linear  arch  at  A  and  C  re- 
spectively, 


3675 , 


352  ,  1350, 


32  ,  16875, 


32  ,  75^32  ,  16875,  M 


^y»"8192'^        13  "'"1024^       5  "^8192'^       45 ''"33'*^  15  "^  8192  '*' 25 


and 


3675^,608   ,  1350^224  ,  16875^96  ,  75_32  .   16875, 


32 


^J/s  "  8192  ^  105  "*^  1024  ^  45  ''"8192  ""^ZS  "'"32  "^  IS"*"  8192  ""^   45' 


Hence 


y,-  -0.9136  ft. -il^' 


1    +  r"~  ,-'»■>- 


I 


-t_* 


./' 


/ 


Fiu    029. 
2.6 


and  y,-2..'5035  ft. -CC. 

Taking  pqrutv  as  the  line  of  load*, 
then 

.4p  =  i?c- 3.4  tons 

and         Av^Ra^&.Q  tons. 

ALso,  the  {jolar  distances  OA^ 
8.'.'.30(i<. 

Join  Op,  Oq . .  .Ov  and  draw  the 
linear  ar(  h  A'I),D, .  .  .  /)/,",  with  its 
sides  A'l),.  /),/>,  .  .  .  parallel  to  the 
rorrewponding  liiicM  Or,  Ot,  ...,  re- 
spectively. ])('Nignating  by  o  /8,  j 
. . .   the    hil()|M-s   of   those   lines, 

f).0  ,    4.6 

tan<r=-^y,    tan^J-— , 


0.6 


14  3  4 

tan,-"-rr,    ton  8^"'"-,    tani-    y,    and    tanO--J=-. 
II  n  H  H 


PARABOLIC  AND  CIRCULAR  RIBS. 


925 


Henre 


A^^.  =  C,F,  +  10tan.  =  16'.r,897      "       D,E  ^D,F,-E,F  =0  5H97  " 

AtE„    //x0.8o88  =  B.M.  =  7.0ft85ft.-ton8,    and  therefore 

the  max.  bending  stress  in  tons  per  sq.  in.=  ±— x 7.0685 X 12- i^^^?:^ 


If  t  is  the  slope  at  E„ 


tan  «=  1^=0.2    and    c  =  ll°19'. 


Therefore  sec  c  =  1 .02    and    7/  sec  .  =  8.3952<. 

Hence  the  total  maximum  skin  .stress  per  square  inch  in  tons  at  E 

S. 3952     1272^ 

tan^ll?'''  ''*™"*''  *"  '*'''  **"*^°*  **  ^"  *"'^  '^*  ^  **  *''•'  «'°P« «»»  ^.-    Then 
Just  oftowe  <Ae  weight,  at  J?„ 


and 


the  axial  thrust -=0<,-8.3486<, 
the  shear  -p/,  -0.243<, 


and  just  Mow  the  weight,  at  A'„ 


and 


the  a  xial  t  hrust  -  O/,  -  S.74 1  <, 
the  shear  -  9/,  - 1 .719<. 


The  absolute  maximum  B.M.  is  tt  C  and  is 

-////.  =  20.00.54  ft  .-tons. 

The  slope  at  C  -tan-- 0.8  ~  {SMO'.  and  //^r  38°  40' -10.434/     Hen(* 
the  absolute  maxm.um  flanp.  stn-ss  n-r  square  inch 

^10.434     1.<x'20.a'}42<  12     10-434    3700 
tho?x;"nhf  r"  ^"^"  "'  "'"  '"'"*"  '^■^^"'  "^-^  ""-  -h  interacts 


Si 


it- 


i(M 


»  ,1 


1  lii;  I' 


I 


926  THEORY  OF  STRUCTURES. 

Let  a  welgnt  IT'  be  concentrated  on  the  rib  at  a  point  ^  whose  horizontal 

distance  from  the  middle  point  of  the 
span  is  a. 

Let  the  radius  NE  make  an  angle  ,1 
with  the  vertical. 

The  "  linear  arch"  consists  of  two 
^C      straight  lines  DA',  DC". 

U-t  .l.l'  =  t/„DF  =  z,  CC'  =  v,. 

Draw  any  ordinate  E'F'  intersect- 
ing the  linear  arch  in  D'. 

Let  the  radius  A'£^'  make  an  angle  9 
with  the  vertical.    Then 


X    0«    V 


M 
Fig.  930. 


£.''F'  =  r(cos5-coeo). 


2-y. 


Also,         AF'^l-rmiO    and    Z)'F'=(Z-rsin  0)— ^+1/, 


if  F'  is  on  the  left  of  F, 

while 
if  F'  is  on  the  right  of  F. 

Also, 

Applying  condition  (2), 


2-y» 


i4F'  =  /  +  rsinO    and    /)'F'=a-rsin  (^)— -^+y. 


ds~rd6. 


r  \(l-rs\nO)~-^^+yAdO+  f '  I  (I- r  Bin  6)^+ y,]d9 

-rf"  (cos  0- COB  a)dd.     (1> 
Applying  condition  (5),  and  assuming  /i-O, 

/' (cos  0  -  coe  a)  { a  -  r  sin  0)^-^i  +  y,  I  do 

(i-r  sin  ^f-f^  +  y^  -r  f  (cos  ff-cos  «)W.     (2) 

I— a  I         J— a 


4-    /     (cos  C— cos  o) 
Applying  condition  (9), 


J°  a-r8ine){(Z-r»inO)y^4-y,}d« 

+  y°(/  +  r  sin  9)  I  (i-r  sin  9)^'+y,|<i« 

-r/""  (COB  e  -  voB  a )  {I  -  r  sin  0  jdO -i- r  J    (i.0B  6- COS. \)il-i^r  sin  e)dd.     (3) 


TEMPERATURE  EFFECT. 


92? 


Eqs.  (1),  (2),  (3)  may  be  easily  integrated,  and  the  resulting  eqtutions 
will  give  the  values  of  y„  z,  and  y,. 

The  corresponding  horizontal  thrust,  h,  may  now  be  obtained  from  the, 
equation  hDE=-M=h{z-  EF) . 

Note. — If  the  axis  is  a  semicircle,  and  if  W  is  at  the  crown, 

o-O,     a  =  90°,    /9-0, 
and  eqs.  (1),  (2),  (3)  reduce  to 

2(»t-2)+y,+y,-2r; 

5«+3i/.-7r. 


Therefore 


«-|r    and    yi-4r=-yi. 


II.  Effect  of  a  Change  of  Temperature.— The  variation  in  the 
span  2/  of  an  arch  for  a  change  of  <°  from  the  mean  temperature  is 
approximately  =  2f</,  t  being  the  coefficient  of  expansion. 

Hence,  if  Ht  is  the  horizontal  force  induced  by  a  change  of  tem- 
perature, the  condition  that  the  length  AC  is  invariable  is  expressed 
by  the  equation 

„    fDE  EFds    „  ,    ^ 
V  Ei ±2e«=0. 


%: 


'a 


If  the  rib  is  of  uniform  section,  /  is  constant;  and  since  E  is  also 
constant,  the  equation  may  be  written 


EU 


DEEFds±2ttl'=0. 


E*.  11.  Let  the  axis  ABC  of  a  rib  of  uniform  sedion  be  a  parabola  of  span 
21  And  rite  k. 


Fio.  031. 


First,  let  the  rib  be  kinged  at  both  ends. 

The  straight  line  AC  is  the  linear  arch.    Then 


i  'I 


|f  f 


928  THEORY  OF  STRUCTURES. 

J*DEEFda^f^EF^{\  +2yx'jdx 

/'    /      x'Y/,     2Jt'  ,\^      16,,,     32  Jfc* 


and  hence 


n^.^mh^-o- 


Second,  let  the  rib  be  fixed  at  Iwth  ends. 

The  linear  arch  is  the  line  A'C  at  a  distance  z(  "DF)  from  AC  given  by  the 
equation 

JDEds^Q=j{DF^EF)d8, 
ot  DFJds^JEFds. 

Hence  zfjx  +f  x')dx=/)(l  -fj  (l  +2|x')<fa, 

Also, 
y^E  EFds'fDFEFds  ^fEF'da 

-z/EF...-/^F.«i.=2H.(|.lf)^({|../+f^). 

Hence  ^/T  j  ^(^  +  5  W     1  V"*"  7  /'/'  (  ^'^''^    ®- 

/?rrrtflrA-.— The  roeflRcient  of  expansion  per  tlegrce  of  Fahrenheit  is  0.0000062 
and  0.0000067  for  cast-  and  wrought-iron  Ix'ams  respectively.  Hence  the 
rorrcsfKJnding  total  expansion  or  contraction  in  a  length  of  100  ft.,  for  a  range 
of  00°  F.  from  the  mean  temperature,  is  0.0372  ft.  (  "  /o")  and  0.0402  ft.  ( =  i"). 

In  [iractice  the  actual  variation  of  length  rarely  exceeds  one  half  of  these 
amounts,  which  is  chiefly  owing  to  structural  constraint. 

Ex.  i'J.  I.(i  tkf  ajtix  AEC  vj  a  rib  0}  iinijorm  /-ection  be  the  arc  of  o  circle  of 
radiun  r  tuhtending  an  angle  '2a  at  the  centre. 


TEMPERATURE  EFFECT. 

First,  let  the  rib  be  hinged  at  both  ends 

It  is  evident  that  the  straight  line  AC  is  the  "  linear  arch."    Then 


900 


jDE.EF.ds^fEF'ds^r'fyo.e-c.s  a)»d^-r.|«(2  +  ^  2«)  -|  sin  2«). 
Also,  /  =  r  sin  a.    Therefore 

£7wn^  '''^^"^*'°^  2a)  -}  sin  2a|  ±2.«-0. 
JVote. — If  the  axis  is  a  semicircle,  a  =  90°,  and 

Second,  let  the  rib  be  fixed  at  both  ends. 

The  "linear  arch"  is  now  a  straight  line  A'C  at  a  distance  z  {-.DF)  from 
AC  given  by  the  equation  ^  ' 


fDEds-0. 


^ 


SI 


Fio    932. 


Then 


or 


or 


Also, 


fDFds'jEFds, 

*fds~r*r  (cos  0-CO6  a)dB, 
««-r(8in  a  -a  C06  a). 

fl>E-EF.ds.f(DF.EF^EF')ds-zfEFds^fEF^ 

-2«-«(8in  a-a  COS  a)w"|a(2+co«2a)  -|  gin  2a!. 


3  •; 

'j  ' 

if 


3' 


930 


THEORY  OF  STRUCTURES. 


Hence 

^!  bir'Csin  a-a  COS  a)  -r'|a(2  +C08  2a)  -}  sin  2a!  [  ±2ttl-0, 
hi '  ' 

^nd  l  =  r  sin  a. 

12.  Deflection  of  an  Arched  Rib. 

Let  the  abutments  be  immov- 
able. 

Let  ABC  be  the  axis  of  the  rib 
in  its  normal  position. 

Let  ADC  represent  the  position 
of  the  axis  when  the  rib  is  loaded. 
Let  BDF  be  the  ordinate  at  the  centre  of  the  span;   jom  AB, 
AD.    Then 


Fio.  933. 


But 


arc  AB=aic  AD     f 


f  being  the  intensity  of  stress  due  to  the  change  in  the  length  of  the 
axis.    Then 


and 


DF^  =  AB^{l-^y-AF-^  =  BF^-AB'^2^-(^y\, 
AB2 1  2^  -  (I)  ^  [  =  BF^  -  DF^  =  (,BF  -  DF)  {BF + DF) 


=  2BF{BD) ,  approximately . 
(-^)    is  also  sufficiently  small  to  be  disregarded.     Hence 

AB2  f     k2+12  f  .      ^  , 

BD,  the  deflection,  =  |^-|  =  —jP  g,  approximately. 

13.  Elementary  Deformation  of  an  Arched  Rib. 

The  arched  rib  represented  by  Fig.  934  springs  from  two  abut- 
ments and  is  under  a  vertical  load.  The  neutral  axis  PQ  is  the 
locus  of  the  centres  of  gravity  of  all  the  cross-sections  of  the  rib^ 


DEFORMATION  OF  AN  ARCHED  RIB. 


931 


and  may  be  regarclod  as  a  linear  arch,  to  which  the  conditions  govern- 
ing the  equilibrium  of  the  rib  are  equally  applicable. 

Let  AA'  be  any  cross-section  of  the  rib.     The.  segment    iA'P 
13  kept  in  equiUbrium  by  the  external  forces  which  act  upon  it 
and  by  the  molecular  action  at  AA'.  * 

The  external  forces  are  reducible  to  a  single  force  at  C  and  to  a 


couple  of  which  the  moment  M  is  the  algebraic  sum  of  the  moments 
with  respect  to  C  of  all  the  forces  on  the  right  of  C. 

The  single  force  at  C  may  be  resolved  into  a  component  T  along 
the  neutral  axis,  and  a  component  S  in  the  plane  AA'.  The  latter 
has  very  Uttle  efifect  upon  the  curvature  of  the  neutral  axis,  and 
may  be  disregarded  as  compared  with  M. 

Before  deformation  let  the  consecutive  cross-sections  BB'  and 
A  A'  meet  m  R;  R  m  the  centre  of  curvature  of  the  arc  CC  of  the 
neutral  axis. 

Aft^  deformation  it  may  be  assumed  that  the  plane  AA'  re- 
mams  unchanged,  but  that  the  plane  BB'  takes  the  position  B"B"' 
Ui  AA'  an<l  B"B"'  meet  in  R';  R'  is  the  centre  of  curvature  of 
the  arc  CC  after  deformation. 

Let  ahc  be  any  layer  at  a  distance  z  from  C. 

Let  CC'^ds,  CR  =  R,  CR'  =  R',  and  let  da  be  the  sectional  area 
of  the  layer  abc. 

By  similar  figures, 


ac 
ds 


R'  + 


,    ab    R+z 


i 


i1 


'  i 


f  ,' 


I  u„ 


i..^i. 


932  THEORY  OF  STRUCTURES. 

Therefore  hc  =  ac-ab  =  dszy-n,—n\. 

The  tensile  stress  in  abc 

"2  (^--^jEda,  very  nearly. 
The  moment  of  this  stress  with  respect  to  C 

Hcr.ce  the  moment  of  resistance  at  AA' 

the  integral  extending  over  the  whole  of  the  section. 
Therefore  M  =  El(^,-^) 


(1) 


Again,  the  effect  of  the  force  T  is  to  lengthen  or  shorten  the 
element  CC,  so  that  the  plane  BB'  will  receive  a  motion  of  trans- 
lation, but  the  position  of  R'  is  practically  unaltered. 

Corollary  1.  Ix't  A  Ix*  the  area  of  the  section  AA'. 

The  total  unit  stress  in  the  laver  abc 


T    Mz 


(2) 


the  sign  being  jAua  or  minus  according  as  M  acts  towards  or  from 
the  edge  of  the  rib  under  consideration. 

From  this  expression  may  be  deduced  (1)  the  position  of  the 
point  at  which  the  intensity  of  the  stress  is  a  maximum  for  any 
given  distribution  of  the  load;  (2)  the  distribution  of  the  load  that 
makes  the  intensity  an  absolute  maximum;  (3)  the  value  of  the 
intensity. 


fl . 


GENERAL  EQUATIONS. 


933 


Cor.  2.  Let  u»  be  the  total  intensity  of  the  vertical  load  per  hori- 
zontal unit  of  length. 

I^t  wi  be  the  portion  of  w  which  produces  only  a  direct  com- 
pression. 

I--et  H  be  the  horizontal  thrust  of  the  arch. 

Let  P  be  the  total  load  between  the  crown  and  AA'  which  pro- 
duces compression. 

Refer  the  rib  to  the  horizontal  OX  and  the  vertical  OPY  as 
the  axes  of  x  and  y  respectively. 

Let  X,  y  be  the  coorilinates  of  C. 


Then 
Therefore 
and  also 


^^^%''    "^"^    dP-^Widx. 


ax 


(3) 

(4) 


i4«  General  Equations. 

Let  /  be  the  span  of  the  arch. 

Let  X,  y  be  the  coortUnates  of  the  point  C  before  deformation 
Let  X',  y'  be  the  coordinates  of  the  point  C  after  deformation 
I^t  5  be  the  angle  between  tangent  at  C  and  OX  before  deforma- 


tion. 


tion. 


Let  6'  be  the  angle  between  tangent  at  C  and  OX  after  deforma- 


Let  ds  h,  the  length  of  the  element  CC  before  deformation 
I^t  ds'  be  the  length  of  the  element  CC  after  deformation. 


dO' 


1 


Egecolfl^ure.    ^.i    .„a    f-^ 


Hence 


J^_J._i_fr   dd  dd'~dd 

El     R'    /i;-(/s'~ds=—rfs—>  very  nearly. 


I^t  t  be  the  change  of  slope  at  C.    Then 


EI      Eldx"^- 


IM 


i 


li  r  : 
'  J 

I.  "ll 


934 


Integrating, 


THEO.'Y  OF  STRUCTURES. 


(5) 


to  being  the  change  of  slope  at  P,  and  a  quantity  whose  value  has 
yet  to  be  determined. 

Again,  the  general  equations  of  equilibrium  at  the  plane  A  A' 
are 

cPM    dS        .  ,         /       „(Py\  ,„. 

for  the  portion  nh,  Cor.  2,  Art.  13,  produces  compression  only  and 
no  shear.    Therefore 

..5.-/'»*.«(g-g) (7) 

So  being  the  still  undetermined  vertical  com^v^aent  of  the  shear  ai 
P,  and  -r-  the  slope  at  P.    Also, 

(IXq 

M  =  Mo  +  SoX-J^'f^'wdx!'  +  H(y    J/o-a^),   •    •     (8) 

Mo  Ix'ing  the  still  undetermined  bending  moment  at  P. 

Ivjuations  (5),  (6),  (7),  and  (8)  contain  the  four  undetermined 
constants  //,  Sg.  Mo,  J'o- 

I,et  .Ui,  Si,  and  Vi  l)e  the  values  of  M,  S,  and  j,  respectively,  at  Q. 

Equations  of  Cotidition.— In  practice  the  ends  of  the  rib  are  either 
fred  or  free. 

If  they  are  fixed,  io  =  0;  if  they  are  free,  .Vo^O.  In  either  case 
the  numlxT  of  undetermined  constants  is  reduced  to  three. 

If  the  abutments  are  immovable,  Xi-/=-0.  If  the  abutments 
yield.  i\  -/  must  U'  found  by  expt'riment.  Let  ix-l^fiH,  fi  being 
some  cfxlhcient.    The  first  equation  of  condition  ia 


Xi-/  =  0,    or    Ji-/  =  /<//.      .     .    . 


m 


.\gain.  Q  is  immoxable  in  a  vertical  direction,  and  the  second 
eq..,ai'iii  of  eomlition  is 

yi-yo-O ;   :   .    (10) 


TEMPERATURE  EFFECT. 


935 


Again,  if  the  end  Q  is  fixed,  ti=0;  and  if  free,  Mi=0;  and  the 
third  equation  of  condition  is 


ti=0,    or    Mi=0. 


......    {IV 

Substituting  in  equations  (7)  and  (8)  the  values  of  the  three 
constants  as  determinod  by  these  conditions,  the  shearing  force 
and  bending  moment  may  be  found  at  any  section  of  the  rib. 

Agam,  cos  6'  =  cos  (0  -  i)  =  cos  fl  +  i  sin  6, 

sine'  =  an  {0-i)  =  s\n  0-icosd. 


Therefore       ^=J+tf    and    %-^-i^. 
as'     ds      as  ds'    ds      ds 

Hence,  approximately. 


(    i    I 


(12> 


.dx 


|(x'-.)-.f    and    ^(/-,)-i^. 

Thus,  if  X  and  y  are  respectively  the  horizontal  and  vertical 
displacements, 

.dx 


dX  .dy  ^  dY 
-j-  =  ^-r  and  -j- 
ds      ds  ds 


V«' 


or 


dX 


dY 


dy  dx 


(13) 


15.  Effect  «f  7  and  of  a  Change  of  t°  in  the  Temperature. 

T 


*'-*('-si)- 


Also,  if  there  is  a  change  from  the  mean  of  <**  in  the  temperature, 
tho  length  rfs^l-^jmust  Ix;  multiplied  by  (l±tO,  «  being  the 
coefficient  of  linear  expansion.    Hence 

ds'^ds(l-~j){l±U, 

"•d«(l  ~;gj  ±  *')»  approximately.    .    (14j 


I: 


k  , 


f  '^ 


A 


936  THEORY  OF  STRUCTURES. 

By  equations  (12), 


and 


dx'^(dx  +  idy)-^  =  (di  +  idy)  (l  -]^  ±  «<) 
dt/'  =  {jdy  -  idx)-^  =  idy  -  idx)  (l  -  ]^  ±  e<) . 


Therefore 

dX  =  d{x'  -x)  =  idy  -  (ttj T  tt\  dx, 

and 

/  T        \ 
dY'-d{yf-y)=  -idx-  \gj^  dj  dy,  approximately. 

Hence 


and 


(16) 


ae) 


Note. — A  nearer  approximation  than  is  given  by  the  preceding  results 
ma}'  be  obtained  as  follows : 

Let  x+dx,  y+dy  be  the  coordinates  of  a  point  very  near  C  before  defor- 
mation. 

Let  x'+dx',  y'+dy*  be  the  coordinates  of  a  point  very  near  C  after  defor- 
mation. 


Then 


and 
or 

or 

and 


da^-dx'+dy^    and    ds'*'-dx"+dy'* 
d«"  -  d»'  -  <ix"  -  dx*  +  (iy"  -  <fy", 
(ds'-ds)(d»'  +da)  -  {dx'-dx)(dx'  +dx)  +  (dy'-dyHdy*  +dy). 
Hence     (,ds'-da)d»-'(dx'-dx)dx  +  {dy'-dy)dy,  approximately, 

dy'-dy-Oh'-dJ"  ''/  -{dx'-dx)i~. 
dx  dy  'dy 

Hence,  by  equations  (12)  and  (If, 

'^-*-^'"-i,ti)'^"(t.)''^ 


RIB  OF  UNIFORM  DEPTH  AND  STIFFNESS.  937 

Integrating, 

"-'-/•f--X'f.(£)--'r(i)- 

These  equations  are  to  be  used  instead  of  eqs.  (15)  and  (16),  the  remainder 
of  the  calculations  being  computed  precisely  as  l)efore 

The  following  problems  are  in  the  main  the  same  as  those  given  in 
Ilankine  s  CivU  Engineering,  20th  edition. 

.1,  ^k/^"    ^f  "^  ^yi""^  Stiffnes«.~Let  the  depth  and  sectional  farm  of 

the  rib  be  uniform,  and  let  its  breadth  at  each  point  vary  as  the  secant  of  the 

inclination  of  the  tangent  at  the  point  to  the  horizontal. 

Let  ^„  /,  be  the  sectional  area  and  moment  of  inertia  at  the  crown 

Let  ^,  /  be  the  sectional  area  and  moment  of  inertia  at  any   point    C 

rig.  934.  J    ir^  t 

Then 

A  -A^Bec^^A,- '.    .    .    .    (17) 

Also  since  the  moments  of  inertia  of  similar  figures  vary  as  the  breadth 

/-/.sec<>-/,^ ^,S) 

.     .     T    HaecO    H       _,  ^ 

'^■'"'  I "  Vi^  ~  1^'  *"*^  ^^^  intensity  of  the  thrust  is  consUnt  through- 
out. 

Hence  eqs.  (5),  (16),  and  (16),  respectively,  become 

'''^'ET,X    ^^'^■' (19) 

'^~"J„  'di'^'-Avr,'**^; (20) 

*'"^"-/^-(A^'')^»-y.) (21) 


^;i 


r. 


i  . 


938 


THEORY  OF  STRUCTURES. 


P'q.  (19)  shows  that  the  deflection  at  each  point  of  the  rib  is  the  same 
as  that  at  corresponding  points  of  a  straight  horizontal  beam  of  a  uniform 
section  eejual  to  that  of  the  rib  at  the  crown,  and  acted  upon  by  the  same 
bending  moments. 

Ribs  of  uniform  stiffness  arc  not  usual  in  practice,  but  the  formula  deduced 
in  the  present  article  may  Ik  applied  without  sensible  error  to  flat  segmental 
rite  of  uniform  section. 

Ex.  13.  Parabolic  rib  of  unijorm  depth  and  stiffness,  with  rolling  load;  the 
ends  fixed  in  direction;   the  abutments  immovable. 


Fio.  93.5 

Let  the  axis  of  y  be  a  tangent  to  the  neutral  curve  at  its  summit.  I(t 
/t  be  the  rise  of  the  curve.  Let  x,  y  he  the  coordinates  at  any  point  C  with 
respect  to  O.    Then 

y-f(2-^)' ''"^ 

and 

dx'      i'Vo     ')'     dx,         r     dx^  I'     dx'°p-     '    '     ^^•'^ 

Let  w  be  the  dead  load  per  horizontal  unit  of  length. 
<.    „,/  <<    ..    livo      "      "  "  "     "       " 

Let  the  live  load  cover  a  length  DE,  =r/,  of  the  span. 
Denote  by  (.V)  formuhr  relating  to  the  unloaded  division  OE,  snd  by  (B) 
formul.T  relating  to  the  loaded  division  DE. 
Eqs.  (7)  and  (S),  resix-ctively,  beaome 

(A)  S  =  S„4  ("-y-u^x; (24) 

(H)  S-S„+  {^''^l'  -n)x-ir'\x-(l-r)l] (25) 

(.\)  Af-.U,  +  .Vff*',^^-»)^'; (26) 

(B)  .V-.U.f.SVf(^"-u-)^--'^'|x-(l-r)n'.    .    .    .    (27) 

Since  the  ends  are  fixed, 

t-O-t (28) 


ir 


RIB  OF  UNIFORM  DEPTH  AND  STIFFNtSS.  939 

Hence  by  eqs.  (19)  and  (26), 

<A,  .-i{«...|,(?^_„)|-}, ,^, 

and  by  eqs.  (19)  and  (27), 

V/hen  x-l,  i-H-O,  and  therefore,  by  the  last  equation, 

Agam,  lett=-3^.    Then 
ax 

*^Q    dx       Jo  dxdx        ^dx    Jq  ^lP^' 
But^-O,    and    ^  =  -.    Hence 

J.'dx^'--TJo'^"-TJoJo^' (32) 

^I  ^^"^  fr.**'*'T  "^  **""  P'"*'''''"  ''  -*  and  y'  -y  are  each  zero  at  Q.    Hence 
equations  (20)  and  (21),  respectively,  become 

'--T'foT'^'-dA^'^y' (33) 

°--./o"^ .    (34) 

Substitute  in  eqs.  (.3.1)  and  (34)  the  value  of  Vgiven  byeq.  (30),and  integrate 
between  the  limits  0  and /.    Then  ;,»"u  imcgraw 

and 

°--^i    -2-^Tn/'    -'^)24-"'''-2l}. 
which  may  be  written 


I 


.J 
^  f 


i!^ 


ii 


I-, 


940 
and 


THEORY  OF  STRUCTURES. 


Hence,  by  eqs.  (31).  (35),  (36), 

wl'     wVr'r     3  \      2,„ 

''^•=-12— 3-(i-r)+3'=^5  • 


•  •  • 


•  •  • 


H- 


(36) 

(37) 
(38) 

(39) 


When  x~l,  M'-Mi,  and  S  =  S,. 
Hence,  by  eqs.  (25)  and  (27), 


and 


S,=-S,  +  (^-^-trji-u>'rf, 

,      „       /SAW       \J'     v/rV 


Substituting  in  these  equations  the  values  of  S„  M„  given  above,  we  have 

AkH 


wl 


and 


(40) 


(41) 


To  find  the  greatest  intensilt/  of  stress,  etc, — The  intensity  of  the  stress  due 

T     H 
to  direct  compression  =■  -j  ~  "T-- 

The  intensity  <if  the  stress  in  the  out«ide  layers  of  the  rib  due  to  bending 
is  the  same  us  that  in  the  outside  layers  of  a  horizontal  beam  of  uniform  sec- 
tion .4,,  acted  upon  by  the  same  moments  as  act  on  the  rib,  for  the  deflections 
of  the  iH'am  and  rib  are  equal  at  every  point  (eq.  (19)  ).  Also,  since  the  rib 
is  fixed  at  both  ends,  the  bending  moment  due  to  that  portion  of  the  load 
which  produces  flexure  is  a  maximum  at  the  loaded  end,  i.e.,  at  Q.     Hence 

//  ! 

the  maximum  intensity  of  stress  (p,)  occurs  at  Q,  and  p,--T-±Jl/,^,  «,  being 

the  distance  of  the  layers  from  the  neutral  axis. 


RIB  OF  UNIFORM  DEPTH  AND  STIFFNESS. 


941 


H  and  Mi  are  both  functions  of  r,  and  therefore  pi  is  an  absolute  muTimnTn 
when 


But 


and 


dr  ~    "a   dr     /,  dr  ' 


dH    15wV  r'(l-r)» 
dr°4    k         45  /.  ' 
■•"  4  iljfc' 


-5;r--«;7V(l-r)'+5-*- 


3"dr- 


(42) 
(43) 

(44) 


Hence  p,  is  an  absolute  maximum  when 


f       (±  ±1  ^\ 
*V^  "^  4  A.W 


The  roots  of  this  equation  are 
r-l 


and 
d*p 


r-± 


45    /, 
2        4  A,k' 
5  3/, 


±1 


(45) 


2  ^,«,A; 
r-l  makes  j^  zero,  so  that  the  maximum  value  of  p  corresponds  to  one 

of  the  remaining  roots. 
Thus, 

the  max.  Mn«<-j(w+^'Af,)-p,' (46) 


and 


t]iemax.teTuion<-~{-H+~M,)-p,",.    .    .    .     (47) 
the  values  of  H  and  Af«  being  found  by  substituting  in  eqs.  (39)  and  (41) 

1  +1'  A- 
2        4  A,k' 

'^"  5  ,     3     /, 


2  A.a.ik 


2         4  .4,ifc' 

2  /l,a,ifc  J 
according  as  the  stress  is  a  thrust  or  a  tension. 


(48) 


I  1 


,  if; 


942  THEORY  OF  STRUCTURES. 

If  eq.  (47)  pves  a  negative  result,  there  is  no  tension  at  any  point  of  the 

rib. 

JVo<e.— The  moment  of  inertia  may  be  expressed  m  the  form 

I  =  qz,'A„ 

q  being  a  coefficient  depending  upon  the  form  of  the  section. 
Hence 

the  maximum  intensity  of  stress  =—/±H+—^j  .   .    •    •    (49) 

Cor.  1 .— If  the  depth  of  the  rib  is  small  as  compared  with  k,  the  fraction 

^  will  be  a  small  quantity,  and  the  maximum  intensity  of  stress  will  approxi 

mately  correspond  to  r  =  |.     The  denominator  in  eq.  (39>  may  be  taken  to  be 
k,  and  it  may  be  easily  showni  that  the  values  of  p,',  p,"  .ire 

P-    -i;(T(j^irF'j^4-^-^31259Z.  f-      •    •    •    ^50) 

Cor.  2.— If  the  numerator  in  eqs.  (48)  is  greater  than  the  denominator, 
then  r  must  be  unUy.    Hence,  by  eq.   (39)  and  putting 

..     ,     4.i    /.       ,  ,  459«i^ 
'"^+TIX'~^^4    k" 


/» u-Hi/     45  ri£/i 
"°S     6fc        4    6ik'  • 


(52) 


and  by  eqs.  (38)  and  (41), 

I'  ,A-b    15  tlEI, 


15„U'+»A''  «,'  ,  5  ttEI, 


(53) 


Thus,  p/,  p,"  can  be  found  by  substituting  these  values  of  H  and  Af,  in 

eqs.  (40)  and  (47).  ^         .     p  j 

Ex.  U.— Parabolic  Rib  of  Uniform  Stiffness,  hinged  at  the  bnds. 
Lot  the  rib  h<>  similar  to  that  of  the  preceding  article. 
Since  the  ends  are  hinged,  .V„-0- A/„  while  i  is  an  undetermmed  constant. 
The  following  equations  apply: 


i:i 


RIB  OF  UNIFORM  STIFFNESS.  943 

(A)  S'8.+  (^-w)z: (54) 

(B)  S-S,+  {^-t,'jx-u^\x-(l-r)l\l     ....    (55) 

(A)  M-S^+(^-^-w)~;     .     .......     (56> 

(B)  A/-S.x+(-^5 „,j___,;P_(l_r)^,,,    .    .    .    (57> 

<-^-£7}%+l-7i--«')-e}; (58> 

1    (  „  a;'     /8fci/       \3^    v/  ■) 

*"'»"£7]%  +  l~i7"-'^J6— e'''"^^"'")'''}-    •   •   (59) 

Assume  that  the  horizontal  and  vertical  displacements  of  the  loaded  end 
are  nil. 

Substitute  in  eqs.  (20)  and  (21)  the  value  of  i  given  by  eq.  (59).     Integrate 
and  reduce,  neglecting  the  term  involving  the  temperature.    Then 

-     .       1    (  „  i'      /SkH      \l'  t*  )      H    \     I 

°-*-A/,l.^°6+W'    -'"^J24-«'^24f (61> 

From  (57),  since  3/,=0, 

„     „       /8Aff       \  /  r* 

0-S,+  \-^-w)^-w'l-^ (62> 

Equations  (60),  (61),  and  (62)  are  the  equations  of  condition. 
Subtract  (61)  from  (60).    Then 

°-:^i  ^»l2^h»-«')40-«'''(24-60)  \  -lEAj' 
which  may  be  written 


Henoe, 


H- 


(65) 


944  THEORY  OF  STRUCTURES. 

Subtract  (63)  from  (62).    Then 

2'|ir+^(5r'-5r«+2r*)i 

«(-fi;i)   ■ 

Eliminating  S,  between  (61)  and  (62), 

<--^i  (""-)s-«<-f:) } <"' 

Also,  by  (55), 

-S,-So+(  7, — to  H - M)'r/ - -P,  suppose (67) 

Eliminating  S^  between  (62)  and  (67), 

-P=S,=  (-*?-«^)|-u-7(r~') (68) 

Eqs.  (62),  (65),  (66),  and  (68)  give  the  values  of  H,  S,,  jS„  and  v 
Again,  the  maximum  bending  moment  M'  occurs  at  a  point  given  by 

•^-Oin  (57),  i.e., 

0-So+(^J?-u.)x-u'lx-(l-r)n (69) 

Subtract  (69)  from  (67).     Then 

-P,-S,-  {^-w)  (J-x)-tp'(I-*). 

Hence,  the  distance  from  the  loaded  end  of  the  point  at  which  the  bending 
moment  is  greatest  is 

l-X ^-;,TTV (70) 


V)+W'- 


HkH' 


Substitute  this  value  of  x  in  (57),  and,  for  convenienoe,  put 


,    »kH 
w+vf — -— -»m. 


■I     I 


Then 


M'=5 


RIB  OF  UNIFORM  STlFFffESS. 


■('-£) -^('-£)  •-?'(-£-')' 


945 


-,(. 


"•  -m.     w 


U-'    \ 
TV 


P'lvf- 


2/' 


"  "  '"  m'\     2  2t 

But  by  (62),  0  =  S„  +  «'' -:'"z_|;.;.    Therefore, 

TO  w'\     2/     2w»' 

Hence,  Af',  the  maximum  bending  moment, 

P' 

°„/    ^    ,    8AAA (71) 

As  before,  the  greatest  stress  (a  thrml) 

-ii^-f''')"^'''' (72) 

and  the  value  of  r  which  makes  ;,,'  an  absolute  r..  um  is  given  by  ^' =0. 
But  by  (71),  M'  involves  r'o  in  the  numerator  and  .^  in  the  denominator,  so 
that  -^  =  0  will  be  an  equation  involving  r'«. 

DivSlLtv  '  T:V'  ''=''.^^'*>'*'h  generally  gives  a  minimum  value  of  «,' 
s^  J  fa^^n^  '-1,  the  equation  reduces  to  one  of  the  thirteenth  order,  bufis" 
sfll  far  too  complex  for  u*.  It  is  found,  however,  that  r=i  gives  a  ll« 
approximation  to  the  a6so/«/e  maximum  thrust 

With  this  value  of  r,  and,  for  convenience,  putting. 

15/,    1 

By  (65), 

""itS'"'-?) (73V. 

By  (62), 

MiH^^-ff <«, 

By  (68), 

*  ^-2ir-^2;-7mf (75) 


Hi 


,'  I. 

1' 


946  THEORY  OF  STRUCTURES. 

By  (66,, 

^''•=24£7l("'^2)  — -ref <7«> 

By  (70), 

I  \  /       u\n-l     w'l 

("'+2)  — ^2 
By  (71), 

^,_8lW_n4l 

("'^2)— ^2- 

JVo<e. — If  the  rib  is  merely  supported  at  the  ends  but  not  fixed,  the  hori- 
zontal displacement  of  the  loaded  end  may  be  represented  by  ftH  (Art.  9). 
Thus  the  ierir  —  iiH  mast  be  added  to  the  right-hand  side  of  eq.  (15). 

Ex.  15. —  Parabolic  Rib  of  Unijorm  Stiffness,  hinged  at  the  Crown  and 
alxo  at  the  Ends. —  In  this  case  M  =  0  at  the  crown,  which  introduces  a  fourth 
equation  of  condition. 

By  (57), 

„     „  I      /8kH      \l'     wV/     1      \' 

which  may  be  written 

0  =  S,+  (-f-w)-^-w'l(r'-r--^) (79) 

Eliminating  S,  between  (79)  and  (62), 

~-u;  =  u''(-2r'+4r-l). 
Hence, 

//  =  ^(u;-u''(2r'-4r  +  l)| (80) 

Hy  (79), 

S„=~(^r'-4r-^l) (81) 

By  (G8), 

P=S,=^(r-l)» (82) 

By  (66), 


DEFLECTION  OF  AN  ARCHED  RIB. 
By  (70)  and  (82), 


l-x- 


By  (71), 
Whenr-i, 


w'l 
2«)'(r-l)>~4' 


AT-fV-.,.. 


947 

(84) 
(85) 

(86) 


8k\        2/'    '^o         8'     ^     ^^'— g' 

1      w'l'  yj'll 

'•""384  ^T;-        «nd        ^'-"M- 

These  results  agree  with  those  of  (73)  to  (78)    if  n-1 
In  general,  when  n-1,  '  ' 


tt +-2(5r» -5r*  4-2r»)  -  u) -t»'(2r» -4r  + 1), 
!)y  (65)  and  (80).    Hence, 

2r'-5r«+9r'+8r+2-0-(2r-l)(r-l)>(r'-2), 
a'ld  the  roots  are  r  =  J,  r  =  l,  r-  ±V^. 

.rJHr'^'  "°^  °"'^  Tu^^  ^^"^  expre'ssions  in  (86)  identical  with  the  corre- 
spending  expressions  of  the  preceding  article  when  n  -  i  or  1 

Again  the  intensity  of  thrust  is  greatest  at  the  outer  flange  of  the  loaded 
and  the  inner  flange  of  the  unloaded  half  of  the  rib,  and  is         ''^^'^^ 

I'   \z,w'     1  /       w'\  J 

The  intensity  of  tension  is  greatest  at  the  inner  flange  of  the  loaded  and 
the  outer  flange  of  the  unloaded  half  of  the  rib,  and  is 

8j,l7;-8-/fcr+2")r- 

The  greatest  total  horizontal  thrust  occurs  when  r=l,  and  its  value  is 

^(u'+t.-). 

i6.  Maximum    Deflection    of    an   Arched    Rib.-The    deflection 
must  necessarily  be  a  i  aximum  at  a  point  given  by  t=0.    Solve 


Hi 


948  THEORY  OF  STRUCTURES. 

for  X  antl  substituto  in  (IG)  to  fiml  the  dcfloction  y'-y;  the  deflec- 
tion   is   an   ab'iohitc  niaxiinuin  when   ^(2/'-?/)=0.     The  rcsuUing 

(•(juation  involves  r  to  a  hi{;h  power,  and  is  too  intricate  to  Ix'  of 
USI-.  It  has  Ix-en  found  by  trial,  however,  that  in  all  ordinary  oases 
the  absolute  inaxunuin  defieetion  occurs  at  the  middle  of  the  rib, 

when  the  live  loail  covers  its  ».hole  length,  i.e.,  when  J=^,  and  r=l. 

Kx.   U}.—Rih  of  Ex.  13.     For  convenience,  put  1  +  4'^>  =  *' 

Then,  by  (39), 

„     /'  w  +  w'     \')ttEli  ,Q_v 

"sk-i-^^si^ ^^^) 

Hy  (38)  and  (41), 

P  ,  ••*  ~  1        •">   it  Ell  ,,  ,c,o\ 

-.V„=j^,(..  +  u-')--T5--^=-M. (88) 

By  (36)  and  (38), 

.S„=-C.-^-^ (89) 

Hy  (30),  (38),  (80), 

r^-j,^{M^-3Mjj+'2M,j) (90) 

Hence,  the  maximum  deflection 

-  -J,  "'^=  -ElJo    i'-'l  "-7^)'^"  -F/32 

/'    ir  t-ir'.<i-l        ."i    ill'  ... 

°=384-l7;-  V^^T28^/^='^'-^"'''I^'^ ('"^ 

The  central  dcdcrlion  tl^  of  .i   uniform  straijiht   horizontal  Invim  of  the 
same  span,  of  the  sum"  .s»-ctioii  as  the  ril)  .-it  the  crown,  and  with  its  ends  fixed, 

is 

rl  ^    ''     "-—  (92) 

Hence,  ncRleclinf;  the  term  involving  the  temfx-rnture, 

d,--'d, (93) 

a 

Kx.  \7.—  Rihol  Ex.  11. 


DEFLECTION  OF  AN  ARCHED  RIB. 


940 


By  (65). 


By  (66)  and  (62) 


//  = 


P  IV  +  ur' 


(94) 


By  (30),  vai),  nn.l  ^95), 


(96) 


Hence,  the  maximum  deflection 

If  the  ends  of  the  beam  in  Case  I  are  free,  its  central  deflection 

5  IHw+u'') 


384       EI 


-d/. 


and 


«/.'■ 


n-1 


d' 


(98) 


Thus,  the  deflection  of  the  arched  rib  in  both  cases  is  less  than  that  of 
the  bi'am. 

Ex.  IH.— Arched  Rib  of  Uniform  Stiffne.is  fixed  rU  the  Ends  and  ctmnected 
at  the  Crown  with  a  Horizontal  Distributing  Girtlrr— The  load  is  transmitted 
to  the  rib  by  vortiiiil  struts  »u  tluit  the  vertical  (lispl.iccmr-iils  of  corrcsixindinji 
points  of  the  rib  and  jrinler  arc  the  same.  The  horizontal  thru.«t  in  the  loaded 
is  not  necessarily  e(|ual  to  that  in  the  unloadcil  division  of  the  rib,  but  the 
ex(v.s.s  of  the  thrust  in  the  loaded  division  will  ix-  borne  by  the  distributinf^ 
girder,  if  the  rib  and  girder  are  connected  in  such  a  manner  that  the  horizontal 
disnlacvmcnt    of  each  ut    the   crown  is  the  same. 

The  fornnilu'  of  Kx.  13  ;ire  applicable  in  the  pre-H-nr  case  wth  the  miKli- 
ficHtion  that  /,  is  to  include  the  moment  of  inertia  of  the  girder. 

The  maximum  thrust  and  tension  in  the  rii)  are  ^'ivcn  by  cquutions  (64i 
and  Mm).  ' 

Let  t'  lie  the  depth  of  the  pirder,  A'  it.«*  Kectional  an'a. 

The  >:reatesi  thrust  in  the  girder   = — ^-   +       -  (nn\ 

The  greatest  tension  in  the  girder-  .    /.   .     .     .  (100^ 

//  and  ilf,  bc-ing  given  by  e<iuution.s  (6<i)  and  (67),  rei<|)ePtively. 


(    j 


I  li 


950 


THEORY  OF  STRUCTURES. 


The  girder  must  have  its  ends  so  supported  as  to  be  capable  of  transmitting 
a  thrust. 

17.  Stresses  in  Spandril  Posts  and  Diagonals. — Fig.  936  represrnts 
an  arch  in  wliicli  the  spandril  consists  of  a  series  of  vertical  posts 
and  diagonal  braces. 


Fig.  936. 

I.iet  the  axis  of  a  ciirvo<l  rib  l)o  a  parabola.  The  arch  is  then 
equilibrated  undi-r  a  iniiforniiy  distributed  load,  and  the  diagonals 
will  Ik'  onl\     idled  into  |)lay  under  a  passing  load. 

Ij't  X,  y  Ix'  the  coordinates  of  any  point  F  of  the  parabola  with 
respect  to  the  vertex  C.     Then 

Let  the  tangent  ;it  F  meet  Cli  hi  L,  and  1..0  horizontal  BE  in  (7. 
LetZiC'  =  A'.     Tiien 

HI.    nC-CL=  BC-CX  =  k'-y. 

T/'t  .V  Ih"  the  total  iMiiiil)er  of  panels. 

Consider  any  iliagonal  Kl)  In'tween  the  Hth  and  (n+l)th  posts. 

Let  I"'  U'  the  greatest  panel  lire  load. 

The  greatest  conipression  in  K!)  cMTurs  when  the  passing  load 
is  eone<'!itrated  at  the  first  ii-\  i>anel-points. 

Imagine  a  vertical  section  a  little  on  the  left  of  FF. 

Tiie  portion  of  the  frame  on  tlir-  right  of  this  section  i.s  kept  in 
eriuiiibriuin  by  the  reaction  l{  at  /*,  and  by  the  stresses  in  the  three 
menilKTs  met  bv  the  s«'cant  plain'. 

Taking  moments  about  (!. 

/>f;/i'cos/?     RAG, 
D  being  the  stress  in  l)K.  and  0  th«  angle  DFF. 


U 


Now, 

Also, 
jind  hence 


CLERK  MAXWELL'S  METHOD. 

w'  Ji(n  —  1) 


951 


R^ 


X 


x  +  GB_k'  +  y     ^,„    k'x-xy 
GB       k'-y'     '^~      2y     ' 


k'x  +  ly 


GE=GB  +  x^^-:r^'i,    and    GA^-^  + 


2y 


f_     k'x  —  xy 
2^' 


Hence 


w' n(n -  I)  ly  + k'x-xy 


The  stresses  in  the  counterbraces  (shown  by  dotted  lines  in  the 
figure)  may  be  obtained  in  the  same  manner. 

The  greatest  thrust  in  EF  =  w'  +  w. 

The  greatest  tension  in  EF  ^  D  cos  0-w,  w  being  the  dead  load 
upon  EF. 

If  the  last  expression  is  negative,  EF  is  never  in  tension. 

i8.  Clerk  Maxwell's  Method  of  determining  the  Resultant  Thrusts 
at  the  Supports  of  a  Franie>^  Arch.— ]>et  Jt<  be  the  change  in  the 
length  ,s  of  any  menilx'r  of  the  frame  under  the  action  of  a  force  P, 
and  let  o  be  the  sectional  area  of  the  member.    Then 

P 

the  sign  depending  upon  the  character  of  the  stress. 

As.«ume  that  all  the  meinlx-rs  except  the  one  under  consideration 
are  jjerfectly  rigid,  and  let  J/  Ik>  the  alteration  in  the  span  /  corre- 

spoiuling  to  J.S.     The  ratio  --  is  eciual  to  a  constant  m,  which  depends 
only  upon  the  geometrical  form  of  the  frame. 


1 


n 


Therefore 


ha 


Again,  P  may  be  supi)osed  to  consist  of  two  parts,  viz.,  /.  due 
to  a  horizontal  force  //  between  the  springings,  and  /a  due  to  a  ver- 
tical force  V  applied  at  one  springing,  wliile  ihe  other  is  hrmly 
secured  to  keep  the  frame  from  turning. 


,4 


952  THEORY  OF  STRUCTURES. 

By  the  principle  of  virtual  velocities, 

/i     Jl 


H^Ts-^"'- 


h   :. 


Similarly,  y  is  equal  to  some  constant  n,  which  depends  only 
upon  the  form  of  the  frame.     Also 


P  =  /i+/2  =  w//  +  nr. 
Therefore 
Hence  the  total  change  in  /  for  all  the  memlx'rs  is 


ha 


If  the  abutments  yield,  let  1'Jl^/tH,  /t  bnng  some  coefficient 
to  be  determined  by  exix'riment.     Then 


H 


-h'io) 


If  the  abutments  are  immov;il>le,  I'Jl  is  zero,  and 


(C) 


//=- 


4-^5) 


:  :  '.    (D) 


r  is  tlie  same  ,is  (lie  enrp'simnili:!'  vt  ;icti()!i  at  the  end  of  a  pirder 
of  the  ^aiiie  s|i:iii  and  siiiiilail'.  loiidi'd  The  nM]uired  flirust  istlie 
nsiiltiuit  of  //  ;iiiil  1  ,  ;iiul  the  stri>-<  its  each  inemlHT  may  \)o  com- 
puted crapliieallv  or  iiy  tiie  iiietliod  (if  moments.     In  any  particular 

e,''^c   |ii(ii'ced  ar-   IciiiiAVS. 

( 1 1   I'repan'  tables  of  (he  values  of  m  and  v  for  each  memlxT. 


EXAMPLES. 


953 


(2)  Assume  a  cross-section  for  each  member,  based  on  a  probable 
assumed  value  for  the  resultant  of  V  and  //. 

(3)  Prepare  a  table  of  the  value  of  m^~  for  each  member,  and 
form  the  sum  2'{m^-^). 

(4)  Determine,  separately,  the  horizontal  thrust  between  the 
springings  due  to  the  loads  at  the  different  joints.  Thus  let  t'l,  fa  be 
the  vertical  reactions  at  the  right  and  left  supports  due  to  any  om 

of  these  loads.     Form  the  sum  l(ninV-~),  using  i-,   for  all  the 

memlx'rs  on  the  right  of  the  load  and  ?2  for  all  those  on  its  left. 
The  corresponding  thrust  may  then  be  found  by  eq.  (C)  or  eq.  (D), 
and  the  total  thrust  //  is  the  sum  of  the  thrusts  due  to  all  the  weights 
taken  separately. 

(5)  Repeat  the  process  for  each  combination  of  live  and  dead 
loail  so  as  to  find  the  maximum  stresses  to  which  any  member  miiy 
be  subjected. 

(6)  If  the  assumed  cross-sections  are  not  suited  to  these  maxi- 
mum stresses,  make  fresh  assumptions  and  -epeat  the  whole  calcu- 
lation. 

The  same  method  may  be  applied  to  determine  the  resultant 
ten.'sions  at  the  supports  of  a  framed  suspension  bridge. 

Xotc.—  Tlte  jormnhv  jor  a  paraMic  rib  may  he  applied  ivithout 
material  error  to  a  rib  in  the  form  of  a  sefjmcnt  of  a  circle.  More  exact 
formula-  may  he  obtained  for  the  latter  in  a  manner  jyrerisely  similar 
to  that  described  in  Exs.  13-17,  hut  the  inteiirations  ivill  he  much  sim- 
plified by  usintf  i>olar  coordinates,  the  centre  of  the  circle  beiny  the  pole. 


EXAMPLES. 

I.  The  nrrh  roprow-ntrd  in  tho  fieurc  is  const rucfrd  of  mafwinry  woighinR 
lbs.  cf.  Tli.>  sp:in  is  40  ft.,  the  ri.si"  S  ft.,  tho  depth  of  irMsnnrv  nUn;- 
centre  .1  fl..  tile  thickness  of  the  iibutnients  fi  ft. 
Tlio  centre  of  resistance  iit  /t.tis  nt  tiie  iniiiiile 
[Hiint,  iiii.l  ;it  the  i<ey  is  1  ft.  Inflow  H.  IVdllcc 
(fl)  the  resuhiint  pressun-  in  the  verticu!  joint  at 
the  key,  (h)  the  resuit:int  pressun'  in  the  horizontal 
joint  at  the  spnntjmiu',  (n  tlie  niaxinuun  stress  in  p,j.    g;,^ 

the  vertical   hne  coincidini;  with  the  .xidc  of  the 
abutment.    .!««.  (u)  23  :ttK)li)..;  ^/,;  .js.ruu  lbs. :  ^,  !  !::J0  Ibs./sq.  in.  if  9-J. 


thr> 

B 


954 


THEORY  OF  STRUCTURES. 


.2.  A  masonry  arch,  of  48  ft.  span  and  17  ft.  rise  and  having  an  intrados 
consisting  of  two  platie  faces,  springs  from  abutments  0.2  ft.  thick  and  with 
vertical  fares.  The  outer  thrnnts  of  the  abutments  are  produced  to  meet  the 
extrados,  which  is  horizontal;  the  depth  of  the  key  is  3  ft.  and  the  specific 
weipht  of  the  masonry  is  LW  lbs.  jxr  cubic  foot.  The  centre  of  resistance 
at  the  sprinsjiiif;  is  at  the  middle  j)oint  and  one  foot  below  the  extrados  at  the 
crown.  Find  the  resultant  pressures  at  the  crown  and  springing.  Also  find 
the  maximum  stresses  of  the  vertical  joint  ahgning  with  the  inside  of  the 
abutment.      Am.  26,400  lbs.,  65,577  ii>s. ;  2646  Ibs./sq.  ft. 


3.  The  intrados  of  an  arch  of  100  ft.  span  and  20  ft.  rise  is  the  segment  of 
a  circle.  The  arch  ring  has  a  uniform  thickness  of  3  ft.  and  weighs  140  lbs. 
per  cubic  foot;  the  suiH'rincumbent  load  may  be  taken  at  480  lbs.  per  lineal 
foot  of  the  ring.  iVtermine  the  mutual  pressures  at  the  key  and  springing, 
their  points  of  aj)])lication  Ix'ing  2  ft.  and  1 J  ft.  res|x?ctively  from  the  intrados. 
Also  find  the  curve  of  the  centres  of  pressure.       .Ins.  56,546  lbs. ;  76,676  lbs. 

4.  Assuming  that  an  arch  may  b«'  divided  into  elementary  portions  by 
imaginary  joint  jjanes  parallel  to  the  direction  of  the  load  upon  the  arch,  find 
the  limiting  span  of  an  arch  with  a  horizontal  upper  surface  and  a  parabolic 
soffit  (latus  rectum  =  40  ft.),  the  depth  over  the  crown  being  6  ft.  and  the 
six'cific  weight  of  the  load  120  lbs.  |x'r  cubic  foot;  the  thrust  of  the  crown  is 
horizontal  ( =  P)  and  4  ft.  above  the  soffit. 

Atix.   l.-.A'*  +A''(21,600-  /')=-  120P,  2X  l)eing  the  s]r.<n. 

5.  Fig.  !)I5S  represents  diic  h;ilf  of  a  masonry  arch  of  8  ft.  rise  and  weighing 
120  ll)s.  |)er  cubic  foot.      The  centres  of  resistance  S  and  T  are  at  the  middle 

|>(>int  of  -l/iand  at  1  ft.  Ix'low  D.  Find  the  resultant 
tlirustsat  .Sand  T,  and  determine  the  maximum  intensity 
of  stress  in  the  vertical  joints  .U' and  AV-'. — Ans.  19,945 
lbs.;  (iSSOlbs.;  1251  Ibs./sq.  ft.;  4587lbs./sq.  ft. 

6.  \  masonry  arch  of  90  ft.  span  and  24  ft.  rise, 
with  a  p.'irabiilic  intrados  and  a  horizontal  extrulos, 
springs  from  abutments  with  vertical  faces,  the  out8i(i(> 
faces  l)eing  carried  up  to  meet  the  extrados.  The  depth 
of  the  masonry  at  the  key  is  6  ft.  The  centre  of  pressure 
is  2  ft.  from  the  extrados  at  tli(>  key  and  at  the  mi('dle  of  the  joint  at  the 
si)ringiiic.  The  masonry  weighs  l.")Oi  lbs.  ]X'T  vnU'iv  foot  ;  width  of  abutment  - 
12  ft.     Find  the  resultant  i)ressurcsat  the  key  and  at  the  spri.iging. 

.1/I.S-.  83,592  lbs. ;  177,10011)8. 
7.  A  masonry  anh  for  a  si)an  of  40  ft.  and  a  rise  of  10  ft.  springs  from 
abutments  with  vertii'al  faces  and  10  ft.  thick.  The  masonry  has  a  dej)th 
n(  ■\\  ft.  at  the  crown,  is  li'Vel  from  aliutment  to  abutment,  and  weighs  l.W  lbs. 
per  <'nbic  foot.  The  intrados  is  a  parabola.  The  centres  of  resistance  at  thi" 
springing  and  at  the  crown  are  4i  ft.  from  the  inside  face  of  the  abutment  and 
2  ft.  above  the  crown  res(H'('tively.  Find  the  resultant  pressures  at  the  crown, 
at  the  springing,  uiid  in  the  veriical  aligning  with  the  inside  face  of  au  abut- 
ment. An.'<.  19,168  lbs  ;  44.3.54  lbs.;  27,702  lbs. 


<-- a-« 

»--li-» 

I 

/I:. 

B 

~S" 

Fig    a'JS, 


EXAMPLES. 


955 


8.  A  masonry  arch  for  a  span  of  40  ft.  and  a  rise  of  10  ft.  sprines  from 

3J  ft.  at  the  crown,  .s  level  from  abutment  to  abutment,  and  weighs  150  lbs 
per  cubic  foot.  The  intrados  is  a  circular  arc.  The  centres  of  resSfanceft  t^l 
springing  and  at  the  crown  are  4J  ft.  from  the  inside  f^rofSZent  Id 
2  ft.  above  the  crown  respectively.  Find  the  resultant  pressures  at  the  ci^l  at 
the  springing,  and  in  the  vertical  aUgning  with  the  in^de  face  of  an  abX'nt! 

9.  The  soffit  of  an  arch  of  30  ft.  span  and  12  ft.  rise  is  a  transformed  catenary. 
The  masonry  rises  12  it.  over  the  crown,  an.l  the  specific  weight  of  thTS 
upon  the  arch  may  be  taken  at  120  lb,,  per  cubic  foot  Determine  the  d.re.t  on 
and  amount  of  the  tlirust  at  the  springing.  Ans.  32,408  lbs.     6?°  10 

10    A  concrete  arch  has  a  clear  .spring  of  15  ft.  and  a  rise  of 'lO  ft  •   the 

height  of  masonry  over  crown  =  Io  ft.;   the  weight  of  the  concrete"  144  lbs 

Zt    X  T .  T'''^''''  ''"'  ^■•-«^--«'d  catenary,  the  amount  and  d  nttbn 

of  the  thrust  at  the  springing,  and  the  curvatures  at  the  crown  and  sp  rng  ng 

Am.  TO=fi.83  ft.-  20,773  11. ■  i  =  71=9'-  3  n  ft    V-  ofl^' 

I..  Determine  the  transformed  catenary  f.  r'  an  arch  of'eO  ft'.'  'span  and 
4o  ft.  rise,  the  m^onry  rising  18  ft.  over  the  crown  and  weighing  l^o  lbs  ,ir 
cubic  foot.    Also  find  the  amount  and  direction  of  the  thrust  at  the  abu.n;;nu! 

'4n«-  w  =  15.586;   116,290  lbs:  75-29'. 

12  A  concrete  arch  for  a  span  of  15  ft.  and  a  rise  of  0  ft.  has  a  depth 
of  9  ft    of  masonry  over  the  crown.     The  concrete  weighs  144  lbs   per  cubic 

Xndi^'""'"  '  "  *""^'"'"^'  'T''''  ^"'^  '""^  ^^-*^  -»  the  cr^;n  and 
^     *'"'^-  ^'"^  »"  =  6.82  ft. ;  6695  lbs. ;  13,552  lbs. 

tnJ^'^i"  ''"""^*''  """'^  "^  ^t  ^*-  'f'''"  """^  '^  ^*-  "'*  ^-eiRhs  144  lbs.  per  cubic 

^j:^=z^  -l  :s:;rs.e^:rT/r  -^--^ 

«..„.     If  the  temperature  r.^  .50o,  find  the   corresponding  horizontal 
^"'^"^-  ,  ^"■''-  -l-^O/  lbs.  If  pmnod  ends;  .39.90/  if  fived  ends 

15.  A  concrete  arch  has  a  clear  spring  of  75  ft.  and  a  rise  of  37J  ft  •  the 
height  of  the  ma.sonry  ov^r  the  crown  is  25  ft.  and  ,hc  weight  of  the 'con! 
Crete  ,s  144  lbs.  per  cubic  foot.  Determine  the  transformed  catenarvtl 
amount  an.l  .hrcct.on  of  the  thrust  at  the  springing  and  the  curvatu.^  at'  he 
crown  and  s|)ringing.  >aiuixa  m  ine 

I'/.s-.  TO  =2.3.9.34;  214,0.53  lb.s.;  07°  20'-  •"'914  ft  •   160  ft 

16.  .\  3-pin  arch,  100  ft.  span.  20  ft.  rise,  loaded  with  200o"lbs.  iK-r'horizontal 

IS  r n;s.'r  r  ;.:rr'.  '^^ ''- '- ''  ^"-^-  '^  '^^-  ^-  ^^  ;= ; 

Divide^the  load  into  eight  parts  and  draw  the  line  of  resistance.     Also  deter- 
mme  the  horizontal  thrust  and  the  maximum  B.M.  at  any  point  of  the  arcb 

x7.  A  semicircular  rib.  pivoted  at  the  crown  and  springings.  i,  loaded 


956 


THEORY  OF  STRUCTURES. 


uniformly  per  horizontal  unit  of  length.  Determine  the  position  and  magni- 
tude of  the  maximum  bending  moment,  and  show  that  the  horizontal  thrust 
on  the  rib  is  one  fourth  of  the  total  load.       Aiu.  ^wr'  at  .8f)6r  from  support. 

i8.  Draw  the  linear  arch  for  a  semicircular  rib  of  uniform  section  under  a 
load  uniformly  distributed  |x'r  horizontal  unit  of  length  (a)  when  hinged  at 
both  ends ;  (b)  when  liiuged  at  both  ends  and  at  the  centre ;  (c)  when  fixed  at 
both  ends. 

Am.  (a)  ,\  parabola  «  being  ||r;   (6)  a  parabola  through  the  three  hinges; 

19.  A  semicircul.ir  rib  of  "JS  ft.  span  carries  a  weight  of  \  ton  at  10  ft.  and 
a  weight  of  J  ton  at  21  ft.  (measured  horizontally)  from  the  left  support.  Find 
the  thrusts  and  shears  at  the  centre  of  the  rib  and  at  the  point  at  which  the 
weights  are  concentrated  when  both  ends  are  hinj^ed. 

Am  Thrusln,  .295  and  .19  ton,  .402  and  .148  ton;  shcar.t,  .183  and  .04 
ton,  .313  and  .13  ton. 

JO.  An  arch  (Fig.  939)  of  r,2  ft.  span^  and  lOJ  ft.  ri.se  has  a  depth  at  the 
crown  of  2}  ft.  and  at  the  springing  line  of  3§  ft.  The  loads  pj-r  foot  of 
breadth  beginning  with  a  are  23.04,  19.45,  l(i.35,  13.94,  12.01,  10.35,  8.98. 


7.93,  0.32,  5.70,  5.52, 5.39  cwts. 


Find  the  horizontal  thrust  per  foot  of  breadth, 

/*n,v.  129  cwt. 


Fig.  0:?9 


Fio.  910. 


21.  The  arch  (Fig.  940)  has  a  clear  span  of  90  ft.  and  radius  of  50  ft. 
Thickness  of  arcli-ring  is  4  ft.  Draw  the  line  of  resistance,  the  load  being  500 
lbs.  i«T  foot  run. 

22,  Determine  the  stability  of  a  segmental  arch  of  40  ft.  span  and  25  ft. 
radius  (Fig.  941);    the  loads  in  hundredweights  l)eing  as  shown.    There  is 


also  a  concentrated  load 
Dejith  of  arch      '_'  ft. 


if  Ji  tons  ]K'r  foot  of  width  at  10  ft.  from  the  centr«. 


EXAMPLES. 


.04 


957 


23.  A  semi-elliptir  rib  (axes  2a  and  26)  is  pivoted  at  the  springings.  Find 
the  position  and  magnitude  of  the  maximum  bending  moment,  the  load  being 
uniformly  distributed  per  horizontal  unit  of  length. 

«'a'172rt'-2.36» 

ir2     4a  +6^  -''**""'^"- 


Ans. 


How  will  the  result  be  affected  if  the  rib  is  also  pivoted  at  the  crown? 

Ans.  ^-  at  .8G6a  from  centre. 

wei^'J;  Pa^'"Tfff  ^T^^"r  "'^  "^  ^^  ^'^  ^P'^"  ""^  ^5  ft.  rise  carries  a 
N^oight  P  at  >o  ft.  from  the  left  support.     Find  the  horizontal  thrust. 

25.  A  pin-ended  parabolic  arch  of  100  ft.  span  an.l  2.5  ft.  rise  is  Icted  u^on 
by  a^horizontal  force  Q  at  2.-,  ft.  from  the  left  support.     Find  The  holla; 

«  J^'i!"  r  Tl^^'"  "''^  "^  ^  ^'-  ^P''"  ""'I  10  ^f-  "««.  hinged^aT'b;;trends 

th^'fownt  loTt"  r"??;T'  r  K^  '"'"'  "•'^"^'^  '>""-"^'  distance  frot' 
the  crown  -s  10  ft.     Find  the  total  thrust  along  the  axis  of  the  rib  on  each 

r=  .«m '"'"'    ' '''' '  ''""^''  °' "'° '""  ^'^^  ""^^^  ^^-j^-" - 

2,    SoK^'th'^^'"^^  "i^^"''^ "^"^  -^'^^  -^^"^  ♦""'''  -^  '^'"S  12,500  tons/sq.  in. 

27.  holve  the  preceding  example,  a.ssumingl)oth  ends  to  be  fixed. 

,0     4  u  J     u     .  ,  '^"*-  ••03=F.41.m/ and  .71.5^.297/ tons 

i,l„nH  t  r^.  "  /u'*'  '^  P'"""''""^  '*-^'^'  "^  100  ft.  span  and  12J  f t  r  se 
-s  loaded  with  ton  at  the  centre  and  1  ton  at  20  ft.  from  the  centre,  ineasS 
horizontaly  Determine  the  thrusts  and  shears  along  the  rib  at  thcTatter 
point,  and  show  how  they  will  be  affected  by  a  change  of  lio°  F  from  th! 
mean     the  coemdcnt  of  linear  expansion  lx,ing  .OOli  forlSO^F.     Take  bl 

"-.Sl^^'m'^m ''■'";?'"  .109.^/and2.,S2,  .1018/  tons;  ^wtoLt 
.U_-.)/ ana  ..{JT  .0126/ tons,  £  being  12,500  tons/s()  in 

?■  ^i?  ^^'  Pf  ;?'"»«  e^^'nple,  assuming  both  ends  to  1x3  fixed. 
.024/'!!'  ^•'"''''''^^^  3.2^.2/  tons;  shear.,  .024 ^  .09/ and  .38 ^ 

30.  A  parabolic  arched  rib  hinged  at  the  ends,  of  Gl  ft.  span  and  16  ft 
ISO  is  loaded  with  1  ton  at  each  of  the  points  of  division  of  eight  equal  ho   ^ 

f  (.0    1.    f.on.  the  mean  ten.,K>rature.     Also  find  the  maximum  flange  .stress,^! 
the  nb  l.>,ng  of  double-tee  section  an.l  12  ins.  deep  throughout.     fcoeSn 
of  linear  expansion  per  1°  F.  =  l  -144000.)  v^oemcieni 

An...  3..Wf  .036.5/  tons;  J(3.94St  .0365/)±i-(60-.o55/)    tons,  E  U-Ing 
12,S0()tons/s<i.  in. 

31.  The  axis  of  an  arched  rib  of  .50  ft.  span,  10  ft.  rise  and  hinged  at  both 
ends  IS  a  paralK,la.     Draw  the  linear  arch  when  the  rib  is  loaded  with  t  to 

^n«.  // =3.176  tons;  ?:^-y±lZ?. 


95S 


rnSORY  OF  STRUCTURES. 


3.3075     292 
.1         ^ 


If  the  arch  is  loaded  so  as  to  produce  a  stress  of  10,000  lbs.  per  square  inch 
in  the  metal,  show  that  the  rib  will  deflect  .029  ft.,  E  being  25,000,000  lbs. 

32.  Solve  the  preceding  example,  a.ssuming  both  ends  to  be  fixed. 

Ans.  // =3.3075  tons; 

33.  A  steel  parabolic  arched  rib  of  50  ft.  span  and  10  ft.  rise  is  hinged 
at  both  ends  and  loaded  at  the  centre  with  a  weight  of  12  tons.  Find  the  hori- 
zontal thrust  on  the  rib  when  the  temperature  varies  00°  F.  from  the  mean, 
and  also  find  the  maximum  flange  stresses,  the  rib  being  of  double-tee  section 
and  12  ins.  deep.  ,        „  teI        H  H 

^  ^««.  //  =  11J^T    „    — •   — ±2010^ 

936960'    A  '^''"'•'^• 

34-  Solve  the  preceding  example,  assuming  both  ends  to  be  fixed. 
Ans.  H  =  U^gT-  '^' 


^±j(G24W-4.32). 


493056     ..     . 

35.  A  parabolic  'ib  of  48  ft.  span  and  12  ft.  rise  carries  a  weight  of  1  ton 
at  the  centre  and  at  two  points  each  12  ft.  (measured  horizontally)  from  the 
centre.     The  rib  is  pin-ended. 

Determine  the  horizontal  thrust  on  the  rib  and  draw  the  linear  arch.  Find 
the  thrust  and  shears  at  the  point?  at  ^vhich  the  weights  are  concentrated. 

Ans.  1.8945  tons;  thrusts,  1.89,  1.72,  and  2.2  tons;  shears,  .395  and 
.474  tons. 

36.  An  arch  with  fixed  ends  of  100  ft.  span  and  25  ft.  rise  carries  a  weight 
P  at  25  ft.  from  the  left  support.  Determine  the  reactions,  the  horizontal 
thrust,  and  the  B.M.  ordinates,  at  the  ends  and  at  P. 

.\ns.  .844P;  .156P;  .52G5P;   - 10',  7'.78,  .30'. 

37.  In  the  preceding  example  determine  how  the  results  are  modified  if  P 
is  replaced  by  a  horizontal  force  Q. 

Ans.   +.1053Q;   -1053(?;  .G33(?;  8'33,  7'.97. 

38.  An  arch  with  fixed  ends  of  100  ft.  span  and  25  ft.  rise  is  constructed 
of  metal  having  a  modulus  of  elasticity  of  28,000,000  lbs.  per  square  inch,  and 
a  coefficient  of  expansion  of  .00<J0055.  If  the  temperature  rises  .^O",  determine 
the  ordinates  of  the  linear  arch  at  the  sj)ringings  and  at  the  centre,  and  also 
find  the  horizontal  thrust.  Ans.  16?  ft.;   :l:85.10.-)/. 

39.  An  arched  ])arabolic  rib  of  04  ft.  span  and  8  ft.  rise  carries  a  load  of 
2  tons  at  the  centre  and  at  S  and  10  ft.  from  the  centre  measured  horizontally. 
Determine  the  axial  thrusts  and  shears  at  the  points  at  which  the  weights  are 
concentrated,  and  also  the  nhx.  max.  H.M.,  (a)  when  the  arch  is  pin-ended; 
(6)  when  the  arch  is  fixed  at  tM)tli  ends. 

Ans.  (a)  r/ir(/,s^.,  s..-)3,  8.25;  8.245;  8.09,  8.25  tons;  shears,  1.69,  ..32;  .75, 
1.75;   .00,  .75  tons;   Il.M.s  88.934,  28  ft.-tons. 

40.  If  a  ilat  parabolic  arched  rib  of  00  ft.  span  and  lOJ  ft.  rise  is  loaded 
at  10  ft.  from  the  centre,  measured  horizontally,  with  050  lbs.,  draw  the  linear 
arch  and  fmd  the  H..M.  and  shears  at  the  point  at  which  the  weight  is  con- 
ceiitiated,  (<()  if  pin-ended;  (h)  if  v\mU  are  fixed.  What  weight  at  the  centre 
of  the  rib  will  give  the  same  horizontal  thrust? 

Am.  in)  2782  ft.-lbs.;  208  and  385.2  lbs.;  565.0  lbs.;  {b)  2247  ft.-lbs.: 
244  and  ;M2  lbs.;  514  lbs. 


m. 


EXAMPLES. 


059 


41.  A  parabolic  arched  rib  of  80  ft.  span  and  13J  ft.  rise,  with  both  ends 
fixed,  carries  three  weights  of  2  tons,  4  tons,  and  6  tons  at  10,  20  and  30  ft 
from  one  end.    Draw  the  equihbrium  polygon  and  determine  the  thrust  and 
shear  on  each  side  of  the  point  at  which  the  6-ton  load  is  concentrated 

Am   H^m.UH  tons;  thrusts,  11.3.5 and  10.8  tons;  shears,  2.4  and  3  5  tons 

42.  A  semicircular  arched  rib  of  40  ft.  span  is  loaded  at  the  centre  and  at 
two  points,  each  12  ft.  from  the  centre,  measured  horizontally,  with  a  weight 
of  1  ton.  Find  the  axial  thrusts  on  the  rib  at  the  centre  and  at  the  points 
where  the  weights  are  concentrated,  both  ends  being  pinned 

Ans.  Thrusts,  79,  1.563,  and  .63  tons;  shears,  0,  .74  and  .49  tons. 
A^^'  r!^c       *''l"'''^""'"  P«'y«on  for  a  flat  parabolic  arch,  hinged  at  both 
ends,  of  100  ft.  span  and  20  ft.  rise,  loaded  with  a  weight  of  2  tons  at  25  ft 
measured  horizontally  from  the  centre.      Ans.    ?=26.9  ft  •  W  =  1  395  tons   '' 

44.  A  flat  parabolic  arched  rib  of  100  ft.  span  and  18  ft.  rise  is  hinge  'at 
the  springing  and  carries  a  load  of  4000  lbs.  concentrated  10  ft.,  measured 
horizontally,  from  the  centre.  Find  the  horizontal  thrust  on  the  rib,  and  also 
find  the  axial  thrust  and  shears  at  the  point  at  which  the  load  is  concentrated. 
Determine  the  points  at  which  the  B.M.  is  nil. 

iQ  •^-"'■J*151V?'";  ""■""^'  ^"^^  ''"'*  ^^  ^^^■'   «*^«"'  1800  and  2200  lbs.; 
ly.oo  and  46.24  ft.  from  end. 

45.  A  semicircular  arched  rib  of  36  ft.  span  is  loaded  with  1  ton  at  the 
crown.     Find  the  horizontal  thrust  on  the  rib  and  draw  the  equilibrium  polygon 
assuming  that  both  ends  are  fixed.         .4ns.  «  =  24',  y,  =2'  ^y  U  =  j'V  tons 

46.  A  flat  parabolic-arch  rib  of  100  ft.  span  and  18  ft.  rii'  is  fixed  at  the 
springing  and  carries  a  load  of  4000  lbs.  concentrated  10  ft.,  measured  hori- 
zontally, from  the  centre.     Find  the  horizontal  thrust  on  the  rib,  and  also 
find  the  axial  thrust  and  shears  at  the  point  at  which  the  load  is  concentrated 
Determine  the  three  points  at  which  the  B.M.  is  nil 

.8.«"^r.n';80  ";'fe„r  ^'^  ""'^    *"''  ""'  "^  ^'  ""'^ 

r  olfi"  ?^''^-'  ^^^  ""^"''  ^^"^  ^"''  *  ^""^  parabolic  arch  with  both  ends  fixed 
of  22J  ft.  rise  and  90  ft.  span,  loaded  at  the  centre  and  at  one-quarter  span 
^^•lth  a  weight  of  1  ton  If  both  ends  were  hinged,  what  should  be  the  weight 
at  the  centre  and  at  the  <,uarter  span  to  produce  the  same  horizontal  thrust' 
.947  tons^'^  "         '  ^^°  ^''••**'  «.  =1.344 and  ft,  =  .656  tons;  //  =  1.46  tons'; 

48.  Find  the  skin  stress  due  to  change  of  curvature  in  a  two-hinged  arch 

tr'S  TrT'  °^  '*'  "'™  ^'^i  '"''^^''  ''^''^  P'-*'d"'"'^«  ^  ">««"  compressive 
stress  of  6  tons  per  square  inch.     (£  =  12,000  tons  per  square  inch.)    Span 

5o0  ft.;  rise,  114  t.;  depth  of  rib,  15  ft.  Also  find  the  deflection  due  to  a 
stationary  test  load  which  produces  a  further  mean  compressive  stress  /.  of  one 
ton  per  square  inch.  ' 

49.  A  pin-ended  arch  of  100  ft.  span  and  25  ft.  rise   is  in  the  form  of  a 
circular  arc.      Pind  the  horizontal  thrust  due  to    (a)  a  weight  of  100  lbs 
at  _.j  ft.  from  th;^^  left  support;    (6)  a  horizontal  for.-.-  of  IQO  lbs.  .at  the  same 
''"'"'    II  11     1.  ,  "*'""'•   (")<"'"  lbs.;   (?))  55.7  lbs 

endfare  fLxJd'  '''         '"  '''"  '''^'''^'"^  ''"''"P''-'  ^  "'"'^''^"^  '^  both 

Ans.  (a)  54.0  lbs. 


a 


960 


THEORY  OF  STRUCTURES. 


51.  Draw  the  pqiiilibriiim  poIvRon  for  a  parabolic  arch  of  100  ft.  span  and 
20  ft.  rise  when  loaded  with  weights  of  3,  2,  4,  and  2  tons,  rcopcctivdy,  at  the 
end  of  the  third,  sixth,  cij^hth,  and  ninth  divi.sion  from  the  left  support,  of  ten 
e<Hial  horizontal  divisions.  (N'eglect  the  weight  of  the  rib.)  If  the  rib  con- 
si-;f  of  i\  web  nntl  of  two  flanges  2J  ft.  from  centre  to  centre,  determine  the  maxi- 
mum llaiige  stress.  Find  the  flange  stresses  at  the  ends  of  the  rib,  and  also 
.■\t  the  points  at  which  the  weights  are  concentrated.     Both  ends  are  absolutely 

fi^f"'-                  A               ,  n«              -j^Q.     "    co«^4  *         8.793.3  ,  24.tO.25 
Am.  ?/,  =  1.9S;  !/j=  -3.434;  //  =  6.8644  tons;  — j — ± -. — , 

7.2070     743.4    6.96    620.28    7.6147    3236.76    8.1494    310.3.56 
A     "^    I    '    A    -"'  I     'A      *      /      •      A      *       /      • 

53.  .\  parabolic  arched  rib  of  100  ft.  span  and  20  ft.  rise  is  fixed  at  the 
spriiigings.  The  uniformly  distributed  load  upon  one  half  of  the  arch  is  100 
tons,  and  ufion  the  other  200  tons.  Find  the  bending  moment  and  shearing 
force  a  I  2.")  ft.  from  each  end. 

Am.  B.M.-|//+»Va,-SW  +  v»,u      S.F.  87i-.4W,-112J +.4H. 

53.  .\  wTought-iron  parabolic  rib  of  90  ft.  span  and  16  ft.  rise  is  hinged  at 
the  (wo  abulnunts;  it  is  of  a  double-tee  section  uniform  throughout,  and  24 
in.  deep  from  centre  to  centre  of  the  flanges.  Determine  the  compression  at 
the  centre,  and  also  the  position  and  amount  of  the  maximum  bending  moment 
((;i  wlieii  a  load  of  4S  tons  is  concentrated  at  the  centre;  (6)  when  a  load  of 
90  tons  is  uniformly  distributed  jK'r  horizontal  unit  of  length. 

IX'termine  (r )  the  deflection  of  the  rib  in  each  case. 


Ans. 


.-)0.2.-)     302SS  „  „  .  , 

(a)  — 1 J — ;  max.  B.M.  is  at  crown  and 


252  ft.-tons. 


{h) 


S0.4 
.1  ■ 


54.  Design  a  parabolic  arched  rib  of  100  ft.  span  and  20  ft.  rise,  hinged 
at  both  ends  and  at  the  middle  joint ;  dei!d  lof  d  =  40  tons  uniformly  distributed 
jxT  luH-izontal  unit  of  length,  and  hve  load^,  u>n  j)cr  horizontal  foot. 

55.  Show  how  the  calculations  in  the  preceding  question  are  affected  when 
lx)th  ends  are  absolutely  fixed. 

56.  In  the  framed  arch  represented  by  the  figure,  the  span  is  120  ft.,  the 
rise  12  ft.,  the  depth  of  the  truss  at  the  crown  5  ft.,  the  fixed  load  at  each  top 

joint   10  tons,  and  the   moving  load    10  tons. 

Determine  the  maximum  stress  in  each  member 

„       „,„  with  anv  distribution  of  load.     Show  that,  ap- 

r  If"       M-l  ' 

proximately,  the  amount  of  metal  required  for 
the  arch:  the  amount  required  for  a  bowstring  lattice  guder  of  the  same  span 
and  17  ft.  deep  at  the  centre:  the  amount  required  for  a  girder  of  the  same 
span  and  12  ft.  deep  ::  100:155:175. 


57.   Weights  n^i,  W.,   11%....  are  concentrated   at    a   series  of  points  in 
horizontal  girder  of  s])aii  2/  resting  upon  supports  at  the  ends,  the  correspond- 


BXAMPLES. 


961 


iSCh'"'  H^'"*r  ^l"  ^^"  '^^"  ■  ■  •  "'«P*"-tivnly.    If  a  flat  parabolic  arch,  fixed 
at  both  ends,  of  the  same  span  and  of  rise  i,  carry  the  same  weights  at  Joints 

58.  Show  that  a  weight  at  the  crown  of  a  flat  parabolic  arch  fixed  at  both 
ends  will  produce  the  same  horizontal  thrust  as  a  weight  at  any  other\Sit 

^rJ^  ^'T'!T}  P''™!^"'^  "bs  for  one  of  the  Harlem  River  bridges  has  a  clear 
openmg  of  olO  ft.,  a  rise  of  90  ft.,  a  depth  of  13  ft.,  and  are  spaced  14  ft  centl 

uve  load  at  S(XK)  lbs.     a  variation  in  temneratun>  of  710  F  f,.««,  n. 
also  to  be  allowed  for.     Determine  the  maSuI  ^X^Zn^^^Z^, 
I  constant)  and  the  maximum  deflection.     (£^  =  26  000  000  IbsT^Ln      ^'^ 
deduce  the  play  at  the  hinges.  -«>,<JW,000  lbs.)     Show  how  to 

«o.  A  c"-t-iron  arch  (see  figure)  who.se  cross  sections  are  rectaneular  anrf 
uniformly  .  ins.  wide  has  a  straight  horizontal  ex-  *"*  "^.".^^^e^"  «°d 
trados  and  is  hinged  at  the  centre  and  at  the  abut- 
ments. Calculate  the  normal  intensity  of  stress  at  the 
top  anu  bottom  edges  D,  E  of  the  vertical  section,  dis- 
tant .5  ft.  from  the  centre  of  the  span,  due  to  a  ver- 
tical  load  of  20  tons  concentrated  at  a  point  distant 
5  ft.  4  in    horizontally  from  B.    Also  find  the  maximum  intensity  of  the 

uZiiTlZ    "'"'  "'^'""'  ""^  ^*^^  '"^  p«'"*  ^*  which  iLcc,;^! 

„  r-*''  J'jn  ".r''"^  *""  ""''^""^  "•"  '^'"^'^^  *t  both  •^"ds,  for  a  span  of  50  ft  and 
a    ise  of  10  ft.,  ,s  a  parabola.     Draw  the  equilibrium  polygorwhen  the  arch 
s  loaded  with  tw.  equal  weights  of  2  tons  concentrated  aMwr^tslS  ft 
from  the  centre  of  the  span.     Also  determine  the  maximum  LTstrerin 
the  rib,  which  IS  a  double-t«e  section  2  ft.  deep.  ^ 

62.  Solve  the  preceding  example  when  both  ends  are  fixed. 
Am.    y,=y,=V^; 
ft.-tons. 


D«'o" 


//  =  3.3075tons;     3.3075    29U;, 

A  I     • 


max.    B.M- 2.025 


63.  The  load  upon  a  parabolic  rib  of  50  ft.  span  and  15  ft.  rise  hineed  at 
^th  ends   consists  of  weights  of  1,  2,  and  3  tons  at  ,K>ints  15   25  and ^  ft 

Jins.  rt  =2.9915  tons;  axial  thrusU,  3,2.595   3  69'>4   ''  «wan   nnH  a  -tZ 
shears,  0.1231,  0.8179,  1.1.S06,  and  1.25:^6  tons.  '  '  '    ^  ^"^^  '°'^' 

64.  Solve  the  preceding  example  when  both  ends  are  fixe 


II 


11 


fcV 


INDEX. 


Abutment,  Minimum  thickness  of,  894 

Acceleration,  M 

Accumulators,  358 

Advantage,  Mechanical,  379 

Advantages  of  2, 3,  or  4  main  trusses,  079 

Airy,  32."> 

Allowance  for  weight  of  beam,  472 

Anchorage,  843 

Anderson,  388 

Angle  of  friction,  378 

'  •     "   repose,  .30,5,  378 
Angular  velocity,  1(H} 
Annealing,  239,  246 
Anti-friction  pivots,  395 
Arc  of  approach,  404 
"    "  recess,  405 
Arch,  Three-hinged  braced.  .59 
Arrh  formulic.  Knipirical,  895 
Arched  ribs,  883 

"        "    ,  Oeflection  of,  9.30 

^'         "    ,  Deformation  of,  9,30 

"    ,  General  e<i nations  for,  9.33 
"    hinged  at  crown  and  ends, 
94H 

"        "   with  fixed  ends,  919 
,  ''         "      "     pin-ends,  910 
Archen,  883 
Area,  Hedu'-tion  of,  2.30 

Baker,  .325,  891 

' '       on  wind  pressure,  747 
BalanciiiB,  ISO 
Hall.  377 

Ratter  for  walls,  ,32<i 
BauMcliinger.  237,  2.38,  24ti 
Heam   bcIp<I  on  liy  ni>li,|ue  forces,  473 

"     .  Kiiiiililirium  of,  102 

"     .  Collnr,  32 
BeaniH      of      approximately      uniform 

strength.  417 
Ueaiiis   fixed  iit  Imth  ends.  4.3(5 

^']  "      '    one  end.  4.39 

'I        on  more  than  two  supports,  512 
restmg  on  two  supports,  434 
,  T*ii)uiar,  4.^1 
^'       of  unifoiin  strength,  44fl 


Hearings,  327 

of  shafts.  Distance  betwem, 

l<elt  transmission.  395 
Hending  moment,  102,  124 

,  Graphical  determina- 
,,  ,,         .    tion  of,  130 

>n   a  plane  which   is 
not     a      principal 
„  plane,  478 

Hevel-wheels,  408 
Blows,  243 
Bochet,  .377 

Boiler-joints,  Efficiency  of,  2.59 
Boiler-plate,  Thickness  of,  261 
Boiler-shells,  2.58 
Boussinewiue,  .324 

l|oNey's  neutral-axis    experiments,  424 
Bow's  method,  12 
Howstring  suspension  bridge  721 
tniss,  49,  717,  726 

Bmke,  .396'""'  ''^'^''"'  ^"^''^^  ^^^ 
Breaking  stress,  239 

strength  of  pillacs,  .59.5,  596, 

Brickwork,  241 
Bridge  stresses,  ('.89 

•  '•       Tu"    .•.*'™pf>    determ.  of,  696 

,  Three-hmged,  731 
"       truss.  Depth  of,  678 
"        Inisses,  46 

,.'.'.     -Types  of,  077 
Mri<lgps,  ('.26 
MruncI,  ,327 

liucklitig  of  pillars.  895 
niiiit  beams,  488 
Hulging  of  pillars,  ,594 
Bulk  modulus,  3.32 

Cable.  rur\e  of,  845 

"     .  Deflection  of,  8.53,  8.54 
'     ,  Length  of,  8.53 
' '     ,  Weight  nf  s.vi 

with  sloping  suspenders,  ^ur^■« 
of,  861 

963 


9G4 


INDEX. 


C'aiiihiT,  OSti 
("antilevcr,  102,  7r>l 

boom  cur'-e,  757 
trusses,  7,")1 
Cast  iron.  2:«» 
Catt'tiary.  ,S4,"> 
Caiioliy,  CkS" 
(ciMctit,  2U 

(■("iitriluiral   whirling  of  shafts,   (i.")7 
Clionls,  7."i(» 

<'(H'tIicipnt  of  elasticity,  217.  2;<2 
Coellicients  for  pillars."  TaMc  of,  (101 

"  "  sha|H's    in    compression, 

('■04 
"  of  friction.  377.  4()S 

■'    rupture.  4)0 
Collapsing  pre.ssiire  of  tul)es,  2f)2 
Collar  lieain,  :{2 
Cohinins,  flexure  of,  (i2.5 
ComiMiuiul  stress.  2ilO 
Coiulitions  of  eipiilihrium  for  areh,  SIK) 
Conical  pivots.  :{'.•;( 
Conjugate  stn-sses,  200 
Constant  coinjK)Meiits  of  pressure,  202 
Continuous  girders,  ."il.'i 

''  ''      ,  .Vilvantages  and  (iis- 

ail\  aiitagesof.  Xii\ 
"  "      ,  Max   M  M  sat  points 

of  supjKirt  of,  ,")37 
Contrarv"  flexure.  Points  of.  ."il.5 
Couliiml)'s  la«sof  friction,  37(i,3SH,  Ml, 
(117 
' '  wedge,  324 

Countcr-ellicicncy,  401,  40ti,  407 
(  nuiitcrforls.  ,320 
Crane,  lU'tit,  3.") 
"     ,  Derrick,  -  , 
"      ,  .III).  10 
■•     .Pit.  21 
"     .  Til  row  of.  21 
Creeping  clTcit.  23S 
Crown  of  arch,  vs.'i 
Culiic  strain.  3t.'> 
Cur\  ature  of  a  U'lim.  410.  420 
Curves  of  dellection,    slope,  and    H.Af., 
,'.03 
"       "    norm   and  laiig   stress,  4,>S 
"    pi-tiui  \  elocity,  lH,"i 
Cylinders.   I'liin.  2.'N 
Cylindriial  pivots.  3".H) 

I'arn,  Kipiililirium  of,  314 
llelleclion  due  ti>  sheiir.  4.^7 
of  arclicd  ril..  030 
"  ■'  caiili|c\<r  ti-u-se«,  7til 

"  tnisse-.  ,><tatical.  701.  SIO 
Iteforiuation  of  arched  ril),  030 
I  >errick  crane.  21 
Design  of  an  1  -  'tion.  400 

•  ft 'ti-fi  |.i;Ui- »£ird.  r.  4!»ii 
"  "  a  ."i.'O-ft  stt  iiiK-span.  702 
"       "  I27-ft    riveted  span.  770 


Oil.itation,  Coeffleient  of,  346 

I  >ii:es'  wind-pressure  formula,  746 

Distortion,  33.">,  ,33,s 

Doulile-iiiter.section  truss,  608,  715> 

Ductility,  230 

Dupuit,"3.S,S 

Dynamometer,  Prony's,  400 

Kartii  foun(hition,  ,308 

"      pressure,  :{07 
Karthwork  theory,  30,") 
Kaston,  3.S.S 

i:tTect  of  high-speed  l)elting,  398 
Ktlective  lielt  tension,  30.S 

length  and  depth  of  a  girder. 
4S(; 
Elliriency,  1(12,  3S0,  404,  408 

of  shafting,  «,j2 
Elastic  constants,  332 
"       curve,  405 
moment,  418 
Elasticities,  Tables  of,  203,  204,  266  271 
Elasticity,  21.') 

"      ',  Coefficient  of,  217,  230 
"         ,  .Mwlulus  of,  217 
"        ,  Uange  of,  240 

,  Torsional  coefficient  of,  234 
Ellipse  of  stress,  20H 
Ellipsoid  of  st-ain,  344 
"si  ivss.  343 
Elliptic  integrals.  Tables  of,  031,  032 
Endless  .screw,  384 
Energy,  1,')8 

,  Con.servation  of,  101 
"      ,  f^irves  of,  1,S7 
•'      ,  Fluctuation  of,  187 

,  Kinetic,  1,V),  lt>2 
"      ,  Potential,  l,--.,-),  162 
' '      ,  Transmission  of,  .354 
K>|uali/ation  of  stres.s,  423 
E<|iiations  of  motion  and  energy,  IM 

"  stress,  General,  ;W9 
E(|iiililirafed  [nilygon,  S.H.S 
Ivpiilibrium  of  relaitiing-walls,  313 
Ei|ui\alent  elongation,  236 
Erection  of  contimioiis  girtler,  .'537 
I'.uler's  si  rut  fonniila',  008 
Examples,   00,   141,    100-170,   180-184, 
ISO  2il,  302-.37:).  400-11.'),  ,')4(W,192, 
032  (-.40.  0(H-07,'),  820-841,  840-882 
I'.xtrados,  S8.'. 

I!ve-bar  projiortionH.  704,  765 
Eve-burx,  tlHS,  702.  704 

,  S1res«<-s  in,  430 

Factor  of  safety,  242 

Kairbnim,  237 

I'aiisliHwe.  325 

I  uliKue,  '^4.~i 

I'ink  Iniw,  M 

Flange  and  web  rivet  eonncttiou,  687 


INDEX. 


9GS 


Flanged  girders,  479 

"      ,  Classification  of,  479 
\\  \'      ,  Equilibrium  of,  480 

"     .  Moments  of  resistance 
of,  482 
Flexure  of  column,  025 

' '      ,  Work  of,  423 
Floor-beams,  795 
Flow  of  lead,  257 

"      "  solids,  235,  254 
Fluctuation  of  energy,  187 

"    stress,  228,  244 
Fluid  pressure,  254 
Fluidity,  Coefficient  of,  254 
Force,  Shearing,  102 

' '     ,  Moments  of,  122 
Forced  oscillation,  225 
Forces,  Polygon  of,  3,  6 
Foundations  in  earth,  20%,  327 
Frames,  1,  19 

' '      ,  Incomplete,  28 
Franke's  formula,  377 
Friction,  370 

' '       ,  Coefficients  of,  377,  405 
"       .Journal,  388 
"       ,  Laws  of,  376 
"       .Rolling,  385 
"       ,  Sliding,  376 
Fuller's  metluKT  898 
Funicular  curve,  99 

polygon,  3,  7 

Galton,  377 

General  equations  for  arched  rib,  933 

"  "       of  stress,  339 

Gin,  22 

Girders,  Hinged,  137 
GofJon's  lorniulu",  Rankine'a  modified, 
km 
' '        pillar  fonnultt,  5^9 
Graphical   detennination  of  slope  and 

deflection,  r>0.") 
Graphical     determination     of     bridge 

stresses,  ri9«> 
Grashof,  34fi,  (.57 
Gravity,  Centres  of,  10 

"       .Specific.  1  of. 

"       ,  Values  of,  lii5 
Gun-barrels,  l-Vf 

Hardness,  roefficioiit  of,  256 

Hartnell,  06 1 

Helical  springs,  lU'iO 

Highway-bridge  loads.  Tal)le  of,  820 

Hingp<l  girders,  137,  .Wd 

Him.  390 

Hodgkinmn'n  pillar  formuhc,  .WS,  ,"590 

Hooke's  law,  217,  237 

Hr>nk«.  StrM!*!'  in.  430 

Horse-power.  159 

Howe  truss.  47 

HiatoQ'fl  foruiula,  16 


Hydraulic  presses,  358 
Hysteresis,  237 

Inclined  plane,  377 
Incomplete  frames,  28 
Impulse,  1.58 
Inertia,  180 

' '      ,  Moments  of,  7 

,  Variable  moment  of,  462 
Inflexion,  Points  of,  516 
Internal  stress,  290 
Iiitradns,  885 
Isotropic  bodies,  332,  345 

Jenkin,  Fleeming,  377 

Jib  crane,  19 

Joint  of  rupture  in  arch,  892 

Joule,  l.TO 

Journal  friction,  388 

K  ystone,  885 
Kilowatt,  l.')9 
Kinetic  energy,  15.'),  100 
King-post  truss,  27,  34 

Lame,  341 

Lateral  bracing,  T.V).  7S3.  784,  789,  797 

Lattice  truss,  (ii      "'i|,  707 

Launhardt's  forti        .  247 

Lenticular  trusx,  i 

Levy,  324 

Limit  of  ela.st  icity,  i  t)2,  236,  240 

Line  of  resistance,  K.H.S 

of  d.iin,  314,  316,330 

"  rupture,  31' 
Linear  arch,  888 

"    ,  Catenary,  899 
"    ,  Circle,  903 
"         "   ,  Ellipse,  903 
"         "    ,  General.  «»08 
"         "    ,  Geostatic,  907 
"         "    ,  HydroKlatic,  906 
"    ,  Par„!x)la,  902 
Live  load,  15 
Load.  Dead,  232 
Loads,  .\ccidental  (live),  15 
"     ,  Distribution  of,  17 
"     ,  Li\e,  116,  2.32 
"       on  bridges  (dead),  680 

"       "        (live).  (W2 
"     ,  Permanent  (dead),  15 
Lock-joint  truss,  38 
Long  struts,  616 

Mansard  roof,  6 

.Masonry,  241 

Maxinumi  MM  at  points  of  supimrt  of  a 

f-ontiiiiiotir:  Kinirr,  .t."," 
.Maxwell's  framed  arch. 
Mechanical  «d\antngc.  379,  382  i 

Middle-third  ttieofy,  891 


966 


INDEX. 


Modulus  of  liulk,  332 

■'         "  cubic  c'oinpre.ssibility,  332 

"         "  clasticitv.  217 

"         "  rigi<}ity,"29S,  338 

"         "  section,  121 
Moiiient,  ISeiidiiig,  102 
of  force,  122 

"  "  inertia,  \'ariai)le,  402 

"  "  resistance,  410 

"  "  stability,  3! 4 

Moments  of  resistance  of  Hanged  girders, 

1S2 
Momentum,  l.Vt,  100 

•'  ,  Conservation  of,  101 

Morin,  377,  388,  389 
Mortar.  241 

Moselcy's  principle,  897 
Muir,  238 

Neutral  axis,  -llO 

"        surface,  410 
Kormal    and    tangential   stress.   Curves 
of,  4,-)8 

<)blii|ue  resistance,  1.57 
<)blii|uity,  2'.Ml 
Oscillation,  1-orced,  22."i 

of  elastic  rod,  22.'>-231 
Overturning  moment,  314 

Pnnillcloeram  of  velocity,  103 

I'auli  truss,  713 

J'eurson,  212 

Perrj-,  .'i-l.^ 

Petit  tniss,  4<t,  715 

Piers,  .")2.  S.").') 

Pillars.  .-.93 

"      ,  Cla-sification  of,  .•93 

"      ,  Manner  of  failure  of,  i)93 
Pin-lH'ariiig  pillar,  0()1 
Pin-connected  trusses,  (iSO 
Pins,  (iss,  702,  S07 
Piston  velocity.  ( 'urves  of,  IS.") 
Pivots,  Coniciil,  .393 

"      ,  Cvlindrical.  39() 

"      ,  S."4iiele's.  391 

"      .Wear  of,  392 
Planes  of  iirinciind  stress.  292,  297    312 
Plastic  defornuitiou  ol  bar,  2.".<i 
Plasticity.  21''. 

Plates,  tliickiiess  of  Hat,  202 
Platform.  Position  of  bridge,  070 
Pole.  7 

Polygon.  I'uiiicular,  3 
"        ,  Nun  I  Icisiiig,  r> 
' '  of  forces.  3,  ti 

Pols' oil's  ratio,  217,  310 
P(;n<«let,.321,  3H1 
I'orlal  bracing,  787 
PolcMiiai  energy,  lo'>,  162 
Power   l.V.t 
Pntt  tniKS,  093 


Pre.sscs,  Hydraulic,  3.'>S 
Pressure  of  earth,  ;J07 
Pressure  of  wind,  1-i 
Princip  r,es,  292,  297,  342 

CSS,  292,  312 
Princij       .  13 
Pronys  dynamometer,  400 
Purcliase  of  niacliinc,  378 
Purlins,  13 

Queen-post  truss,  30 

Rafters,  13 

Rankine,  201 ,  .3.S9,  m\ 

Rankine's  earthwork  theory,  305 

Redtcnbacher,  4(K) 

Reinforced  concrete  beams,  443 

Relations  between  deflection,  slope,  and 

U-nding  moment,  ".03 
Relative  velocity,  104 
Rennie,  ,377 
Rejiose,  Angle  of,  305 
Resenoir  walls,  328 
Resilience,  218 
Resilience,  Table  of,  .>15 
Resistance,  Line  of,  314,  310 
Retaining-walls,  313 

,  Practical  rules  for,  324 
Reuleaux,  202,  392,  4(K) 
Rigidity,  Flexural,  ()47 

"       ,  Torsional,  047 

"       ,  Modtilus  of,  298,  ,339,  347 
Ri>ef  connection  of  flange  and  web,  687 
Riveted  joints,  707 

"  "     ,  KMiciency  of,  773 

"  "     ,  Friction  of,  777 

"  "     ,  Tlieoretii'al      deduction, 

770 

"        plates,  Strength  of,  766 

"        trusses,  0,S0 
Riveting,  Chain,  7()8 
,  Cold,  705 
,  Hot,  705 

• '       ,  Zig/ag,  7(»8 
Rivets,  704 
Itollers,  702,  791,  823 
Polling  friction,  385 
!fiM>f,  Nlansurd,  0 
truss,  30 
weights.  15,  05 
lio|ie  transmission,  395 
Rouse'.s  wind-pressure  fonnula,  746 

Saddles,  SU 

SI    Venant,  210.  324,  ,3,53,  647,  649 
Salta.sh  bridge,  721 
Saiilt  Ste   .Marie  bridge,  683 
Schiele  pivots,  391 
Schwedier  trusM,  712 
Screw,  3H1 
"     ,  KndlesB,  384 


INDEX. 


967 


Semigirder,  102 
Set.  218,  237 
Shafting,  303 

"       ,  Efficiency  of,  652 
Shafts,  Distance   between   bearings   of, 
<i51 
' '     ,  Non-circular,  647 
' '     ,  Torsional  strength  of,  642 
"       under  bending  and  torsion,  654 
„  "  "    variable  resistance,  649 

Shear,  216,  298 

' '    ,  Deflection  due  to,  457 
Shearing  force,  102 

stress.  Distribution  of,  453 
Shear-legs  (shears),  22 
Shells,  Spherical,  260 
Shocks,  243 
Shoes,  790 
Simple  shear,  298 
Single-intersection  bridge,  709,  714 
Skewback,  885 
Slip  of  Mts,  399 

Slope  and  deflection,  Oraphical  deter- 
mination of,  ."iOS 
Snow,  Weight  of,  15 
Spandrel.  885 

Specification  for  pillars,  606 
Specifications,  783 
Specific  weight,  231 
Springing,  885 
Springs,  222-225 
' '      ,  Carriage,  470 
' '      ,  Deflection  of,  467,  468, 469,  472 
' '      ,  Energy  of,  467,  468,  469,  472 
"      ,  Flat.  464 
' '      ,  Helical,  (KiO 
' '      ,  Paral>olic,  469 
' '      ,  Rectangular,  468 
"      .Spiral,  466 
' '      ,  Triangular,  468 
Square-bearing  pillars.  tiOl 
Stanton  on  wind  pressure,  747 
Statical  deflection  of  trusses,  761 
Steel,  240 
StifTeners,  782 
Stiffening  truss.  86.1 
Stiffness  of  beams,  451 

"  l)eltH  and  ropes.  400 
Stoken  on  wind  pressure.  746 
Straight-line  formula,  600 
Strain,  215,  216.217 

"    ,  Compound,  290 
Struts,  15 

' '     ,  Rending  of,  608 

"     ,  Long.  616 

"       in  torsion,  611 

I'       under  a  lateral  load,  610 

with  one   s(|uare   and   one  pin 
end,  (;t;5 
'*         ' '     pin  ends.  »iG8 
'         "     (Uluare  ends,  012 
Strength  modulus,  421 


Stress,  21.5,  216,  217 
Strj'ss,  Internal,  290 
Stresses  in  bridge  members,  689 
Stress-strain  cur\e,  240 
Stringers,  (i85,  782,  793,  795 
Suspenders,  844 
Suspension-bridge  loads,  873 

modifications,  875 
bridges,  842 

Table  of  coeff.  of  axle  friction,  408 
\\      "     "      "   linear  expansion,  268 
w      .,     .',•     .".  s'»ding  friction,  377 
^^  elliptic  integrals,  631,  632 

^^      '    eye-bar  proportions,  764,  765 

factors  of  safety,  243 
^'      "  highway-bridge  loads,  820 

"  moments  of  inertia,  422 
"      "  roof  weights,  67 
"      "  resiliences,  .545 
"      "  section  moduli,  422 
"      "  shearing  strength  of  timbersv 

"      "  strengths,  elasticities,  etc  : 

of  alloys,  263 

of  cements,  270 

of  concrete,  270 

of  copper  wire,  265 

of  iron  and  steel,  264 

of  manila  rope,  265 

of  rocks,  268-270 

of  timbers,  26(i 
^^      ^        of  wire  rope,  264 

"  transverse  strength  of  timberK. 
265,  2(i7 
''      "  values  of  A,  A.,  and  /,  601 
::      ;;       .''"£,«./£•.  and  ,,,271 
^  ,,     :  weights  of  bridges,  814-819 
Telford,  891 
Temperature  effect  on  arched  ribs.  927, 

Theorem  of  three  moments,  515 

Thick  cylinders,  3,55 

Thomson,  246 

Three-hinged  braced  arch,  .59,  731 

XI       i'  bridge  truss,  62,  731,  739 

Thurston.  389,  390 

Toot  lied  gearing,  404 
Torsion,  216.  349,  tV41 
"      .Angle  of,  641 
Torsional  rigidity,  647 

' '        St  rength  of  shafts,  642 
Tower,  390 
Traction,  .348 
Tract  rix,  ,394 

Transformed  catenary,  899 
Transmission  of  energy.  354 
TraiiHvei'w)  eiuHiicity,  Coeiiictont  of,  347 

strength,  265 
Trellis  truss.  698 
Treses,  2M,  2Sft 


•  I 


IF 


968 


INDEX. 


Triangle  of  velocity,  163 

TriiKid,  13,  23 

Truss,  Haltimore,  714,  71.') 

•  •  ,  Kowstriiig,  44,  49,  50,  717,  720 

"  ,  Bridge,  23 

"  ,  Double-intersection,  714 

' '  ,  Kink,  ,'>4 

"  ,  Howe,  48 

"  ,  King-i)ost,  27,  34 

"  ,  I,t'iilii'\ilar,  721 

"  ,  l,ock-ja\v,  38 

"  ,  I'auli,  713 

"  ,  Petit,  40,  71.5 

"  ,  Pin-tonnected,  680 

"  ,  Pratt.  18 

"  ,  (Jueen-iK)Ht,  30 

"  ,  Hiveted,  <)80 

"  .Roof,  13,  23 

"  ,  Scliwedlor,  712 

"  ,  Single-intersection,  709,  714 

"  ,  Three-hinged,  731 

' '  ,  Warren,  47 

Uniform  strength,  Iteams  of,  446 
Uniformly  varying  presssure,  618 
riiwin's  formula,  2.'>1 
■('seful  work,  Itil 

A'alup  of  IT  for  arched  rib,  SS,"),  922 

A'ahies  of  A.  .1,,  and  /  for  shapes,  004 

A'auban,  .320 

Velocity,  l.'v? 

,     ' '       ,  Cur\e8  of  piston,  185 


\'elocity,  Parallelogram  of,  163 

,  Relative,  104 

,  Triangle  of,  103 
\ibration,  Theory  of,  221 
\olunie  strain,  345 
V'oussoir,  88.5 

Waddell,  681 ,  682 

Walls  of  uniform  strength,  328 

Warren  truss,  47,  OIK) 

Wear  of  pi\ots,  392 

Wedge,  378 

Weight  of  beam.  Allowance  for,  472 
"  snow,  1,") 

Weights  of  roofs,  l.'i,  04 

,  Tables  of,  2C3-26S 

Weyrauch's  formula,  248 

theorj'  of  buckling,  620 

Wheel  and  axle,  402 

Whirling  of  shafts,  0,57 

Wind  loads  on  tnisses,  783 
"      pressure,  10,  745,  783 
"  "       ,  Dine's  formula,  746 

"  "       ,  Empirical  regulation,747 

"  "       ,  Kng.  Commission,  740 

"  "      ,  Rouse's  formula,  746 

Wiihler's  experiments,  237,  243 

Wo.k,  1.55,  101,  102 
of  flexure,  423 

Wrought  iron,  240 

Yield-point,  241 


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Donlittle's  Treatise  on  Practical  Astronomy 8vo,' 

G     ■  '8  Elements  of  Geodesy gvo 

Bayford's  Text-book  of  Geodetic  Astronomy 8vo, 

Herriman's  Elements  of  Precise  Surveying  and  Geodesy 8vo, 

•  Michie  and  Harkiw's  Practical  Astronomy gvo, 

•  White's  Elements  of  Theoretical  and  Descriptive  Astronomy lamo 


50 
00 
00 
00 
00 
00 
50 
00 
00 


3 
3 

3 
I 
I 

a 
a 
3 

4  00 
3  00 
I  50 
«  50 


50 
50 
00 
oo 
SO 
00 
50 


3  00 
00 


BOTANY. 

Davenport's  SUtistical  Methods,  with  Special  Reference  to  Biok>gical  Variation. 

i6mo,  morocco,    t  as 

Thom<<  and  Bennett's  Structural  and  Phytiological  Botany i6mo,    a  as 

Westermaisr's  Compendium  of  General  Botany.    (Schnsidar.) •*»,    a  oo 

t 


I   ' 


CHEMISTRY. 

*  Abegg's  Theory  of  Electrolytic  Dissociation.    iVon  Ende.)                     i2mo,  i  25 

Adriance's  Laboratory  Calculations  and  Sp«rilic  Gravity  Tables iimo,  i  25 

Alexeyeff's  General  Principles  of  Organic  Synthesis.     (Matthews.)         .       8vo,  3  00 

Allen's  Tables  for  Iron  Analysis 8vo,  3  00 

Arnold's  Compendium  of  Chemistry.     (Mandel.t Small  8vo,  3  50 

Austen's  Notes  for  Chemical  Students umo,  i  50 

Bernadou's  Smokeless  Powder.-  Nitro-cellulose,  and  Theory  of  the  Cellulose 

Molecule 1  amo,  a  50 

*  Browning's  Introduction  to  the  Rarer  Elements 8vo,  i  50 

Brush  and  Penfield's  Manual  of  Determinative  Mineralogy. 8vo,  4  00 

*  Claassen's  Beet-sugar  Manufacture.     (Ball  and  Rolfe.1. 8vo,  3  00 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.    (Boltwood.).   8vo,  3  00 

Cohn's  Indicator  and  Test-papers iimo,  2  00 

Tests  and  Reagents 8vo,  3  00 

Crafts's  Short  Course  in  Qualitative  Chemical  Analysis.  (Scbaefler.).  .    lamo,  i  50 

*  Danneel's  Electrochemistry.     iHerriam.^ iimo,  i  25 

Dolezalek's  Theory  of  the  Lead  Accumulator  (Storage  Battery).        (Von 

Ende. ) iimo,  3  50 

Drechsel's  Chemical  Reactions.     (Merrill.) lamo,  1  25 

Duhem's  Thermodynamics  and  Chemistry.     (Burgess.) 8vo,  4  00 

Eissler's  Modern  High  Explosives 8vo,  4  00 

Effront's  Enzymes  and  their  Applications.     (Prescott.) 8vo,  3  00 

Erdmann's  Introduction  to  Chemical  Preparations.     (Dunlap.) i2mo,  i  2s 

Fletcher's  Practical  Instructions  in  Quantitative  Assaying  with  the  Blowpipe. 

i2mo,  morocco,  i  50 

Fowler's  Sewage  Works  Analyses i2mo,  2  00 

Fresenius's  Manual  of  Qualiutive  Chemical  Analysis.     (Wells.) 8vo,  5  00 

Manual  of  Qualitative  Chemical  Analysis.  Part  I.  Descriptive.  ( Wells.)  8vo,  3  00 
System  of   Instruction   in   Quantitative    Chemical  Analysis.      (Cohn.) 

»  vols 8vo,  13  50 

Fuertes's  Water  and  Public  Health ismo,  i  50 

Furman's  Manual  of  Practical  Assaying 8vo,  3  00 

*  Getman's  Exercises  in  Physical  Chemislry i3mo,  2  00 

Gill's  Gas  and  Fuel  Analysis  for  Engineers i2mo,  i  25 

*  Gooch  and  Browning's  Outlines  of  Qualitative  Chemical  Analysis.  Small  8vo,  i   25 

Grotenfelt's  Principles  of  Modern  Dairy  Practice.     ( WoU.) i3mo,  3  00 

Groth's  Introduction  to  Chemical  Crystallography  (Marshall) i3mo,  i  25 

Hammarsten's  Text-book  of  Physiological  Chemistry.     (MandeL) 8vo,  4  00 

Helm's  Principles  of  Mathematical  Chemistry.     (Morgan.) ismo,  i  50 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco,  a  50 

Hind's  Inorganic  Chemistry 8vo,  3  00 

*  Laboratory  Manual  for  Students i2mo,  i  00 

HoUeman's  Text-book  of  Inorganic  Chemistry.     (Cooper.) 8vo,  2  50 

Text-book  of  Organic  Chemistry.     (Walker  and  Mott.) 8vo,  2  50 

*  Laboratory  Manual  of  Organic  Chemistry.     (Walker.) i2mo,  100 

Hopkins's  Oil-chemists'  Handbook 8vo,  3  00 

Iddings's  Rock  Minerals 8vo,  5  00 

Jackson's  Directions  for  Laboratory  Work  in  Physiological  Chemistry.  .8vo,  i  25 

Keep's  Cast  Iron 8vo,  2  50 

Ladd's  Manual  of  Quantitative  Chemical  Analysis i2mo,  i  00 

Landauer's  Spectrum  Analysis.     (Tingle.) 8vo,  3  00 

*  Langworthy  and  Austen.        The  Occurrence  of  Aluminium  in  Vegetable 

Products,  Animal  Products,  and  Natural  Waters   8vo,  2  00 

Lassar-Cohn's  Application  of  Some  General  Reactions  to  Investigations  in 

Organic  Chemistry.     (Tingle.) lamo,  i  00 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control 8vo,  7  50 

Lob's  Electrochemistry  of  Organic  Compounds.     (Lorenz.) 8vo,  3  00 

4 


.  iimo, 

.  umo, 

lamo. 


.iimo, 
.  iimo, 
lamo. 


oo 
oo 
50 
50 
60 

00 
35 
50 
00 
00 
00 
50 
50 
oc 
no 
00 
50 
00 


todge's  Botes  on  Asuying  and  HeUUurgical  Laboratory  Exp«rimenti 8to, 

Low's  Technical  Method  of  Ore  Analysis.  gy^ 

Lunge's  Techno-chemical  Analysis.     1  Cohn.  1 1  jmo 

•  McKay  and  Larsen's  Principles  and  Practice  of  Butter-making 8vo, 

Mandel's  Handbook  for  Bio-chemical  Laboratory ismo 

•  Martin's  Laboratory  Guide  to  QuaUtative  Analysis  with  the  Blowpipe .    i amo," 
Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint.) 

3d  Edition,  Rewritten gy^ 

Examination  of  Water.     (Chemical  and  Bacteriological.) lamo, 

Matthew's  The  Textile  Fibres '      gyo' 

Meyer's  Determination  of  Radicles  in  Carbon  Compounds.     (Tingle.).  .  lamo. 
Miller's  Manual  of  Assaying 

Cyanide  Process 

Minet's  Production  of  Aluminum  and  its  Industrial  Use.     (Waldo.).       

Mixter's  Elementary  Text-book  of  Chemistry ^  _  timo, 

Morgan's  An  OutUne  of  the  Theory  of  Solutions  and  its  Results   . . 

Elements  of  Physical  Chemistry 

•  Physical  Chemistry  for  Electrical  Engineers 

Morse's  Calculations  used  in  Cane-sugar  Factories i6mo.  morocco, 

•  Muir's  History  of  Chemical  Theories  and  Laws gyo, 

MuUiken's  General  Method  for  the  Identification  of  Pure  Organic  Compc  nds' 

„.„  .     .  ^oL  I Large  8vo, 

O  Brine  s  Laboratory  Guide  in  Chemical  Analysis gyo, 

O'DriscoU's  Notes  on  the  Treatment  of  Gold  Ores gyo.' 

(Ostwald's  Conversations  on  Chemistry.     Part  One.     ( Ramsey. )!  , .    . .    i  imoj 

Part  Two.     (TurnbuU.) lamo, 

•  PauU's  Physical  Chemistry  in  the  Service  of  Medicine.     (Fischer.)         lamo 

•  Penfleld's  Notes  on  Determinative  Miiteralogy  and  Record  of  Mineral  Tests. 

PJctefs  The  Alkaloids  and  their  Chemical  Constitution.     (Biddle.) ....   8to! 

Pinner's  Introduction  to  Organic  Chemistry.     (Austen.) tamo' 

Poole's  Calorific  Power  of  Fuels gyo' 

Prescott  and  Winslow's  Elements  of  Water  Bacteriology,  wiUi  Special  Refer- 
ence to  Sanitary  Water  Analysis. ,,n,o      i  as 

•  Reisig's  Guide  to  Piece-dyeing. gyo'  ,j  ^ 

Richards  and  Woodman's  Air.  Water,  and  Food  from  a  Sanitary  Standpoint  8vo'.    a  00 
Hicketts  and  Russell's  Skeleton  Notes  upon  Inorganic  Chemistry.     (Part  I 

Non-metalUc  Elements.) gy^^  morocco 

Ricketts  and  Miller's  Notes  on  Assaying. '  g     ' 

Rideal's  Sewage  and  the  Bacterial  Purification  of  Sewage. ,  gyo' 

Disinfection  and  the  Preservation  of  Food. gvo' 

Riggs's  Elementary  Manual  for  the  Chemical  Laboratory gyo' 

Robine  and  Lenglen's  Cyanide  Industry.     (Le  Clerc. ) gyo' 

Ruddiman's  Incompatibilities  in  Prescriptions.  s     ' 

•  Whys  in  Pharmacy ^'2°' 

Sabin's  Industrial  and  Artistic  Technology  of  Paints  and  Varnish.            '  ^o' 
Salkowski's  Physiological  and  Pathotogical  Chemisti^y.     (Orndorff.).  gyo' 

Schimpf 's  Text-book  of  Volumetric  Analysis ' 

Essentials  of  Volumetric  Analysis 

•  QualiUtive  Chemical  Analysis g 

Smith's  Lecture  Notes  on  Chemistry  for  DenUl  Students  gyo ' 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Houses i6mo.  mwocco" 

c.     ,^^°f^°V"  '^""  ^"*"  Manufacturers ,6mo,  morocco,' 

Stockbndge's  Rocks  and  Soils gj^' 

•  Tillman's  Elementary  Lessons  in  Heat gy  ' 

•  Descriptive  General  Chemistry g     ' 

Treadwell's  QuaUutive  Analysis.     (HaU.) g     ' 

OuantiUtive  Analysis.     (HaU.) g     * 

Tunieaure  and  RusseU's  PubUc  Water-supplies. gy^' 

5  ' 


5  00 
a  00 
a  00 
I  50 
a  00 
I  as 

SO 
5  00 
I  50 
i  oa 


75 
00 
00 

00 
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00 


.lamo, 
tamo. 


I  00 
3  00 
a  so 
a  50 
I  as 
'  as 
a  50 
3  00 
3  00 
a  50 
I  so 
3  00 

3  00 

4  00 

5  00 


Van  DcTcntcr't  Physical  Chemistry  (or  Beginners.     (Bottwood.) lamo, 

*  Walke's  Lectures  on  Explosives 8vo, 

Ware's  Beet-sugar  Manufacture  and  Refining    Small  8to,  cloth, 

Washington's  Manual  of  the  Chemical  Analysis  of  Rocks 8to, 

Weaver's  Military  Explosives 8vo, 

Wehrenfennig's  Analysis  and  Softening  of  Boiler  Feed- Water 8vo, 

Wells's  Laboratory  Guide  in  Qualiutive  Chemical  Analysis 8vo, 

Short  Course  in  Inorganic  Qualitative  Chemical  Analysis  for  Engineering 
Students lamo. 

Text-book  of  Chemical  Arithmetic lamo, 

Whipple's  Microscopy  of  Drinking-water 8vo, 

Wilson's  Cyanide  Processes lamo, 

Chlorination  Process lamo, 

Winton's  Microscopy  of  Vegetable  Foods 8vo, 

WuUing's  Elementary  Course  in  Inor.auic,  Pharmaceutical,  and  Medical 
Chemistry lamo, 


1 

50 

00 

00 

00 

00 

00 

50 

50 

15 

SO 

50 

50 

50 

BRIDGES   AND    ROOFS. 


CIVIL  ENGINEERING. 

HYDRAULICS.       MATERIALS   OF    EHGIREERIHG. 
RAILWAY   E5GIIIEERI1IG. 


Baker's  Engineers'  Surveying  Instruments lamo, 

Bixby's  Graphical  Computing  Table Paper  ig)\a4l  inches. 

Breed  and  Hosmer's  Pr  nciples  and  Practice  of  Surveying  8vo, 

*  Burr's  Ancient  and  Modern  Engineering  and  the  Isthmian  Canal  ...     8vo, 

Comstock's  Field  Astronomy  for  Engineers 8vo, 

Crandall's  Text-book  on  Geodesy  and  Least  Squares 8vo, 

Davis's  Elevation  and  Stadia  Tables 8vo, 

Elliott's  Engineering  for  Land  Drainage umo, 

Practical  Farm  Drainage lamo, 

•Fiebeger's  Treatise  on  Civil  Enkineering 8vo, 

Flemer's  Phototopographic  Methods  and  Instruments. 8vo, 

Folwell's  Sewerage.     (Designing  and  Maintenance.'' 8vo, 

Freitag's  Architectural  Engineering,     ad  Edition,  Rewritten 8vo, 

French  and  Ives's  Stereotomy 8vo, 

Goodhue's  Municipal  Improvements lamo. 

Gore's  Elements  of  Geodesy 8vo, 

Bayford's  Text-book  of  Geodetic  Astronomy 8vo, 

Bering's  Ready  Reference  Tables  (Conversion  Factors') i6mo,  Tiorocco, 

Bowe's  ReUining  Walls  for  Earth lamo, 

*  Ives's  Adjustments  of  the  Engineer's  Transit  and  Level i6mo,  Bds. 

Ives  and  Hilts's  Problems  in  Surveying i6mo,  morocco, 

Johnson's  (J.  B.  1  Theory  and  Practice  of  Surveying Small  8vo, 

Johnson's  (L.  J.)  Statics  by  Algebraic  and  Graphic  Methods 8vo, 

Laplace's  Philosophical  Essay  on  Probabilities.    (Truscott  and  Emory.).  lamo, 

Mahan's  Treatise  on  Civil  Engineering.     (1873.)     (Wood.) 8vo, 

»       Descriptive  Geometry 8vo, 

Merriman's  Elements  of  Precis.  Surveying  and  Geodesy 8vo, 

Merriman  and  Brooks's  Handbook  for  Surveyors i6mo,  morocco, 

Rugent's  Plane  Surveying 8vo, 

Ogden's  Sewer  Design lamo, 

Parsons's  Disposal  of  Municipal  Refuse 8vo, 

Patton's  Treatise  on  Civil  Engineering 8vo  hall  leather, 

Reed's  Topographical  Drawing  and  Sketching 4to, 

R.deal's  Sewage  and  the  Bacterial  Purification  of  Sewage 8vo, 

Siebert  and  Biggin's  Modern  Stone-cutting  and  Masonry 8to, 

A 


3 

00 

as 

00 

50 

50 

00 

00 

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00 

00 

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50 

50 

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50 

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50 

00 

00 

00 

00 

50 

50 

00 

50 

00 

00 

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00 

00 

SO 

Smitli'i  Manual  of  Topoinpliical  Drawinc.     (McMillan.) Sro, 

Sondaricktr's  Graphic  Statici,  with  Applications  to ':  runat,  Baama,  and  Arehas. 

8to, 

Taylor  and  Thompaon's  Trtatiia  on  Concrete,  PUin  and  Reinforced 8to, 

•  Trautwine'i  Civil  Engineer*!  Pocket-bo .'-    i6mo,  morocco, 

Venable's  Garbage  Crematories  in  Amerii:'   8»o, 

Wait's  Engineering  and  Architecttiral  Jurisprudence 8to' 

Sheep, 

Law  of  Operations  Preliminary  to  Construction  in  Engineering  and  Archi- 

itttutt 8to, 

Sheep, 

law  of  Contracts gvo, 

Warren's  Stereotomy — Problems  in  Stone-cutting 8to, 

Webb's  Problems  in  the  Use  and  Adjustment  of  Engineering  Instruments. 

lomo,  morocco, 
Wilson's  Topographic  Surveying 8vo, 


BRIDGES  AND  ROOFS. 

Boiler's  Practical  Treatise  on  the  Construction  of  Iron  Highway  Bridges.  Svo, 

•       Thames  River  Bridge 4to,  paper. 

Burr's  Course  on  the  Stresses  In  Er.dgts  ar.d  Roof  Trucscs,  Aiched  Ribs,  and 

Suspension  Bridges 8vo, 

Burr  and  Falk's  Influence  Lines  for  Bridge  and  Roof  Computations 8vo, 

Design  and  Construction  of  Hetall.c  Bridges 8vo 

Du  Bois's  Mechanics  of  Engineering.     VoL  II Small  4to, 

Foster's  Treatise  on  Wooden  Trestle  Bridges 4to, 

Fowler's  Ordinary  Foundations 8vo, 

Greene's  Roof  Trusses 8vo, 

Bridge  Trusses 8vo, 

Arches  in  Wood,  Iron,  and  Stone 8vo 

Howe's  Treatise  on  Arches 8»o, 

Design  of  Simple  Roof-trusses  in  Wood  and  SteeL 8vo, 

Symmetrical  Masonry  Arches 8vo, 

Johnson,  Bryan,^and  Turneaure's  Theory  and  Practice  in  the  Designing  of 

Modem  Framed  Structures.. Small  4to,  lo 

Herriman  and  Jacoby 's  Text-book  on  Roofs  and  Bridges : 

Part  I.    Stresses  in  Simple  Trusses. 8vo,    2 

Part  11.    Graphic  SUtics. 8vo,    a 

Part  ni.  Bridge  Design 8vo,    2 

Part  IV.  Higher  Structures 8vo,    2 

Horison's  Memphis  Bridge 4to,  10 

Waddell's  De  Pontibus,  a  Pocket-book  for  Bridge  Engineers .  .  i6mo,  morocco,    2 

*  Specifications  for  Steel  Bridges ismo, 

Wright's  Designing  of  Draw-spans.     Two  parts  in  one  volume 8vo,    3 


a  50 


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00 

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2 

50 

2 

50 

4 

00 

3 

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2 

so 

SO 
50 
50 
SO 
00 
00 
so 
SO 


HYDRAULICS. 

Barnes's  Ice  Formation 8vo, 

Bazin's  Experiments  upon  the  Contraction  of  the  Liquid  Vein  Issuing  from 

an  Orifice.     (Trautwine. ) 8vo. 

Bovey's  Treatise  on  Hydraulics. 8vo, 

Church's  Mechanics  of  Engineering. 8vo, 

Diagrams  of  Mean  Velocity  of  Water  in  Open  Channels paper. 

Hydraulic  Main's 8vo, 

CofBn's  Graphical  Solution  of  Hydr.'.ulic  Problems i6mo,  morocco, 

Flather's  Dynamometers,  and  the  Measurement  of  Power umo, 

7 


3  00 


00 
00 
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50 
00 
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00 


Folwell's  Wafe.-supply  Engineerins Svo, 

Frizell's  Water-power 8vo, 

Fuertes's  Water  and  Public  Health. .  ....iimo, 

Wat^r-filtration  Works.  .  .  ....nmo, 

GancuiUet  and  Kutter's  General  Formula  for  the  Uniform  Flow  of  Water  in 

Rivers  and  Other  Channels.     (Bering  and  Trautwine.  i 8vo, 

Hazen's  Filtration  of  Public  Water-supply.  8vo, 

Hazlehurst's  Towers  and  Tanks  for  Water-works.  8yo, 

Herschel's  lis  Experiments  on  the  Carrying  Capacity  of  Large,  RiTeted,  Metal 

Conduits.  .  •  8vo. 

Mason's  Water-supply.     (Considered  Principally  from  a  Sanitary  Standpoint. ) 

8vo, 
Merriman's  Treatise  on  Hydraulics. 8vo, 

•  Michie's  Elements  of  Analytical  Mechanics 8vo, 

Schuyler's  Reservoirs  for   Irrigation,  Water-power,  and  Domestic  Water- 
supply.  .  Lar«e  8vo, 

•  Thomas  and  Watt's  Improvement  of  Rivers 4to, 

Turneaure  and  Russell's  Public  Water-supplies 8vo, 

Wegmann's  Design  and  Construction  of  Dams 4to, 

Water-supply  of  the  City  of  Tew  York  from  1658  to  1895 4to, 

Whipple's  Value  of  Pure  Water  Urge  iimo, 

Williams  and  Hazen's  Hydraulic  Table* 8vo, 

Wilson's  Irrigation  Engineering Sma..  8vo, 

Wolff's  Windmill  as  a  Prime  Mover 8vo, 

Wood's  Turbines.  .  . 8vo, 

Elemeats  of  Analytical  Mechanic* Sro, 


4  ro 

5  00 

1  50 

2  50 

4  00 

3  00 
»  50 


4 

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5 

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5 

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10 

00 

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00 

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50 

30 

MATERIALS  OF  ENGINEERING. 

Baker's  Treatise  on  Masonry  Construction ...  8to, 

Roads  and  Pavements.  ....    8vo, 

Black's  United  Sutes  Public  Works Obtong  4(0, 

•  Bovey's  Strength  of  Materials  and  Theory  of  Structure*. 8vo, 

Burr's  Elasticity  and  Resistance  of  the  Materials  of  Engineering. ......  8vo. 

Byrne's  Highway  Construction 8vo, 

Inspection  of  the  Materials  and  Worknunship  Employed  in  Construction. 

i6mo. 

Church's  Mechanics  of  Engineering 8vo, 

Du  Bois's  MechanicS'of  Engineering.     Vol.  I Small  4to, 

•Eckel's  Cements,  Limes,  and  Plaster*  8vo, 

Johnson's  Materials  of  Construction Large  8vo, 

Fowler's  Ordinary  Foundations 8vo, 

Graves's  Forest  Mersuratinn.  ...  Evo, 

•  Greene's  Structural  Mechanic*. 8vo, 

Keep's  Cast  Iron.  ... 8vo, 

Lanza's  Applied  Mechanics 8vo. 

Marten's  Handbook  on  Testing  Materials.     rHenning.  1     >  vols 8vo, 

Maurer's  Technical  Mechanics.  8vo, 

Merrill's  Stones  for  RuiMing  and  Decoration 8vo, 

Merriman's  Mechanics  of  Materials  8vo, 

•  Strength  of  Materials  iimo, 

Metcalf's  Steel.     A  Manual  for  Steel-ujers umo, 

Patton's  Practical  Treatise  on  Foundations.        8vo, 

Richardson's  Modern  Asphalt  Pavements  8vo, 

Richev's  Handbonk  lor  Superintendents  of  Construction i6mo,  mor, 

•  Ries's  Clays:  Their  Occurrence,  Properties,  and  UsM 8vo, 

Rockwell's  Roads  and  Pavtmtnts  in  Franc* iimo, 

8 


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50 

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00 

00 

00 

00 

as 

S«biii'»  Industrial  and  Artirtie  TcchnoloKy  of  Painti  ari  Varnifh.. . .        8to 

Smith's  Materiab  of  Machines. umo' 

Snow's  Principal  Species  of  Wood. . .     g„' 

Spalding's  Hydraulic  Cement. .ismo' 

Tert-book  on  Roads  and  PaTements. '  ^   iimo' 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced 8to! 

Thurston's  Materials  of  Ensineering.     3  Parts [  gyo' 

Part  I.     Ifon-metallic  Materials  of  Enfineering  and  HeUUurgr  8vo' 

Part  n.     Iron  and  Steel 8„' 

Pait  in.    A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents g^^i 

Tillson's  Street  Parements  and  Paving  Materials g^* 

WaddeU's  De  Pontibus.    (A  Pocket-book  for  Bridge  Engineers.).  .i6mo,  mor.,' 

*        Specifications  for  Steel  Bridges ,  j„,o 

Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials,  and  an  Appendix  on 

the  Preservation  of  Timber g^^ 

Wood's  (De  V.)  Elements  of  Analytical  Mechanics. , .  . .  g,o,' 

Wood's  (M.  P.)  Rustless  Coatings:   Corrosion  and  BItctrolTsis  of  bon  and' 

8*~»- 8to. 


3 

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4  00 


RAILWAY  EirGINEERING. 

Andrew's  Budbook  for  Street  Railway  Engineers jzs  Inches,  morocco,  i  as 

Berg's  BuiMings  and  Structures  of  American  Railroada ato  s  oo 

Brook's  Handbook  of  Street  Raib-owl  Location. ,6mi),  m^co!  1  so 

Butf  s  Civil  Engineer's  FieM-book. ,6mo,  morocco,  a  so 

CrandaU-s  Transition  Curve 161110.  moiocco,  i  50 

Railway  and  Other  Earthwork  Tablaa. g^o  1  w 

r4wson's  "Engineering"  and  Electric  Tnctioo  Pocket-book. .  161110,  moroccol  s  «) 

Drcdge'sHistoryof  the  Pennsylvania  Railroad:  (1870).    .                       Paner  «  00 

Fisher's  Table  of  Cubic  Yards CardboaVd '  « 

Godwin's  Railroad  Engineers'  Field-book  and  Eipk>rsrs'  Guide. . .  i6mo,  mor.',  a  50 
Hudson's  Tables  for  Calcufaiting  the  Cubic  ConUnts  of  Excavations  and  Em- 
bankments                      g^  J  j^ 

Molitor  and  Beard's  Manual  for  Resident  Engine    s! . i6mo'  1  00 

Ragle's  FieU  Manual  for  Railroad  Engineers. .  .     ,6mo,  morocco,  3  00 

L""^?" i."'*"  "'°""  '"  S"*'""" i6mo,  morocco,  3  00 

^^fl'*."/"*l°~''°« .6mo,morocco,  300 

T    J*?"^*"':""- i6mo.morocco,  .50 

Tayfer's  Prismoidal  Formula  and  Earthwork 8n>  i  50 

•  Trautwine's  Method  of  Calculating  the  Cube  Contents  of  Excavations  and' 

Embankments  by  the  Aid  of  Diagrams 8vo,  a  00 

The  FieU  Practlca  of  Uying  Out  Circular  Curves  for  Railroads. 

lamo,  morocco,  a  so 

Cross-section  Sheet p,.,,  „ 

Webb's  Railroad  Construction [[[:,,[  ,6no.  morocco!  S  00 

Economics  of  Railroad  Construction.                                        La,,,  „„o  ,  „ 

Wellington's  Economic  Theory  of  the  Location  of  Railways SmaUSvo!  S  00 


DRAWIRO. 

Barr's  Kinematics  of  Machinery 8vo  a  «» 

•  Bartlett's  Mechanical  Drawing '     g„|  3  ^ 

*  "  Abridged  Id. .        »n',  i  so 

CooMdge's  Manual  of  Drawing ,„.  ,,,,,_  ,  ^ 

B 


Coolidte  and  FrMinaii'i  BkiMat*  ot  Oantnl  Draftiiic  for  Mtcbanical  Engi- 
neer!  ObloilB  4to, 

Durley't  Kinematlci  of  Machine*. Sro, 

Emch'i  Introduction  to  ProjectiTe  Geometry  and  its  Applicationa. Sto, 

Hill'i  Text-booV  on  Shades  and  Shadows,  and  PerspectiTs Sro. 

Jamison's  Element*  of  Mechanical  Drawinf. 8to, 

Advanced  Mechanical  Drawing 8to, 

Jones's  Machine  Design: 

Part  I.    Kinematics  of  Machinery. Sro, 

Part  n.     Form,  Strength,  and  Proportions  of  Parts. 8*0, 

MacCord's  Elements  of  Descriptive  Geometry 8yo, 

Kinematics;  or.  Practical  Mechanism. 8yo, 

Mechanical  Drawing 4to, 

Velocity  Diagrams 8to, 

MacLeod's  Descriptive  Geometry.. Small  8to, 

*  Mahan's  Descriptive  Geometry  and  Stone-cutting Sto, 

Industrial  Drawing.     (Thompson.) 8vo, 

Meyer's  Descriptive  Geometry 8vo, 

Reed's  Topographical  Drawing  and  Sketching 4to, 

Reid's  Course  in  Mechanical  Drawing 8vo, 

Teit-book  of  Mechanical  Drawing  and  Elementary  Machine  Design .  8vo, 

Robinson's  Principles  of  Mechanism. 8vo, 

Schwamb  and  Merrill's  Elements  of  Mechanism 8vo, 

Smith's  (R.  S.)  Manual  of  Topographical  Drawing.     iMcMiUan.) 8vo, 

Smith  (A.  W.)  and  Man's  Machine  Design 8vo, 

*  Titsworth's  Elements  of  Mechanical  Drawing Oblong  8vo, 

Warren's  Elements  of  Plane  and  Solid  Free-hand  Geometrical  Drawing.  lamo. 

Drafting  Instruments  and  Operations umo, 

Manual  of  Elementkry  Projection  Drawing umo, 

Manual  of  ElemenUry  Problems  in  the  Linear  Perspective  of  Form  and 

Shadow lamo. 

Plane  Problems  in  Elementary  Geometry lamo. 

Primary  Geometry. lamo, 

Elements  of  Descriptive  Geometry,  Shadows,  and  Perspective 8vo, 

General  Problems  of  Shades  and  Shadows 8vo, 

Elements  of  Machine  Construction  and  Drawing 8vo, 

Problems,  Theorems,  and  Eiamplcs  in  Descriptive  Geometry 8va, 

Weisbach's    Kinematics    and    Power    of    Transmission.        (Hermann    and 

Klein.) 8vo, 

Whelpley's  Practical  Instruction  in  the  Art  of  Letter  Engraving tamo, 

Wilson's  (H.  M.)  Topographic  Surveying 8to, 

Wilson's  (V.  T.)  Free-hand  Persper've 8vo. 

Wilson's  (V.  T.)  Free-hand  Lettering , 8vo, 

WooU 's  Ekmsntary  Course  in  Descriptive  Geometry Larg*  Sro, 

ELECTRICITY  AWD  PHYSICS. 

*  Abegg's  Theory  of  Electrolytic  Dissociation.     (Von  Ende.) lamo, 

Anthony  and  Brackett's  Teit-book  of  Physics.     (Magie.) Small  8vo 

Anthony's  Lecture-note*  nn  the  Theory  of  Electrical  Measurements. . . .  lamo, 
Benjamin's  History  of  Electricity. 8vo, 

VoIUic  Cell.  .  .     8vo, 

Classen's  Quantitative  Chemical  Analysis  by  Electrolysis.     (Boltwood.)  Rvo, 

*  CoUins'i  Manual  of  Wireless  Telegraphy lamo, 

Morocco, 
Crehore  and  Squicr's  Polarising  Photo-chronograph 8vo, 

*  Danneel'i  Electrochemistry.     iMerriam.  1.  lamo, 

Dawson's  "Engineering"  and  Electric  Traction  Pocket-book.  lAmo,  morocco, 

10 


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00 

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00 

as 

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35 

50 

00 

as 

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SO 

00 

SO 

50 

00 

00 

SO 

50 

00 

00 

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I  00 
3  00 
3  00 


Dolezatok'i   Tbaorj  of   tha   Laad  Aecumuktor   (Storact   Batttry).      (Von 

Bnde.) iimo,  a  50 

Duhcin'i  Thtnnodjmaiiiica  and  Chamiitry.    (BurfM*-) Sto,  4  oo 

Plathar*!  DTnamomatan.  and  tha  Maaauramant  of  Powar. iimo.  j  00 

Oilberfs  Da  Matnata.     (Motular) Sto,  2  50 

Banchatf  a  AHarnating  Cuiranta  Ezpkinad. i  jmo,  i  00 

Barinc'i  Raady  Rafaranca  Tabba  (ConTanion  Factor*) i6nio  morocco,  a  50 

Holman'a  Praciaion  of  Mcaauramanta Sto,  a  oo 

Talaacopic  llirror-acala  Mathod,  AdjiutmanU,  and  Taata Large  Sto,  75 

Kinzbrunnar'a  Taatinf  of  Continuoua-currant  Machinaa. Sto,  a  cc 

Landaucr'a  Spactrum  Analyaia.     (Tincla.) Sto,  3  00 

LaChataliar'aBich-tamparaturaIfeaauramanta.(Bottdoiiard — Burgaaa.)  lamo,  3  00 

Lob'a  Elactrochamiitry  of  Organic  Compounda.     (Loranz.) Sto,  3  00 

•  Lrona't  Treatiie  on  Elactromagnatic  Phanomana.  Vola.  I.  and  n.  Sto,  aach,  6  00 

•  Hichie'a  ElauMnta  of  W»  7a  Motion  Relating  to  Sound  and  Light Sto,  4  00 

Riaudet's  Elemcntarr  Traatiaa  on  Electric  Batteriea.     (Fiahback.) lamo,  a  50 

•  Parahall  and  Hobart'a  Electric  Machine  Dcaign 4to,  half  morocco,  la  so 

Reagan'a  LocomotiTei:   Simple,  Compound,  and  Electric.     iHew  Edition. 

Large  lamo,  3  50 

•  Rocenberg'a  Electrical  Engineering.     (Haldana  Gee— Kinzbrunner.).  .  .Sto,  a  00 

Rran,  Horrii,  and  Hozie's  Electrical  Machinery,    VoL  L  . '. Sto,  a  50 

Thunton*!  Stationarr  Steam-enginet. Sto,  a  «o 

•  TiUman'i  Elementary  Leaaon*  in  Beat Sto,  i  50 

Tory  and  Pitcher'a  Manual  of  Laboratory  Phyaica. Small  Sto,  2  oo 

nika'a  Modem  Electrolytic  Copper  Refining Sto,  3  At 


LAW. 

*  DaTit'i  Elamantt  of  Law. Sto, 

*  Treatiae  on  the  Military  Law  of  United  Sutee. Sto, 

*  Sheep, 

*  Dudley'*  Military  Law  and  tha  Procedure  of  Courta-martial . .    .  Large  i  amo. 

Manual  for  Courta-martiaL i6mo,  morocco. 

Wait'*  Engineering  and  Architectural  Juriaprudence Sto, 

Sheep, 
Law  of  Operationa  Preliminary  to  Conatruction  in  Engineering  and  Archi- 
tecture       . .  Sto 

Sheep, 

Law  of  Contract*. Sto, 

Wintbrop'a  Abridgment  of  Military  Law tame. 


5  00 
5  so 
3  00 
a  so 


KAlfUFACTURES. 

Bamadou'e  SmokalaM  Powder— Ritro-ceUnfcMa  and  Theory  of  the  CeDukiaa 

Molecule lamo, 

BoUaad'a  Iron  Pounder. lamo. 

The  Iron  Pounder,"  Supplement tamo. 

Encyclopedia  of  Founding  and  Dictionary  of  Foundry  Terma  Uaed  in  the 
Practice  of  Moulding lamo, 

•  Claaiaen'a  Beet-eugar  Manufacture.    (Hall  and  Rolf  a.) Sto, 

•  Eckel'*  Cement*,  Limoe.  and  Pkutar* Sto, 

Bi**ler'*  Modem  High  RzpteaiTea. Sto, 

Bffront'*  Bnzymea  and  their  AppUcationa.    (Preaeott) Sto, 

Flligerai-'*  Boaton  Machiniat tamo. 

Ford'*  Boiler  Making  for  Boiler  Maker*. iSmo, 

Hopkin'*  Oil-cb*mi*t('  Handbook. Sto, 

Kaep'i  Ca*t  Iron. Sro, 

u 


a  so 
a  50 

a  50 

3  00 
300 
6  00 

4  00 
3  00 
I  oo 
I  00 
3  00 
a  90 


Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  Sute 

CoolroL LargeSvo,  750 

•  McKay  and  Larsen's  Principles  and  Practice  of  Buttei-makins            8vo,  1  50 

Matthews's  The  Teitile  Fibres gyo,  3  50 

Metcalf's  SteeL     A  Manual  for  Steel-users: .......  .umo!  2  00 

Metcalfe'r  Cost  of  Manufactures— And  the  Administration  of  Workshops. 8vo,  5  00 

Meyer's  Modern  Locomotive  Construction 410,  10  00 

Morse's  Calcubtions  used  in  Cane-sugar  Factories i6mo,  morocco,  i  50 

•  Reistj's  Guide  to  Piece-dyeing gy,,^  ,5  ^q 

Rice's  Concrete-block  Manufacture gyo,  3  00 

Sabin's  Iclustrial  and  Artistic  Technology  of  Paints  and  Varnish gyo',  3  00 

Smith's  Press-working  of  Meuls gyo,  3  00 

Spalding's  Hydraulic  Cement umo,  2  00 

Spencer's  Handbook  for  Chemists  of  Beet-sugar  Bouses i6mo  morocco!  3  00 

Handbook  for  Cane  Sugar  Hantifacturers ...  i6mo  morocco,  3  00 

Taylor  and  Thompson's  Treatise  on  Concrete,  Plain  and  Reinforced 8vo,  5  00 

Thurston's  Manual  of  Steam-boilers,  their  Designs,  Construction  and  Opera- 

«'0° 8vo.  5  00 

•  Walke's  Lectures  on  Explosives gyo,  ^  go 

Ware's  Beet-sugar  Manufacture  and  Refining Small  8to,  4  00 

Weaver's  Military  Explosives gyo,  3  00 

West's  American  Foundry  Practice lamo,  3  50 

Moulder's  Text-book umo!  2  so 

Wolff's  Windmill  as  a  Prime  Mover gyo,  3  00 

Wood's  Rustless  Coatings:  Corrosion  and  Electrolysis  of  Iron  and  Steel.  .8vo,  4  00 


MATHEMATICS. 

Baker's  Elliptic  Functions gyo, 

•  Bass's  Elements  of  Diflerential  Calculus lamo, 

Briggs's  ElemenU  of  Plane  Analytic  Geometry lamo! 

Compton'i  Manual  of  Logarithmic  Compuutiooa. lamo 

Davis's  Introduction  to  the  Logic  of  Algebra gyo 

•  Dickson's  College  Algebra jju.,,  umo! 

•  Introduction  to  the  Theory  of  Algebraic  Equations Large  umo, 

Emch's  Introduction  to  Projective  Geometry  and  its  Applications gyo 

Balsted's  Elements  of  Geometry gyo 

Elementary  Synthetic  Geometry gyg 

Rational  Geomatry lamo, 

•  Johnson's  (J.  B.)  Thrie-pUce  Logarithmic  Tables:  Vest-pocket  siie  paper! 

100  copies  for 

•  Mounted  on  heavy  cardboard,  8  X  10  inches, 

10  copies  for 
Johnson's  (W.  W.)  Elementary  Treatise  on  Diflerential  Cakuhis    SmaUSvo, 

Elementary  Treatise  on  the  !::tegral  Calculus SmalPSvo, 

Johnson's  (W.  W.)  Carre  Tracing  in  Cartesian  Co-ordinates     tamo, 

Jobnson's  (W.  W.)  Treatise  on  Ordinary  and  Partiaf  Differential  Equations. 

Small  Svo, 
Johnson's  i  W.  W.)  Theory  of  Errors  and  the  Method  of  Least  Squsres.  umo, 

•  Johnson's  (W  W.)  Theoretical  Mechanics lamo, 

Laplace's  Philosophical  Essay  on  ProbabiUties.    (Truscott  and  Emory. )  umo, 

•  Ludlow  and  Bau.     Elements  of  Trigonometry  and  Logarithmic  and  Ot'ier 

Tables gyo. 

Trigonometry  and  Tables  published  separately Each, 

•  Ludlow's  Logarithmic  and  Trigonometric  Tables 8to 

Manning's  Irrati  :.nal  Nur;ibers  and  their  Representation  by  Sequences  and  Series 

umo, 

u 


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4  00 
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50 
75 

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00 

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00 

00 

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I  50 
3  00 
a  00 

3  00 
a  oc 
I  00 

I  as 


Mathenutical  Monographs.     Edited  by  ManiBeld  Merriman  and  Robert 

S.  Woodward OcUto,  each 

No.  I.  History  of  Modern  Mathematics,  by  David  Eugene  Smith. 
Ho.  2.  Synthetic  Projective  Geometry,  by  George  Bruce  Halsted. 
Ho.  3.  DeterminanU,  by  Lttntf  Gifford  Wald.  Ho.  4.  Hyper- 
bolic Functions,  by  James  McMahon.  Ho.  5.  Harmonic  Func- 
Uons,  by  William  E.  Byerly.  Ho.  6.  Grassmann's  Space  Analysis, 
by  Edward  W.  Hyde.  Ho.  7.  Probability  and  Theory  of  Errors, 
by  Robert  S.  Woodward.  Ho.  8.  Vector  Analysis  and  Qtuternions, 
by  Alexander  Macfarlane.  Ho.  9.  Differential  Equations,  by 
William  Woolsey  Johnson.  Ho.  10.  The  Solution  of  Equations, 
by  Mansfield  Merriman.  No.  1 1.  Functions  of  a  Complex  Variable, 
by  Thomas  S.  Fiskr. 

Maurer's  Technical  Mechanics 8to, 

Merrioun's  Method  of  Least  Squares. gvo. 

Rice  and  Johnson's  ElemenUry  Treatise  on  the  Diflerential  Calculus. .  Sm.  8vo, 

Differential  and  Integral  Calculus.     2  vols,  in  one Small  8vo, 

*  Veblen  and  Lennes's  Introduction  to  the  Real  Infinitesimal  Analysis  of  On* 

Variable  8„, 

Wood's  Elements  of  Co-ordinate  Geometry gyo. 

Trigonometry:  Analytical,  Plane,  and  Spherical lamo. 


4  00 

2  00 

3  00 
a  30 

a  00 
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MECHANICAL  ElfGIlfEERIKG, 


MATERIALS  OF  ERGINEERING,  STEAM-ENGIHES  AND  BOaBRS. 

Bacon's  Forge  Practice ,  ,    lamo, 

Baldwin's  Steam  Heating  for  Buildings lamo, 

Barr's  Kinematics  of  Machinery. 8vo, 

•  Bartlttt's  Mechanical  Drawing 8vo, 

•  "  "  "        Abridged  Ed 8vo, 

Benjamin's  Wrinkles  and  Recipes lamo. 

Carpenter's  Experimental  Engineering 8vo, 

Heating  and  Ventilating  Buildings. gyo. 

Clerk's  Gas  and  Oil  Engine Small  8vo, 

Coolidge's  Manual  of  Drawing 8vo,  paper, 

Coolidge  and  Freeman's  Elements  of  Gennml  Drafting  for  Mechanical  En- 

•ineers Obk>ng  410, 

Cromwell's  Treatise  on  toothed  Gearing lamo, 

Treatise  on  Belts  and  Pulleys lamo, 

Durley's  kinematics  of  Machines 8vo, 

Flather's  Dynamometers  and  the  Measurement  of  Power lamo, 

ftope  Driving lamo. 

Gill's  Gas  and  Fuel  Analysis  for  Engineer* tamo. 

Ball's  Car  Lubrication lamo, 

Bering's  Ready  Reference  Tables  (Conversion  Factors) i6mo,  morocco. 

Button's  The  Gas  Engine 8vo, 

Jamison's  Mechanical  Drawing gro, 

Jones's  Machine  Design: 

Part  I.     Kinematics  of  Machinery 8vo, 

Part  n.     Form,  Strength,  and  Proportions  of  Parts 8vo, 

Kent's  Mechanical  Engineers'  Pocket-book i6mo,  morocco. 

Kerr's  Power  and  Power  Transmission 8vo, 

Leonard's  Machine  Shop,  Tools,  and  Methods   8vo, 

•  Loreni's  Modern  Refrigerating  Machinery.    '  Pope,  Haven,  and  Dean.)     8vo, 
MacCord's  Kinematics;  cr.  Practical  Mechanism gvo. 

Mechanical  Drawing. ^to, 

Velocity  Diagram* gvo, 

13 


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MacFwtond'a  Standard  Reduction  Facton  for  Gaiaa. tn, 

Uahan'i  Indintrial  Drawing.     (Thompcon.) 8to 

Poolc'i  Calorific  Power  of  Fueb. gyo. 

Raid's  Course  in  Mechanical  Drawiaf Sto, 

Text-book  of  Mechanical  Drawinc  and  Elementary  Machine  Desitn.STo, 

Richard's  Compressed  Air lamo, 

Robinson's  Principles  of  Mechanism. Sn, 

Schwaub  and  Merrill's  Elements  of  Mechanism 8to, 

Smith's  (O.)  Press-working  of  Metals gro 

Smith  (A.  W.)  and  Marx's  Machine  Design. . gyo, 

Thurston's  Treatise  on  Friction  and  Lost  Work  in  Machinery  and  Mill 
.  'Wofk 8»o. 

Animal  as  a  Bfachine  and  Prime  Motor,  and  the  Laws  of  Energetics .  lamo, 
Tillson's  Complete  Automobile  Instructor i6mo, 

Morocco, 

Warren's  Elements  of  Machine  Construction  snd  Drawing. 8vo, 

Weisbacli's   Kinematics   and   the    Power   of   Transmission.     (Herrmann — 

Klein.) gyo^ 

Machinery  of  Transmission  and  Governors.     (Herrmann — Klein.).  .8to, 

WoWs  Windmill  as  a  Prime  Mover gyo, 

Wood's  Turbines gyo' 


1  so 
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3  00 
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MATERIALS  OF  EWGIWEERIlfG. 

•  Bovey's  Strength  of  Materials  and  Theory  of  Structures gyo. 

Burr's  EUsticity  and  Resistance  of  the  Materials  of  Engineering.    6th  Edition. 

Reset- gvoi 

Church's  Mechanics  of  Engineering gy^^ 

•  Greene's  Structural  Mechanics  .    gyf, 

Johnson's  Materials  of  Construction gyo 

Keep's  Cast  Iron gyo 

Lanza's  Applied  Mechanics gyo. 

Martens 's  Handbook  on  Testing  Materials.     (Henning.) gyo| 

Maurer's  Technical  Mechanics gyo, 

Merriman's  Mechanics  of  Materials gyo 

•  Strength  of  Materials lamo, 

Metcalf's  Steel.     A  Manual  for  Steel-users lamo", 

Sabin's  Industrial  and  Artistic  Technology  of  Painta  and  Varnish 8vo, 

Smith's  Materials  of  Machines. lamo, 

Thurston's  Materials  of  Engineering 3  vols.,  8yo, 

Part  II..    Iron  and  SteeL gyo. 

Part  III.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents gyo, 

Wood's  (De  V.)  Treatise  on  the  Resistance  of  Materials  and  an  Appendix  on 

the  Preservation  of  Timber gyo. 

Elements  of  Analytical  Mechanics. gyo. 

Wood's  (M.  P.;  Rustless  Coatings:   Corrosion  and  Electrolysis  of  Iron  and 
Steel  gyo. 


7  50 


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50 

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STEAM-ElfGIRES  AND  BOILERS. 

Berry's  Temperature-entropy  Diagram lamo, 

Carnot's  Reflections  on  the  Motive  Power  of  Heat.     (Thurston.) lamo, 

Dawson's  "  Engineering  "  and  Electric  Traction  Pocket-book i6mo,  mor.. 

Ford's  Boiler  Msking  for  Boiler  Makers iBmo, 

Goss's  Loccmotive  Sparks gyo. 

Locomotive  Performance  .  gyo, 

Bemenway's  Indicator  Practice  and  Steam-engine  Economy lamo, 

14 


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50 
00 
00 
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Hntton't  MKhanical  Enginccrinc  of  Power  Plants. Sro,  5  00 

Heat  and  Heat-encine* gvo  j  00 

Kent*!  Steam  boiler  Economy gvo,  4  00 

Kneaw'i  Practice  and  Theory  of  the  Injector 8to,  i  50 

■acCord'i  SUde-vaWei 8yo,  2  00 

Meyer's  Modem  Locomotive  Construction 4to,  10  00 

Peabody's  Manual  of  the  Steam-encine  Indicator. ismo,  i  50 

Tables  of  the  Properties  of  Saturated  Steam  and  Other  Vapors   8to,  i  oo 

Thermodynami:s  of  the  Steam-engine  and  Other  Heat-engines. 8to,  5  00 

Valve-gears  for  Steam-engines g^o,  3  50 

Peahody  and  Miller's  Steam-boilers gvo,  4  00 

Pray's  Twenty  Years  with  the  Indicator Large  8td,  2  50 

Pupin's  Thermodynamics  of  Reversible  Cycles  in  Gases  and  Saturated  Vapors. 

(Osterberg.) „„o,  i  as 

Reagan's  Locomotives:   Simple,  Compound,  and  Electric.     Hew  Edition. 

Large  lamo,  3  -0 

Rontgen's  Principles  of  Thermodynamics.     (Du  Bois.) 8vo,  5  o« 

Sinclair's  Locomotive  Engine  Running  and  Management ismo,  a  oo 

Smart's  Handbook  of  Engineering  Laboratory  Practice lamo,  3  go 

Snow's  Steam-boiler  Practice gvo,  3  00 

Spangler's  Vahre-gears gvo,  a  50 

Ifotes  on  Thermodynamics lamo,  i  00 

Spangler,  Greene,  and  Marshall's  Elements  of  Steam-engineering gvo,  3  00 

Thomas's  Steam-turbines gvo,  3  50 

Thurston's  Handy  Tables gvo,  i  50 

Manual  of  the  Steam-engine a  vols,,  gvo,  10  00 

Part  I.     History,  Structure,  and  Theory gvo,  <i  00 

Part  II.     Design.  Construction,  and  Operation gvo,  6  00 

Bandbooii  of  Engine  and  Boiler  Trials,  and  the  Use  of  the  Indicator  and 

the  Prony  Brake gvo,  5  00 

Stationary  Steam-engines 8vo,  a  50 

Steam-boiler  Expkisions  in  Theory  and  in  Practice lamo,  i  50 

Manual  of  Stram-boilers,  their  Designs,  Construction,  and  Operation  gvo,  5  00 

Wchrenfenning'sAnalysisandSofteningof  Boiler  Feed-water  (Patterson)  gvo,  400 

Weisbach's  Heat,  Steam,  and  Steam-engines.     (Du  Bois.) 8vo,  5  00 

Whitham's  Steam-engine  Design gvo,  j  00 

Wood's  Thermodynamics,  Heat  Motors,  and  Refrigerating  Machines.  ..gvo,'  4  00 


MECHAinCS  AND  MACHINERY. 

Barr's  Kinematics  of  Machinery gvo,  j  ui 

•  Bovey's  Strength  of  MaUriab  and  Theory  of  Structures  '.  isvo!  7  so 

Chase's  The  Art  of  Pattern-making lamo,  a  50 

Church's  Mechanics  of  Engineering gvo,  6  00 

Notes  and  Examples  in  Mechanics 8vo!  a  oo 

Compton's  First  Lessons  in  MeUI-working .  .ijmo!  1  so 

Compton  and  De  Groodt's  The  Speed  Lathe .lamo'  1  «o 

Cromwell's  Treatise  on  Toothed  Gearing '.'.'.'.'.'.  lamo!  i  so 

Treatise  on  Belts  and  Pulleys. '    '    '  ,2ao\  i  so 

Dant's  Text-book  of  ElemenUry  Mechanics  for  Colleges  and  Schools.   lamo!  i  50 

Dingey's  Machinery  Pattern  Making lamo,  a  00 

Dredge's  Record  of  the  Transportation  Exhibits  Building  of  the  World's' 

Columbian  Exposition  of  1893 4to  half  morocco,  s  00 

Du  Bois's  Elementary  Principles  of  Mechanics: 

Vol      I.     Kinematics. «vo,  3  50 

VoL    U.    Statics jvo  4  00 

Mechanics  of  Engineering.    Vol.    I ..!..!.  !sniaU4toi  7  so 

_     .    ,   _,                                    Vol  n. Small  4to,  10  00 

Durley's  Kinematics  of  Machines. gvo  4  00 

16 


Fiugvrald's  Boston  Machinist i6nio, 

Flather's  Dynamometers,  and  the  Heasurement  of  Power iimo. 

Rope  Driving i  jmo, 

Goss's  Locomotive  Sparks 8vo, 

Locomotive  Performance 8vo, 

*  Greene's  Structural  Mechanics. .      . .    8vo, 

Hall's  Car  Lubrication iimo, 

BoUy's  Art  of  Saw  Filins i8mo, 

James's  Kinematics  of  a  Point  and  the  Rational  Mechanics  of  a  Particle. 

Small  8vo, 

*  Johnson's  (W.  W.>  Theoretical  Mechanics umo, 

Johnson's  (L.  J.)  Statics  by  Graphic  and  Algebraic  MethoLS 8vo, 

Jones's  Machine  Design : 

Part    1.     Kinematics  of  Machinery 8vo, 

Part  n.     Form,  Strength,  and  Proportions  of  Parts 8vo, 

Kerr's  Power  and  Power  Transmission 8vo, 

Lanza's  Applied  Mechanics 8vo, 

Leonard's  Machine  Shop,  Tools,  and  Methods 8vo, 

*  Lorenz's  Modern  Refrigerating  Machinery.     (Pope,  Haven,  and  Dean.).8vo, 
MacCord's  Kinematics:  or.  Practical  Mechanism 8vo, 

Velocity  Diagrams 8vo, 

*  Martin's  Teit  Book  on  Mechanics,  VoL  I,  Sutics lamo, 

Maurer's  Technical  Mechanics 8vo, 

.   erriman's  Mechanics  of  Materials 8vo, 

*  Elements  of  Mechanics lamo, 

*  Michie's  Elements  of  Analytical  Mechanics Svo, 

*  ParshaUand  Hobart's  Electric  Machine  Design 4to,  half  morocco, 

Reagan's  Locomotives :  Simple,  Compound,  and  Electric.     New  Edition. 

Large  izmo, 
Reid's  Course  in  Mechanical  Drawing 8vo, 

Text-book  of  Mechanical  Drawing  and  Elementary  Machine  Design. 8vo, 

Richards's  Compressed  Air lamo, 

Robinson's  Principles  of  Mechanism Svo, 

Ryan,  Ilorris,  and  Hoxie's  Electrical  Machinery.     VoL  I Svo, 

Sanborn's  Mechanics:  Problems Large  izmo, 

Schwamb  and  Merrill's  Elements  of  Mechanism Svo, 

Sinclair's  Locomotive-engine  Running  and  Management izmo. 

Smith's  (O.)  Press-working  of  MeUls Svo, 

Smith's  ( A.  W. )  Materials  of  Machines izmo. 

Smith  (A.  W. ,  and  Han's  Machine  Design. Svo, 

Spanglcr,  Greene, and  Marshall's  Elements  of  Steam-engineering Svo, 

Thurston's  Treatise  on  Friction  and  Lest  Work  in    Machinery  and    Hill 
Work Svo, 

Animal  as  a  Machine  and  Prime  Motor,  and  the  Laws  of  Energetics,  i  zmo, 

Tillson's  Complete  Automobile  Instructor i6mo, 

Morocco, 

Warren's  Elements  of  Machine  Construction  and  Drawing Svo, 

Weisbach's  Kinematics  and  Power  of  Transmission.   ( Herrmann — Klein. ) .  Svo, 

Machinery  of  Transmission  and  Governors.      (Herrmann — Klein. ). Svo, 
Wood's  Elements  of  Analytical  Mechanics gyo, 

Principles  of  Elementary  Mechanics tzmo. 

Turbines 8to, 

The  World's  Columbian  Exposition  of  1893 4to, 


00 

00 

00 

00 

00 

50 

00 

75 

00 

00 

00 

50 

00 

00 

50 

00 

00 

00 

50 

35 

00 

00 

00 

00 

la 

50 

00 

00 

00 

SO 

00 

50 

50 

00 

00 

00 

00 

00 

00 

00 

00 

50 

00 

50 

00 

00 

00 

>5 

50 

00 

MEDICAL. 

De  Fursac's  Manual  of  Psychiatry.     (Rosanoff  and  Collini.) Large  izmo, 

Ehrlich's  Collected  Studies  on  Immunity.     1  Bolduan.  I Svo, 

Hammarsten's  Text-book  on  Physiological  Chemistry.     (MandeL).       .     8vo, 

16 


Z    JO 

6  00 
4  00 


Latsar-Cohn'i  Practical  Urinary  Analysis.     (Lorenz.) lamo, 

*  Pauli's  Physical  Chemistry  in  the  Service  of  Uedicme.     (Fischer.)         iimo, 

*  Pozzi-Escot's  The  Toxins  and  Venoms  and  their  Antibodies.     (Cohn. ).  i  jmo, 

Rostoski's  Serum  Diacnosis.     (Bolduan.) i2mo, 

Salkowski's  Physiological  acd  Pathological  Chemistry.     (Crcdorfl.) .         9vo, 

*  Satterlee's  Oatlines  of  Human  Embryology umo. 

Steel's  Treatise  on  the  Diseases  of  the  Dog. 8to, 

Von  Behring's  Suppressfon  of  Tuberculosis.     (Bolduan.) ixmo, 

Wassermann's  Immune  Sera ■  Hsemolysis,  Cytotozins,  and  Precipitins.     (Bol- 
duan.)   iimo.cloth, 

Woodhull's  Notes  on  Military  Hygiene i6mo, 

*  Personal  Hygiene umo, 

Wulling's  An  Elementary  Course  in  Inorganic  Pharmaceutical  and  Medical 
Chemfktry lamo. 


I    CO 

1  »S 

I    03 

1  CO 

2  50 
I    2S 

3  50 
I   oo 

I  oo 
I  so 
I  oo 


METALLURGY. 

Egletton's  Metallurgy  of  Silrer,  Gold,  and  Mercury: 

VoL    I.     Sihrer 8to, 

VoL  n.     Gold  and  Mercury 8to, 

Goetel's  Minerals  and  Metals:    A  Reference  Book i6mo,  mor. 

*  Ilea's  Lead-smelting lamo. 

Keep's  Cast  Iron 8to, 

Kunhardt's  Practice  of  Ore  Dressing  In  Europe 8to, 

Le  Chatelier's  High-temperature  Measurements.  (Bondouard — Burgeas.)i2mo, 

Metcalf's  SteeL    A  Manual  for  Steel-users. lamo, 

Miller's  Cyanide  Process umo, 

Hinet's  Production  of  Aluminum  ana  its  Industrial  Dae.     (Waldo.)... .  lamo, 

Robine  and  Lenglen's  Cyanide  Industry.     (Le  Clerc.) 8vo, 

Smith's  Materials  of  Machines umo, 

Thurston's  Materials  of  Engineering.    In  Three  Parts. 8yo, 

Part   n.    Iron  and  SteeL 8vo, 

Part  m.     A  Treatise  on  Brasses,  Bronzes,  and  Other  Alloys  and  their 

Constituents 8vo, 

Hike's  Modem  Electrolytic  Copper  Refining 8to, 


so 
50 
oo 
50 
50 
50 

00 

oo 
t  oo 
»  50 
4  oo 

1  oo 
8  oo 
3  50 

2  50 

3  oo 


MnrERALOGY. 


Barringer's  I>escription  of  Minerals  of  Commercial  Value.   Oblong,  morocco,  a  so 

Boyd's  Resources  of  Southwest  Virginia 8to,  3  00 

Map  of  Southwest  Virignia. . .     Pocket-book  form.  3  00 

Brush's  Manual  of  Determinative  Mineralogy.     (PenfieU.) . .   Svo,  4  00 

Chester's  Catalogue  of  Minerals 8to,  paper,  i  00 

Cloth,  I  as 

Dictionary  of  the  Names  of  Minerals. Svct  3  so 

Dana's  System  of  Mineralogy Large  8to,  half  leather,  la  50 

First  Appendix  to  Dana's  New  "  System  of  Mineralogy." Large  8vo,  i  00 

Text-book  of  Mineralogy 8vo,  4  00 

Minerals  and  How  to  Study  Them lamo.  i  so 

Catalogue  of  American  Localities  of  Minerals Large  Sto,  i  00 

Manual  of  Mineralogy  and  Petrography .  lamo  a  oo 

Douglas's  Untechnical  Addresses  on  Technical  Subjects. lamo,  i  oo 

Eakle's  Mineral  Tables. 8to,  i  as 

Egleston's  Catalogue  of  Minerals  and  Synonyms. 8vo,  a  so 

Goesel's  Minerals  and  Metals :    A  Reference  Book i6mo,  mor.  3  00 

Oroth's  Introduction  to  Chemical  Crystallography  (Marshall) lamo,  i  as 

17 


Udingt'i  Rock  Minerali g^o, 

MeiriU't  Ron-meUllic  Mincnli:  Their  Occumnct  and  Um*. 8to, 

•  PenfleU'i  Notes  on  DeUrminative  Hineralotr  and  Record  of  Mineral  Taiti. 

8to,  paper, 

*  Richard! '■  Synopiii  of  Mineral  Charaeten. lamo,  morocco, 

•  Riei'i  Clayi:  Their  Occurrence,  Propcrtict,  and  Diet.. 8vo, 

Roicnbuich'i   Microscopical  Phytiofraphy  of  the  Rock-makin(  Mincraia. 

(Iddinct.) 8to, 

*  Tillman's  Text-book  of  Important  Minerals  and  Rocks. 8to, 


oo 
oo 

50 
as 

00 


5  00 

2  00 


MIIIUTO. 


Boyd's  Resources  of  Southwest  Vir(inia 8to. 

Hap  of  Southwest  Virginia. Pocket-book  form 

Douslas's  Ontechnical  Addresses  on  Technical  Subjects. lamo, 

Eissler's  Modern  High  Explosives.      a 

Goescl's  Minerals  and  Metals:    A  Reference  Book i6mo,  mor. 

Goodyear't  Coal-mines  of  the  Western  Coast  of  the  United  States. iimo, 

Ihlteng's  Manual  of  Mining gro, 

*  Ilea's  Lead-smelting iimo, 

Kunhardt's  Practice  of  Ore  Dressing  In  Europe 8to, 

Miller's  Cyanide     rocess. lamo, 

O'Driscoll's  Holes  on  the  Treatment  of  Oold  Ores. 8to, 

Robine  and  Lenglen's  Cyanide  Industry.    (Le  Clerc.) 8vo, 

♦  Walke's  Lecttires  on  Explosives. 8vo, 

Weaver's  Military  Ezplosives. 8to, 

Wilson's  Cyanide  Processes. lamo. 

Chlorination  Process lamo. 

Hydraulic  and  Placer  Mining lamo, 

Ireatise  on  Practical  and  Theoretical  Mine  Ventilation. tamo. 


00 

oo 

00 

TO 

oo 

50 

oo 

so 

50 

oo 

oo 

00 

00 

00 

so 

50 

oo 

as 

SAinTARY  SCIENCE. 


Bashore's  Sanitation  of  a  Country  House lamo, 

•  Outlines  of  Practical  Sanitation lamo, 

Folw*<l's  Sewerage.    (Designing,  Construction,  and  Maintenance.) 8vo, 

Water-supply  Engineering gyo 

Fowler's  Sewage  Works  Analyses lamj 

Fuertes's  Water  and  Public  Health lamo. 

Water-filtration  Works lamo, 

Gerhard's  Guide  to  Sanitary  House-inspection i6mo, 

Hazen's  Filtration  of  Public  Water-supplies 8to, 

Leach's  The  Inspection  and  Analysis  of  Food  with  Special  Reference  to  State 

Control gyo. 

Mason's  Water-supply.  ( Considered  principally  from  a  Sanitary  Standpoint)  Svo, 

Examination  of  Water.    (Chemical  and  BacteriologlcaL) lamo, 

•  Merriman's  Elements  of  Sanitary  Engineering 8vo, 

Ogden's  Sewer  Design lamo, 

Prescott  and  Winslow's  Elements  of  Water  Bacteriok>gy,  with  Special  Refer- 
ence to  Sanitary  Water  Analysis. lamo, 

•  Price's  Handbook  on  Sanitation lamo, 

Richards's  Cost  of  Food.     A  Study  in  Dietaries lamo. 

Cost  of  Living  as  Modified  by  Sanitary  Science. lamo. 

Cost  of  Shelter lamo, 

18 


I  oo 
I  as 

J  U(l 
4  Oo 

a  oo 
1  50 
a  50 
I  oo 

3  oo 

7  SO 

4  00 
I  as 
a  oo 
a  00 


as 

50 


I 
I 

I  00 
I  00 
I  00 


Richuds  ud  Woodman's  Air.  Water,  and  Food  from  a  Sanitary  Stand- 
point   8,0 

•  RUhardi  and  WlUiami't  The  DUtarr  Computer gvo' 

Ridtal't  Se»a|«  and  Bacterial  Purlflcation  of  Sewate. 8to 

Turncaure  and  RuncU'i  Public  Water-iuppilet 8»o' 

Von  Behrlnj't  Snppreulon  of  Tuberculoiii.     (BoMuan.).  ...... .....umo 

Whipple's  Microscopy  of  Orinklnf-water 8vo' 

Wlnton's  Microscopy  of  Vegetable  Foods.    g^' 

WoodhuU's  Rotes  on  MlUtary  Hyaiena ,*«.„' 

•  p«««aH«i«» :::::::;.::::;:;.;'So; 


2  00 

I  50 

4  00 

5  00 

I  00 
.)  so 
7  50 
I  50 
I   00 


nSCELLAlTEOUS. 

Emmons's  Geolocical  Guide-book  of  the  Rocky  Mountain  Excursion  of  the 

International  Concreas  of  Geologists. Larae  8to 

Ferret's  Popular  Treatlsa  on  the  Winds. g^' 

Gannett's  Sutistical  Abstract  of  the  World ',]„_.' 

Haines's  American  Railway  Management j^' 

Ricketts's  History  of  Rensselaer  Polytecbnic  Inatitttte.  1824-1894.. sinall  8»o' 

Rotherham's  Emphaslied  Hew  Tastament Lar«8To 

The  World's  Cohimbian  Exposition  of  1893  .»to 

Winstow's  Elamantt  of  AppUed  Microwopr .' .' .'  umo! 

HEBREW  AMD  CHALDEE  TEXT-BOOKS. 

Green's  Elamantary  Hebrew  Grammar. ,„ao 

Hebrew  Chrsstomathy.    g     ' 

Gesenlns's  Hebrew  and  Chaldea  Lexicon  to  thii'  Old  Testanient  Scriptures." 

,_    .  ,  *Z'??!"^L" Small 4to. half  morocco. 

Lottsris's  Hebrew  Bibla. g^ 

19 


1  so 
4  00 
75 
a  5:. 
3  00 
a  00 
I  00 
I  so 


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a  00 

s  00 
»  as 


